Q6: 6. y=x sin x: x y'=y+x sqrtx^2-y^2(x ≠ 0 and x>y or x<-y)

Given: The function is y = x sin x. The differential equation is xy' = y + x √(x^2 - y^2), for x ≠ 0 and x>y or x y or x y is satisfied. However, LHS = -π^2 and RHS = π^2. Assuming there is a tacit assumption in the problem that the expression is restricted to domains where x cos x ≥ 0, then x cos…

Differential Equations Class 12 Mathematics NCERT Solutions

Q1EXERCISE 9.1

$rac{d^{4} y}{d x^{4}}+\sin \left(y^{\prime \prime \prime}\right)=0$

Solution

Given: The differential equation is

\frac{d^{4} y}{d x^{4}}+\sin \left(y^{\prime \prime \prime}\right)=0

.

To Find: The order and degree of the differential equation.

Solution: The given differential equation is:

\frac{d^{4} y}{d x^{4}}+\sin \left(y^{\prime \prime \prime}\right)=0

This can be written as:

\frac{d^{4} y}{d x^{4}}+\sin \left(\frac{d^3 y}{d x^3}\right)=0

Order: The order of a differential equation is the order of the highest derivative appearing in the equation. In this equation, the highest order derivative is

\frac{d^{4} y}{d x^{4}}

, which has an order of 4. Therefore, the order of the differential equation is 4.

Degree: The degree of a differential equation is the highest power (as a positive integer) of the highest order derivative, after the differential equation has been expressed as a polynomial in derivatives. The given equation cannot be expressed as a polynomial in its derivatives because of the term

\sin \left(y^{\prime \prime \prime}\right)

. The sine function has an infinite series expansion involving powers of its argument. For example:

\sin(x) = x - \frac{x^3}{3!} + \frac{x^5}{5!} - \dots

So,

\sin \left(y^{\prime \prime \prime}\right) = y^{\prime \prime \prime} - \frac{(y^{\prime \prime \prime})^3}{3!} + \frac{(y^{\prime \prime \prime})^5}{5!} - \dots

Since the equation cannot be written as a finite polynomial in the derivatives, its degree is not defined.

Final Answer: The order is 4 and the degree is not defined.

Q2EXERCISE 9.1

$y^{\prime}+5 y=0$

Solution

Given: The differential equation is

y^{\prime}+5 y=0

.

To Find: The order and degree of the differential equation.

Solution: The given differential equation is:

y^{\prime}+5 y=0

This can be written as:

\frac{dy}{dx} + 5y = 0

Order: The order of a differential equation is the order of the highest derivative appearing in the equation. In this equation, the highest order derivative is

y' = \frac{dy}{dx}

, which has an order of 1. Therefore, the order of the differential equation is 1.

Degree: The degree of a differential equation is the highest power of the highest order derivative, provided the equation is a polynomial in its derivatives. The given equation is a polynomial in the derivative

y'

. The highest order derivative is

y'

, and its power is 1. Therefore, the degree of the differential equation is 1.

Final Answer: The order is 1 and the degree is 1.

Q3EXERCISE 9.1

$\left( rac{d s}{d t}\right)^{4}+3 s \frac{d^{2} s}{d t^{2}}=0$

Solution

Given: The differential equation is

\left(\frac{d s}{d t}\right)^{4}+3 s \frac{d^{2} s}{d t^{2}}=0

.

To Find: The order and degree of the differential equation.

Solution: The given differential equation is:

\left(\frac{d s}{d t}\right)^{4}+3 s \frac{d^{2} s}{d t^{2}}=0

Order: The order of a differential equation is the order of the highest derivative appearing in the equation. The derivatives present are

\frac{ds}{dt}

(order 1) and

\frac{d^2s}{dt^2}

(order 2). The highest order derivative is

\frac{d^{2} s}{d t^{2}}

, which has an order of 2. Therefore, the order of the differential equation is 2.

Degree: The degree of a differential equation is the highest power of the highest order derivative, provided the equation is a polynomial in its derivatives. The given equation is a polynomial in the derivatives

\frac{ds}{dt}

and

\frac{d^2s}{dt^2}

. The highest order derivative is

\frac{d^{2} s}{d t^{2}}

, and its power in the equation is 1. Therefore, the degree of the differential equation is 1.

Final Answer: The order is 2 and the degree is 1.

Q4EXERCISE 9.1

$\left( rac{d^{2} y}{d x^{2}}\right)^{2}+\cos \left( rac{d y}{d x}\right)=0$

Solution

Given: The differential equation is

\left(\frac{d^{2} y}{d x^{2}}\right)^{2}+\cos \left(\frac{d y}{d x}\right)=0

.

To Find: The order and degree of the differential equation.

Solution: The given differential equation is:

\left(\frac{d^{2} y}{d x^{2}}\right)^{2}+\cos \left(\frac{d y}{d x}\right)=0

Order: The order of a differential equation is the order of the highest derivative appearing in the equation. The derivatives present are

\frac{d^2y}{dx^2}

(order 2) and

\frac{dy}{dx}

(order 1). The highest order derivative is

\frac{d^{2} y}{d x^{2}}

, which has an order of 2. Therefore, the order of the differential equation is 2.

Degree: The degree of a differential equation is the highest power (as a positive integer) of the highest order derivative, after the differential equation has been expressed as a polynomial in derivatives. The given equation cannot be expressed as a polynomial in its derivatives because of the term

\cos \left(\frac{dy}{dx}\right)

. The cosine function has an infinite series expansion involving powers of its argument. For example:

\cos(x) = 1 - \frac{x^2}{2!} + \frac{x^4}{4!} - \dots

So,

\cos \left(\frac{dy}{dx}\right) = 1 - \frac{1}{2!}\left(\frac{dy}{dx}\right)^2 + \frac{1}{4!}\left(\frac{dy}{dx}\right)^4 - \dots

Since the equation cannot be written as a finite polynomial in the derivatives, its degree is not defined.

Final Answer: The order is 2 and the degree is not defined.

Q5EXERCISE 9.1

$\frac{d^{2} y}{d x^{2}}=\cos 3 x+\sin 3 x$

Solution

Given: The differential equation is

\frac{d^{2} y}{d x^{2}}=\cos 3 x+\sin 3 x

.

To Find: The order and degree of the differential equation.

Solution: The given differential equation is:

\frac{d^{2} y}{d x^{2}}=\cos 3 x+\sin 3 x

Order: The order of a differential equation is the order of the highest derivative appearing in the equation. In this equation, the highest order derivative is

\frac{d^{2} y}{d x^{2}}

, which has an order of 2. Therefore, the order of the differential equation is 2.

Degree: The degree of a differential equation is the highest power of the highest order derivative, provided the equation is a polynomial in its derivatives. The given equation is a polynomial in the derivative

\frac{d^2y}{dx^2}

. The terms

\cos 3x

and

\sin 3x

are functions of

x

only and do not involve any derivatives, so they do not affect the polynomial nature of the equation with respect to its derivatives. The highest order derivative is

\frac{d^{2} y}{d x^{2}}

, and its power is 1. Therefore, the degree of the differential equation is 1.

Final Answer: The order is 2 and the degree is 1.

Q6EXERCISE 9.1

$\left(y^{\prime \prime \prime}\right)^{2}+\left(y^{\prime \prime}\right)^{3}+\left(y^{\prime}\right)^{4}+y^{5}=0$

Solution

Given: The differential equation is

\left(y^{\prime \prime \prime}\right)^{2}+\left(y^{\prime \prime}\right)^{3}+\left(y^{\prime}\right)^{4}+y^{5}=0

.

To Find: The order and degree of the differential equation.

Solution: The given differential equation can be written using Leibniz notation as:

\left(\frac{d^{3} y}{d x^{3}}\right)^{2}+\left(\frac{d^{2} y}{d x^{2}}\right)^{3}+\left(\frac{d y}{d x}\right)^{4}+y^{5}=0

Order: The order of a differential equation is determined by the order of the highest derivative present in the equation. The derivatives in the equation are

y^{\prime \prime \prime}

(third order),

y^{\prime \prime}

(second order), and

y^{\prime}

(first order). The highest order derivative is

y^{\prime \prime \prime}

or

\frac{d^{3} y}{d x^{3}}

. Thus, the order of the differential equation is 3.

Degree: The degree of a differential equation is the highest power (positive integer) of the highest order derivative, provided the equation is a polynomial in its derivatives. The given equation is a polynomial in the derivatives

y^{\prime \prime \prime}

,

y^{\prime \prime}

, and

y^{\prime}

. The highest order derivative is

y^{\prime \prime \prime}

, and its power is 2. Thus, the degree of the differential equation is 2.

Final Answer: The order of the differential equation is 3 and the degree is 2.

Q7EXERCISE 9.1

$y^{\prime \prime \prime}+2 y^{\prime \prime}+y^{\prime}=0$

Solution

Given: The differential equation is

y^{\prime \prime \prime}+2 y^{\prime \prime}+y^{\prime}=0

.

To Find: The order and degree of the differential equation.

Solution: The given differential equation can be written using Leibniz notation as:

\frac{d^{3} y}{d x^{3}}+2 \frac{d^{2} y}{d x^{2}}+\frac{d y}{d x}=0

Order: The order of a differential equation is determined by the order of the highest derivative present in the equation. The derivatives in the equation are

y^{\prime \prime \prime}

(third order),

y^{\prime \prime}

(second order), and

y^{\prime}

(first order). The highest order derivative is

y^{\prime \prime \prime}

or

\frac{d^{3} y}{d x^{3}}

. Thus, the order of the differential equation is 3.

Degree: The degree of a differential equation is the highest power of the highest order derivative, provided the equation is a polynomial in its derivatives. The given equation is a polynomial in the derivatives

y^{\prime \prime \prime}

,

y^{\prime \prime}

, and

y^{\prime}

. The highest order derivative is

y^{\prime \prime \prime}

, and its power is 1 (since

y^{\prime \prime \prime} = (y^{\prime \prime \prime})^1

). Thus, the degree of the differential equation is 1.

Final Answer: The order of the differential equation is 3 and the degree is 1.

Q8EXERCISE 9.1

$y^{\prime}+y=e^{x}$

Solution

Given: The differential equation is

y^{\prime}+y=e^{x}

.

To Find: The order and degree of the differential equation.

Solution: The given differential equation can be written using Leibniz notation as:

\frac{d y}{d x}+y=e^{x}

Order: The order of a differential equation is determined by the order of the highest derivative present in the equation. The only derivative present is

y^{\prime}

(first order). The highest order derivative is

y^{\prime}

or

\frac{d y}{d x}

. Thus, the order of the differential equation is 1.

Degree: The degree of a differential equation is the highest power of the highest order derivative, provided the equation is a polynomial in its derivatives. The given equation is a polynomial in the derivative

y^{\prime}

. The highest order derivative is

y^{\prime}

, and its power is 1. Thus, the degree of the differential equation is 1.

Final Answer: The order of the differential equation is 1 and the degree is 1.

Q9EXERCISE 9.1

$y^{\prime \prime}+\left(y^{\prime}\right)^{2}+2 y=0$

Solution

Given: The differential equation is

y^{\prime \prime}+\left(y^{\prime}\right)^{2}+2 y=0

.

To Find: The order and degree of the differential equation.

Solution: The given differential equation can be written using Leibniz notation as:

\frac{d^{2} y}{d x^{2}}+\left(\frac{d y}{d x}\right)^{2}+2 y=0

Order: The order of a differential equation is determined by the order of the highest derivative present in the equation. The derivatives in the equation are

y^{\prime \prime}

(second order) and

y^{\prime}

(first order). The highest order derivative is

y^{\prime \prime}

or

\frac{d^{2} y}{d x^{2}}

. Thus, the order of the differential equation is 2.

Degree: The degree of a differential equation is the highest power of the highest order derivative, provided the equation is a polynomial in its derivatives. The given equation is a polynomial in the derivatives

y^{\prime \prime}

and

y^{\prime}

. The highest order derivative is

y^{\prime \prime}

, and its power is 1. Thus, the degree of the differential equation is 1.

Final Answer: The order of the differential equation is 2 and the degree is 1.

Q10EXERCISE 9.1

$y^{\prime \prime}+2 y^{\prime}+\sin y=0$

Solution

Given: The differential equation is

y^{\prime \prime}+2 y^{\prime}+\sin y=0

.

To Find: The order and degree of the differential equation.

Solution: The given differential equation can be written using Leibniz notation as:

\frac{d^{2} y}{d x^{2}}+2 \frac{d y}{d x}+\sin y=0

Order: The order of a differential equation is determined by the order of the highest derivative present in the equation. The derivatives in the equation are

y^{\prime \prime}

(second order) and

y^{\prime}

(first order). The highest order derivative is

y^{\prime \prime}

or

\frac{d^{2} y}{d x^{2}}

. Thus, the order of the differential equation is 2.

Degree: The degree of a differential equation is the highest power of the highest order derivative, provided the equation is a polynomial in its derivatives. The given equation is a polynomial in the derivatives

y^{\prime \prime}

and

y^{\prime}

. The term

\sin y

involves the dependent variable

y

, not a derivative, so the equation is still a polynomial in terms of its derivatives. The highest order derivative is

y^{\prime \prime}

, and its power is 1. Thus, the degree of the differential equation is 1.

Final Answer: The order of the differential equation is 2 and the degree is 1.

Q11EXERCISE 9.1

The degree of the differential equation $\left( rac{d^{2} y}{d x^{2}}\right)^{3}+\left( rac{d y}{d x}\right)^{2}+\sin \left( rac{d y}{d x}\right)+1=0$ is

(A)

3

(B)

2

(C)

1

(D)

not defined

Solution

Given: The differential equation is

\left(\frac{d^{2} y}{d x^{2}}\right)^{3}+\left(\frac{d y}{d x}\right)^{2}+\sin \left(\frac{d y}{d x}\right)+1=0

.

To Find: The degree of the given differential equation.

Solution: The order of a differential equation is the order of the highest derivative appearing in it. Here, the highest order derivative is

\frac{d^2y}{dx^2}

, so the order is 2.

The degree of a differential equation is the highest power (positive integer) of the highest order derivative, after the differential equation is made free from radicals and fractions as far as the derivatives are concerned. A differential equation must be a polynomial equation in its derivatives for the degree to be defined.

The given equation is:

\left(\frac{d^{2} y}{d x^{2}}\right)^{3}+\left(\frac{d y}{d x}\right)^{2}+\sin \left(\frac{d y}{d x}\right)+1=0

This equation is not a polynomial equation in its derivatives because of the term

\sin \left(\frac{d y}{d x}\right)

. The Taylor series expansion of

\sin(z)

is

z - \frac{z^3}{3!} + \frac{z^5}{5!} - \dots

. Substituting

z = \frac{dy}{dx}

results in an infinite series of powers of

\frac{dy}{dx}

.

Since the differential equation cannot be expressed as a polynomial in its derivatives, its degree is not defined.

Final Answer: The correct option is (D) not defined.

Q12EXERCISE 9.1

The order of the differential equation $2 x^{2} \frac{d^{2} y}{d x^{2}}-3 \frac{d y}{d x}+y=0$ is

(A)

2

(B)

1

(C)

0

(D)

not defined

Solution

Given: The differential equation is

2 x^{2} \frac{d^{2} y}{d x^{2}}-3 \frac{d y}{d x}+y=0

.

To Find: The order of the given differential equation.

Solution: The order of a differential equation is defined as the order of the highest derivative present in the equation.

In the given equation, the derivatives present are:

$\frac{d^{2} y}{d x^{2}}$ , which is of order 2.
$\frac{d y}{d x}$ , which is of order 1.

The highest order of the derivatives is 2. Therefore, the order of the differential equation is 2.

Final Answer: The correct option is (A) 2.

Q1EXERCISE 9.2

$y=e^{x}+1$ : $y''-y'=0$

Solution

To Verify: That the function

y=e^{x}+1

is a solution of the differential equation

y''-y'=0

.

Given Function:

y = e^x + 1

Solution: First, we find the first and second derivatives of the given function with respect to

x

.

Differentiating

y

with respect to

x

, we get:

y' = \frac{d}{dx}(e^x + 1) = e^x

Differentiating

y'

with respect to

x

, we get:

y'' = \frac{d}{dx}(e^x) = e^x

Now, we substitute the expressions for

y'

and

y''

into the left-hand side (LHS) of the given differential equation,

y'' - y' = 0

.

LHS:

y'' - y' = e^x - e^x = 0

RHS:

0

Since LHS = RHS, the given function

y=e^{x}+1

is a solution to the differential equation

y''-y'=0

.

Hence verified.

Q2EXERCISE 9.2

$y=x^{2}+2 x+\mathrm{C}$ : $y'-2 x-2=0$

Solution

To Verify: That the function

y=x^{2}+2 x+\mathrm{C}

is a solution of the differential equation

y'-2 x-2=0

.

Given Function:

y = x^2 + 2x + C

where C is an arbitrary constant.

Solution: First, we find the first derivative of the given function with respect to

x

.

Differentiating

y

with respect to

x

, we get:

y' = \frac{d}{dx}(x^2 + 2x + C) = 2x + 2

Now, we substitute the expression for

y'

into the left-hand side (LHS) of the given differential equation,

y' - 2x - 2 = 0

.

LHS:

y' - 2x - 2 = (2x + 2) - 2x - 2

= 2x - 2x + 2 - 2 = 0

RHS:

0

Since LHS = RHS, the given function

y=x^{2}+2 x+\mathrm{C}

is a solution to the differential equation

y'-2 x-2=0

.

Hence verified.

Q3EXERCISE 9.2

$y=\cos x+\mathrm{C}$ : $\quad y'+\sin x=0$

Solution

To Verify: That the function

y=\cos x+\mathrm{C}

is a solution of the differential equation

y'+\sin x=0

.

Given Function:

y = \cos x + C

where C is an arbitrary constant.

Solution: First, we find the first derivative of the given function with respect to

x

.

Differentiating

y

with respect to

x

, we get:

y' = \frac{d}{dx}(\cos x + C) = -\sin x

Now, we substitute the expression for

y'

into the left-hand side (LHS) of the given differential equation,

y' + \sin x = 0

.

LHS:

y' + \sin x = (-\sin x) + \sin x = 0

RHS:

0

Since LHS = RHS, the given function

y=\cos x+\mathrm{C}

is a solution to the differential equation

y'+\sin x=0

.

Hence verified.

Q4EXERCISE 9.2

$y=\sqrt{1+x^{2}}$ : $y'=\frac{x y}{1+x^{2}}$

Solution

Given: The function is

y=\sqrt{1+x^{2}}

. The differential equation is

y'=\frac{x y}{1+x^{2}}

.

To Verify: That the given function is a solution to the corresponding differential equation.

Solution: We start with the given function:

y = \sqrt{1+x^{2}}

Now, we differentiate this function with respect to

x

to find

y'

:

y' = \frac{d}{dx} (\sqrt{1+x^{2}})

Using the chain rule,

\frac{d}{dx}(\sqrt{u}) = \frac{1}{2\sqrt{u}} \frac{du}{dx}

:

y' = \frac{1}{2\sqrt{1+x^{2}}} \cdot \frac{d}{dx}(1+x^2)

y' = \frac{1}{2\sqrt{1+x^{2}}} \cdot (2x)

y' = \frac{x}{\sqrt{1+x^{2}}}

This is the Left Hand Side (LHS) of the differential equation.

Now, let's evaluate the Right Hand Side (RHS) of the differential equation, which is

\frac{x y}{1+x^{2}}

. Substitute the expression for

y

into the RHS:

\text{RHS} = \frac{x (\sqrt{1+x^{2}})}{1+x^{2}}

Since

1+x^2 = (\sqrt{1+x^2})^2

, we can simplify the expression:

\text{RHS} = \frac{x \sqrt{1+x^{2}}}{(\sqrt{1+x^{2}})^2}

\text{RHS} = \frac{x}{\sqrt{1+x^{2}}}

Comparing the LHS and RHS:

\text{LHS} = y' = \frac{x}{\sqrt{1+x^{2}}}

\text{RHS} = \frac{x y}{1+x^{2}} = \frac{x}{\sqrt{1+x^{2}}}

Since LHS = RHS, the given function is a solution to the differential equation.

Hence Verified.

Q5EXERCISE 9.2

$y=\mathrm{A} x$ : $\quad x y'=y(x \neq 0)$

Solution

Given: The function is

y = Ax

. The differential equation is

xy' = y

, where

x \neq 0

.

To Verify: That the given function is a solution to the corresponding differential equation.

Solution: We start with the given function:

y = Ax

Now, we differentiate this function with respect to

x

to find

y'

:

y' = \frac{d}{dx} (Ax)

y' = A

This is the derivative of the function.

Now, let's evaluate the Left Hand Side (LHS) of the differential equation, which is

xy'

. Substitute the expression for

y'

into the LHS:

\text{LHS} = x \cdot y' = x \cdot A = Ax

The Right Hand Side (RHS) of the differential equation is simply

y

. From the given function, we have:

\text{RHS} = y = Ax

Comparing the LHS and RHS:

\text{LHS} = Ax

\text{RHS} = Ax

Since LHS = RHS, the given function is a solution to the differential equation.

Hence Verified.

Q6EXERCISE 9.2

$y=x \sin x : \quad x y'=y+x \sqrt{x^{2}-y^{2}}(x \neq 0$ and $x>y$ or $x<-y)$

Solution

Given: The function is

y = x \sin x

. The differential equation is

xy' = y + x \sqrt{x^2 - y^2}

, for

x \neq 0

and

x>y

or

x<-y

.

To Verify: That the given function is a solution to the corresponding differential equation.

Solution: We start with the given function:

y = x \sin x

Now, we differentiate this function with respect to

x

using the product rule,

(uv)' = u'v + uv'

:

y' = \frac{d}{dx} (x \sin x)

y' = (1) \cdot \sin x + x \cdot (\cos x)

y' = \sin x + x \cos x

Now, let's evaluate the Left Hand Side (LHS) of the differential equation, which is

xy'

. Substitute the expression for

y'

into the LHS:

\text{LHS} = x \cdot y' = x(\sin x + x \cos x)

\text{LHS} = x \sin x + x^2 \cos x

Now, let's evaluate the Right Hand Side (RHS) of the differential equation, which is

y + x \sqrt{x^2 - y^2}

. Substitute the expression for

y

into the RHS:

\text{RHS} = (x \sin x) + x \sqrt{x^2 - (x \sin x)^2}

\text{RHS} = x \sin x + x \sqrt{x^2 - x^2 \sin^2 x}

Factor out

x^2

from under the square root:

\text{RHS} = x \sin x + x \sqrt{x^2(1 - \sin^2 x)}

Using the trigonometric identity

1 - \sin^2 x = \cos^2 x

:

\text{RHS} = x \sin x + x \sqrt{x^2 \cos^2 x}

\text{RHS} = x \sin x + x |x \cos x|

For the LHS to be equal to the RHS, we must have:

x \sin x + x^2 \cos x = x \sin x + x |x \cos x|

x^2 \cos x = x |x \cos x|

This equality holds only if

x \cos x \ge 0

. The conditions given in the problem (

x>y

or

x<-y

) do not guarantee that

x \cos x \ge 0

for all valid

x

. For example, if

x=\pi

, then

y=0

. The condition

x>y

is satisfied. However, LHS

= -\pi^2

and RHS

= \pi^2

. Assuming there is a tacit assumption in the problem that the expression is restricted to domains where

x \cos x \geq 0

, then

|x \cos x| = x \cos x

. In that case:

\text{RHS} = x \sin x + x (x \cos x) = x \sin x + x^2 \cos x

Comparing LHS and RHS under this assumption:

\text{LHS} = x \sin x + x^2 \cos x

\text{RHS} = x \sin x + x^2 \cos x

Under the assumption that

x \cos x \ge 0

, LHS = RHS.

Hence Verified (with the assumption that $x \cos x \ge 0$ ).

Q7EXERCISE 9.2

: $y'=\frac{y^{2}}{1-x y}(x y \neq 1)$

Solution

Given: The relation is

xy = \log y + C

. The differential equation is

y' = \frac{y^2}{1-xy}

, where

xy \neq 1

.

To Verify: That the given relation implicitly defines a function that is a solution to the differential equation.

Solution: We start with the given relation:

xy = \log y + C

We differentiate this relation implicitly with respect to

x

. We use the product rule for the term

xy

and the chain rule for the term

\log y

.

\frac{d}{dx}(xy) = \frac{d}{dx}(\log y) + \frac{d}{dx}(C)

1 \cdot y + x \cdot \frac{dy}{dx} = \frac{1}{y} \cdot \frac{dy}{dx} + 0

Let's use the notation

y'

for

\frac{dy}{dx}

:

y + xy' = \frac{1}{y} y'

Now, we need to solve this equation for

y'

. We group the terms containing

y'

on one side.

y = \frac{1}{y} y' - xy'

y = y' \left(\frac{1}{y} - x\right)

y = y' \left(\frac{1 - xy}{y}\right)

Now, we can isolate

y'

:

y' = \frac{y}{\left(\frac{1 - xy}{y}\right)}

y' = \frac{y \cdot y}{1 - xy}

y' = \frac{y^2}{1 - xy}

This is the given differential equation. The condition

xy \neq 1

ensures that the denominator is not zero.

Hence Verified.

Q8EXERCISE 9.2

$:(y \sin y+\cos y+x) y'=y$

Solution

Given: The relation is

y - \cos y = x

. The differential equation is

(y \sin y + \cos y + x)y' = y

.

To Verify: That the given relation implicitly defines a function that is a solution to the differential equation.

Solution: We start with the given relation:

y - \cos y = x

We differentiate this relation implicitly with respect to

x

. We use the chain rule for the term

\cos y

.

\frac{d}{dx}(y) - \frac{d}{dx}(\cos y) = \frac{d}{dx}(x)

\frac{dy}{dx} - (-\sin y) \cdot \frac{dy}{dx} = 1

Let's use the notation

y'

for

\frac{dy}{dx}

:

y' + y' \sin y = 1

Factor out

y'

:

y'(1 + \sin y) = 1

Solve for

y'

:

y' = \frac{1}{1 + \sin y}

Now we have an expression for

y'

. Let's substitute this and the original relation into the Left Hand Side (LHS) of the given differential equation. The LHS is

(y \sin y + \cos y + x)y'

. From the given relation, we can express

x

as

y - \cos y

. Substitute this into the LHS expression:

\text{LHS} = (y \sin y + \cos y + (y - \cos y)) y'

Simplify the term in the parenthesis:

\text{LHS} = (y \sin y + y) y'

Factor out

y

from the parenthesis:

\text{LHS} = y(1 + \sin y) y'

Now, substitute the expression we found for

y'

, which is

y' = \frac{1}{1 + \sin y}

:

\text{LHS} = y(1 + \sin y) \left(\frac{1}{1 + \sin y}\right)

Assuming

1 + \sin y \neq 0

(which is true since

y'

is defined), we can cancel the

(1 + \sin y)

terms:

\text{LHS} = y

The Right Hand Side (RHS) of the differential equation is

y

. So, we have shown that LHS =

y

and RHS =

y

.

Since LHS = RHS, the given relation defines a solution to the differential equation.

Hence Verified.

Q9EXERCISE 9.2

$x+y=\tan ^{-1} y$ : $y^{2} y'+y^{2}+1=0$

Solution

To Verify: The function

x+y=\tan^{-1} y

is a solution to the differential equation

y^{2} y'+y^{2}+1=0

.

Given function:

x+y=\tan^{-1} y

Solution: Differentiating the given function with respect to

x

, we get:

\frac{d}{dx}(x) + \frac{d}{dx}(y) = \frac{d}{dx}(\tan^{-1} y)

1 + \frac{dy}{dx} = \frac{1}{1+y^2} \frac{dy}{dx}

Let

y' = \frac{dy}{dx}

. The equation becomes:

1 + y' = \frac{1}{1+y^2} y'

Multiplying both sides by

(1+y^2)

:

(1+y^2)(1+y') = y'

1 + y' + y^2 + y^2 y' = y'

Subtracting

y'

from both sides:

1 + y^2 + y^2 y' = 0

Rearranging the terms, we get:

y^2 y' + y^2 + 1 = 0

This is the same as the given differential equation.

Since the given function satisfies the differential equation, it is a solution.

Hence Verified.

Q10EXERCISE 9.2

$y=\sqrt{a^{2}-x^{2}} x \in(-a, a): \quad x+y \frac{d y}{d x}=0(y \neq 0)$

Solution

To Verify: The function

y=\sqrt{a^{2}-x^{2}}

for

x \in(-a, a)

is a solution to the differential equation

x+y \frac{d y}{d x}=0

(where

y \neq 0

).

Given function:

y=\sqrt{a^{2}-x^{2}}

Solution: We can write the given function as:

y^2 = a^2 - x^2

Differentiating both sides with respect to

x

:

\frac{d}{dx}(y^2) = \frac{d}{dx}(a^2 - x^2)

2y \frac{dy}{dx} = 0 - 2x

2y \frac{dy}{dx} = -2x

Since it is given that

y \neq 0

, we can divide both sides by 2:

y \frac{dy}{dx} = -x

Rearranging the terms, we get:

x + y \frac{dy}{dx} = 0

This is the same as the given differential equation.

Since the given function satisfies the differential equation, it is a solution.

Hence Verified.

Q11EXERCISE 9.2

The number of arbitrary constants in the general solution of a differential equation of fourth order are:

(A)

0

(B)

2

(C)

3

(D)

4

Solution

Question: The number of arbitrary constants in the general solution of a differential equation of fourth order are: (A) 0 (B) 2 (C) 3 (D) 4

Concept: The order of a differential equation corresponds to the number of arbitrary constants present in its general solution. The general solution of an

n^{th}

order differential equation will have exactly

n

arbitrary constants.

Solution: The given differential equation is of the fourth order. Therefore, its general solution will contain 4 arbitrary constants.

Final Answer: The correct option is (D).

Answer: (D)

Explanation: The number of arbitrary constants in the general solution of a differential equation is equal to its order. Since the order is 4, there are 4 arbitrary constants.

Q12EXERCISE 9.2

The number of arbitrary constants in the particular solution of a differential equation of third order are:

(A)

3

(B)

2

(C)

1

(D)

0

Solution

Question: The number of arbitrary constants in the particular solution of a differential equation of third order are: (A) 3 (B) 2 (C) 1 (D) 0

Concept: A particular solution of a differential equation is a solution obtained from the general solution by assigning specific values to the arbitrary constants. These specific values are usually determined by initial or boundary conditions.

Solution: The given differential equation is of the third order. Its general solution would have 3 arbitrary constants. However, a particular solution is free of arbitrary constants because all the constants have been replaced by specific numerical values. Therefore, the number of arbitrary constants in any particular solution is always zero, regardless of the order of the differential equation.

Final Answer: The correct option is (D).

Answer: (D)

Explanation: A particular solution has no arbitrary constants; it is a specific solution that satisfies given conditions.

Q1EXERCISE 9.3

$\frac{d y}{d x}=\frac{1-\cos x}{1+\cos x}$

Solution

Given: The differential equation

\frac{d y}{d x}=\frac{1-\cos x}{1+\cos x}

.

To Find: The general solution of the differential equation.

Solution: The given differential equation can be solved by the method of separation of variables.

dy = \left( \frac{1-\cos x}{1+\cos x} \right) dx

Integrating both sides, we get:

\int dy = \int \frac{1-\cos x}{1+\cos x} dx

y = \int \frac{1-\cos x}{1+\cos x} dx + C \quad \text{...(1)}

where C is the constant of integration.

To evaluate the integral, we use the half-angle trigonometric identities:

1 - \cos x = 2\sin^2\left(\frac{x}{2}\right)

1 + \cos x = 2\cos^2\left(\frac{x}{2}\right)

Substituting these into the integral:

\int \frac{1-\cos x}{1+\cos x} dx = \int \frac{2\sin^2(x/2)}{2\cos^2(x/2)} dx

= \int \tan^2\left(\frac{x}{2}\right) dx

Using the identity

\tan^2\theta = \sec^2\theta - 1

:

= \int \left[ \sec^2\left(\frac{x}{2}\right) - 1 \right] dx

= \int \sec^2\left(\frac{x}{2}\right) dx - \int 1 dx

Now, we integrate each term:

\int \sec^2\left(\frac{x}{2}\right) dx = \frac{\tan(x/2)}{1/2} = 2\tan\left(\frac{x}{2}\right)

\int 1 dx = x

So, the integral is

2\tan\left(\frac{x}{2}\right) - x

.

Substituting this result back into equation (1):

y = 2\tan\left(\frac{x}{2}\right) - x + C

Final Answer: The general solution of the given differential equation is

y = 2\tan\left(\frac{x}{2}\right) - x + C

.

Q2EXERCISE 9.3

$\frac{d y}{d x}=\sqrt{4-y^{2}}\quad(-2<y<2)$

Solution

Given: The differential equation is

\frac{dy}{dx}=\sqrt{4-y^{2}}\quad(-2<y<2)

To Find: The general solution of the differential equation.

Solution: The given differential equation can be written as:

\frac{dy}{dx} = \sqrt{4-y^2}

This is a separable differential equation. Separating the variables, we get:

\frac{dy}{\sqrt{4-y^2}} = dx

Now, integrating both sides:

\int \frac{dy}{\sqrt{4-y^2}} = \int dx

We use the standard integration formula

\int \frac{1}{\sqrt{a^2 - x^2}} dx = \sin^{-1}\left(\frac{x}{a}\right)

. Here,

a=2

.

\sin^{-1}\left(\frac{y}{2}\right) = x + C

where C is the constant of integration.

This can also be expressed as:

\frac{y}{2} = \sin(x+C)

y = 2\sin(x+C)

Final Answer: The general solution of the given differential equation is

\sin^{-1}\left(\frac{y}{2}\right) = x + C

.

Q3EXERCISE 9.3

$\frac{d y}{d x}+y=1(y \neq 1)$

Solution

Given: The differential equation is

\frac{dy}{dx}+y=1 \quad (y \neq 1)

To Find: The general solution of the differential equation.

Solution: The given differential equation can be written as:

\frac{dy}{dx} = 1 - y

This is a separable differential equation. Since

y \neq 1

, we can divide by

(1-y)

. Separating the variables, we get:

\frac{dy}{1-y} = dx

Now, integrating both sides:

\int \frac{dy}{1-y} = \int dx

-\ln|1-y| = x + C_1

To simplify, we can multiply by -1:

\ln|1-y| = -x - C_1

Taking the exponent of both sides:

|1-y| = e^{-x-C_1} = e^{-x} \cdot e^{-C_1}

1-y = \pm e^{-C_1} \cdot e^{-x}

Let

C = \pm e^{-C_1}

be an arbitrary non-zero constant.

1-y = Ce^{-x}

y = 1 - Ce^{-x}

Final Answer: The general solution of the given differential equation is

y = 1 - Ce^{-x}

.

Q4EXERCISE 9.3

$\sec ^{2} x \tan y d x+\sec ^{2} y \tan x d y=0$

Solution

Given: The differential equation is

\sec^{2} x \tan y dx+\sec^{2} y \tan x dy=0

To Find: The general solution of the differential equation.

Solution: The given differential equation is:

\sec^{2} x \tan y dx+\sec^{2} y \tan x dy=0

This is a separable differential equation. Rearranging the terms to separate variables:

\sec^{2} y \tan x dy = -\sec^{2} x \tan y dx

Assuming

\tan x \neq 0

and

\tan y \neq 0

, we can divide both sides by

\tan x \tan y

:

\frac{\sec^2 y}{\tan y} dy = -\frac{\sec^2 x}{\tan x} dx

Now, integrating both sides:

\int \frac{\sec^2 y}{\tan y} dy = -\int \frac{\sec^2 x}{\tan x} dx

Both integrals are of the form

\int \frac{f'(u)}{f(u)} du = \ln|f(u)|

. For the left side, let

t = \tan y

, then

dt = \sec^2 y dy

. The integral is

\int \frac{1}{t} dt = \ln|t| = \ln|\tan y|

. For the right side, let

u = \tan x

, then

du = \sec^2 x dx

. The integral is

\int \frac{1}{u} du = \ln|u| = \ln|\tan x|

.

So, the equation becomes:

\ln|\tan y| = -\ln|\tan x| + \ln|C|

where

\ln|C|

is the constant of integration.

Using properties of logarithms:

\ln|\tan y| + \ln|\tan x| = \ln|C|

\ln|\tan x \tan y| = \ln|C|

|\tan x \tan y| = |C|

\tan x \tan y = \pm C

Let's denote

\pm C

by a new constant

K

.

\tan x \tan y = K

Final Answer: The general solution of the given differential equation is

\tan x \tan y = K

.

Q5EXERCISE 9.3

$(e^{x}+e^{-x}) d y-(e^{x}-e^{-x}) d x=0$

Solution

Given: The differential equation is

(e^{x}+e^{-x}) dy-(e^{x}-e^{-x}) dx=0

To Find: The general solution of the differential equation.

Solution: The given differential equation is:

(e^{x}+e^{-x}) dy - (e^{x}-e^{-x}) dx = 0

This is a separable differential equation. Rearranging the terms:

(e^{x}+e^{-x}) dy = (e^{x}-e^{-x}) dx

Separating the variables, we get:

dy = \frac{e^{x}-e^{-x}}{e^{x}+e^{-x}} dx

Now, integrating both sides:

\int dy = \int \frac{e^{x}-e^{-x}}{e^{x}+e^{-x}} dx

For the integral on the right-hand side, we use the substitution method. Let

t = e^{x}+e^{-x}

. Then,

dt = (e^{x} - e^{-x}) dx

.

The integral becomes:

\int dy = \int \frac{1}{t} dt

y = \ln|t| + C

Substituting back the value of

t

:

y = \ln|e^{x}+e^{-x}| + C

Since

e^x > 0

and

e^{-x} > 0

for all real

x

, their sum

e^{x}+e^{-x}

is always positive. Therefore, the absolute value sign can be removed.

y = \ln(e^{x}+e^{-x}) + C

Final Answer: The general solution of the given differential equation is

y = \ln(e^{x}+e^{-x}) + C

.

Q6EXERCISE 9.3

$\frac{d y}{d x}=\left(1+x^{2}\right)\left(1+y^{2}\right)$

Solution

Given: The differential equation is

\frac{dy}{dx}=\left(1+x^{2}\right)\left(1+y^{2}\right)

To Find: The general solution of the differential equation.

Solution: The given differential equation is:

\frac{dy}{dx} = (1+x^2)(1+y^2)

This is a separable differential equation. Separating the variables, we get:

\frac{dy}{1+y^2} = (1+x^2) dx

Now, integrating both sides:

\int \frac{dy}{1+y^2} = \int (1+x^2) dx

We use the standard integration formulas:

\int \frac{1}{1+u^2} du = \tan^{-1}(u)

and

\int u^n du = \frac{u^{n+1}}{n+1}

.

Integrating the left side:

\int \frac{dy}{1+y^2} = \tan^{-1}(y)

Integrating the right side:

\int (1+x^2) dx = \int 1 dx + \int x^2 dx = x + \frac{x^3}{3}

Combining the results and adding the constant of integration C:

\tan^{-1}(y) = x + \frac{x^3}{3} + C

Final Answer: The general solution of the given differential equation is

\tan^{-1}(y) = x + \frac{x^3}{3} + C

.

Q7EXERCISE 9.3

$y \log y d x-x d y=0$

Solution

Given: The differential equation is

y \log y dx - x dy = 0

.

Solution: We can rewrite the equation as:

y \log y dx = x dy

To solve this differential equation, we use the method of separation of variables. We group the terms involving

x

and

dx

on one side, and the terms involving

y

and

dy

on the other side.

\frac{dx}{x} = \frac{dy}{y \log y}

Now, we integrate both sides of the equation:

\int \frac{1}{x} dx = \int \frac{1}{y \log y} dy

The integral of the left side is:

\int \frac{1}{x} dx = \log|x|

For the integral on the right side, we use the substitution method. Let

t = \log y

. Then, differentiating with respect to

y

, we get

dt = \frac{1}{y} dy

.

Substituting these into the integral:

\int \frac{1}{y \log y} dy = \int \frac{1}{t} dt = \log|t| = \log|\log y|

Equating the results of both integrations, we get:

\log|x| = \log|\log y| + C

where

C

is the constant of integration. We can write the constant

C

as

\log C_1

for convenience.

\log|x| = \log|\log y| + \log C_1

Using the property of logarithms,

\log a + \log b = \log(ab)

:

\log|x| = \log(C_1 |\log y|)

Taking the antilogarithm on both sides:

|x| = C_1 |\log y|

x = \pm C_1 \log y

Let

k = \pm C_1

be an arbitrary constant. Then the general solution is:

x = k \log y

This can also be expressed as:

\log y = \frac{x}{k}

y = e^{\frac{x}{k}}

Letting

A = 1/k

, another constant, we get:

y = e^{Ax}

Final Answer: The general solution of the given differential equation is

y = e^{Ax}

.

Q8EXERCISE 9.3

$x^{5} \frac{d y}{d x}=-y^{5}$

Solution

Given: The differential equation is

x^5 \frac{dy}{dx} = -y^5

.

Solution: We solve this differential equation using the method of separation of variables. First, we separate the variables

x

and

y

:

\frac{dy}{-y^5} = \frac{dx}{x^5}

-y^{-5} dy = x^{-5} dx

Next, we integrate both sides of the equation:

\int -y^{-5} dy = \int x^{-5} dx

Using the power rule for integration,

\int u^n du = \frac{u^{n+1}}{n+1}

: LHS:

\int -y^{-5} dy = -\frac{y^{-5+1}}{-5+1} = -\frac{y^{-4}}{-4} = \frac{y^{-4}}{4}

RHS:

\int x^{-5} dx = \frac{x^{-5+1}}{-5+1} = \frac{x^{-4}}{-4}

Equating the results and adding the constant of integration,

C

:

\frac{y^{-4}}{4} = -\frac{x^{-4}}{4} + C

To simplify the expression, we can multiply the entire equation by 4:

y^{-4} = -x^{-4} + 4C

Let

K = 4C

be a new arbitrary constant.

y^{-4} = -x^{-4} + K

Rearranging the terms to get the final form of the solution:

x^{-4} + y^{-4} = K

This can also be written as:

\frac{1}{x^4} + \frac{1}{y^4} = K

Final Answer: The general solution of the given differential equation is

x^{-4} + y^{-4} = K

.

Q9EXERCISE 9.3

$\frac{d y}{d x}=\sin ^{-1} x$

Solution

Given: The differential equation is

\frac{dy}{dx} = \sin^{-1} x

.

Solution: This is a first-order differential equation that can be solved by direct integration. We separate the variables:

dy = \sin^{-1} x \, dx

Now, we integrate both sides:

\int dy = \int \sin^{-1} x \, dx

y = \int \sin^{-1} x \, dx + C

where

C

is the constant of integration.

To evaluate the integral

\int \sin^{-1} x \, dx

, we use integration by parts. The formula for integration by parts is

\int u \, dv = uv - \int v \, du

. Let

u = \sin^{-1} x

and

dv = dx

. Then, we have

du = \frac{1}{\sqrt{1-x^2}} dx

and

v = \int dx = x

.

Applying the integration by parts formula:

\int \sin^{-1} x \, dx = (\sin^{-1} x)(x) - \int x \left(\frac{1}{\sqrt{1-x^2}}\right) dx

\int \sin^{-1} x \, dx = x \sin^{-1} x - \int \frac{x}{\sqrt{1-x^2}} dx

To solve the remaining integral,

\int \frac{x}{\sqrt{1-x^2}} dx

, we use substitution. Let

t = 1-x^2

. Then

dt = -2x \, dx

, which implies

x \, dx = -\frac{1}{2} dt

.

Substituting this into the integral:

\int \frac{x}{\sqrt{1-x^2}} dx = \int \frac{1}{\sqrt{t}} \left(-\frac{1}{2} dt\right) = -\frac{1}{2} \int t^{-1/2} dt

= -\frac{1}{2} \frac{t^{1/2}}{1/2} = -t^{1/2} = -\sqrt{t}

Substituting back

t = 1-x^2

:

\int \frac{x}{\sqrt{1-x^2}} dx = -\sqrt{1-x^2}

Now, substitute this result back into the integration by parts expression:

\int \sin^{-1} x \, dx = x \sin^{-1} x - (-\sqrt{1-x^2}) = x \sin^{-1} x + \sqrt{1-x^2}

Finally, substituting this into the expression for

y

:

y = x \sin^{-1} x + \sqrt{1-x^2} + C

Final Answer: The general solution of the given differential equation is

y = x \sin^{-1} x + \sqrt{1-x^2} + C

.

Q10EXERCISE 9.3

$e^{x} \tan y d x+\left(1-e^{x}\right) \sec ^{2} y d y=0$

Solution

Given: The differential equation is

e^{x} \tan y dx + (1-e^{x}) \sec^{2} y dy = 0

.

Solution: To solve this differential equation, we use the method of separation of variables. First, we rearrange the equation:

(1-e^{x}) \sec^{2} y dy = -e^{x} \tan y dx

Now, we separate the variables by moving all terms with

y

to the left side and all terms with

x

to the right side:

\frac{\sec^2 y}{\tan y} dy = \frac{-e^x}{1-e^x} dx

\frac{\sec^2 y}{\tan y} dy = \frac{e^x}{e^x-1} dx

Next, we integrate both sides of the equation:

\int \frac{\sec^2 y}{\tan y} dy = \int \frac{e^x}{e^x-1} dx

For the left-hand side integral, let

u = \tan y

. Then

du = \sec^2 y dy

. The integral becomes:

\int \frac{1}{u} du = \log|u| = \log|\tan y|

For the right-hand side integral, let

v = e^x - 1

. Then

dv = e^x dx

. The integral becomes:

\int \frac{1}{v} dv = \log|v| = \log|e^x-1|

Equating the results of both integrations and adding a constant of integration,

\log|C|

:

\log|\tan y| = \log|e^x-1| + \log|C|

Using the property of logarithms,

\log a + \log b = \log(ab)

:

\log|\tan y| = \log|C(e^x-1)|

Taking the antilogarithm on both sides gives the general solution:

|\tan y| = |C(e^x-1)|

\tan y = C(e^x-1)

where

C

is an arbitrary constant.

Final Answer: The general solution of the given differential equation is

\tan y = C(e^x-1)

.

Q11EXERCISE 9.3

$(x^{3}+x^{2}+x+1) \frac{d y}{d x}=2 x^{2}+x ; y=1$ when $x=0$

Solution

Given: The differential equation

(x^{3}+x^{2}+x+1) \frac{d y}{d x}=2 x^{2}+x

, with the initial condition

y=1

when

x=0

.

Solution: First, we find the general solution of the differential equation using the method of separation of variables.

dy = \frac{2x^2+x}{x^3+x^2+x+1} dx

We need to integrate the right-hand side. Let's first factor the denominator:

x^3+x^2+x+1 = x^2(x+1) + 1(x+1) = (x^2+1)(x+1)

So the equation becomes:

dy = \frac{2x^2+x}{(x^2+1)(x+1)} dx

We use partial fraction decomposition for the expression on the right. Let

\frac{2x^2+x}{(x^2+1)(x+1)} = \frac{A}{x+1} + \frac{Bx+C}{x^2+1}

2x^2+x = A(x^2+1) + (Bx+C)(x+1)

To find

A

, let

x = -1

:

2(-1)^2 + (-1) = A((-1)^2+1) + 0

2-1 = A(2) \implies 1 = 2A \implies A = \frac{1}{2}

Expanding the equation:

2x^2+x = Ax^2+A + Bx^2+Bx+Cx+C

2x^2+x = (A+B)x^2 + (B+C)x + (A+C)

Comparing coefficients of powers of

x

:

$x^2$ : $A+B = 2 \implies \frac{1}{2}+B = 2 \implies B = 2 - \frac{1}{2} = \frac{3}{2}$
$x$ : $B+C = 1 \implies \frac{3}{2}+C = 1 \implies C = 1 - \frac{3}{2} = -\frac{1}{2}$

So, the partial fraction decomposition is:

\frac{2x^2+x}{(x^2+1)(x+1)} = \frac{1/2}{x+1} + \frac{(3/2)x - 1/2}{x^2+1} = \frac{1}{2} \frac{1}{x+1} + \frac{1}{2} \frac{3x-1}{x^2+1}

Now, we integrate both sides of the differential equation:

\int dy = \int \left(\frac{1}{2} \frac{1}{x+1} + \frac{1}{2} \frac{3x-1}{x^2+1}\right) dx

y = \frac{1}{2} \int \frac{1}{x+1} dx + \frac{1}{2} \int \frac{3x-1}{x^2+1} dx

y = \frac{1}{2} \log|x+1| + \frac{1}{2} \left( \int \frac{3x}{x^2+1} dx - \int \frac{1}{x^2+1} dx \right)

y = \frac{1}{2} \log|x+1| + \frac{3}{4} \int \frac{2x}{x^2+1} dx - \frac{1}{2} \int \frac{1}{x^2+1} dx

Integrating the terms:

\int \frac{2x}{x^2+1} dx = \log(x^2+1)

(since

x^2+1 > 0

)

\int \frac{1}{x^2+1} dx = \tan^{-1}x

So, the general solution is:

y = \frac{1}{2} \log|x+1| + \frac{3}{4} \log(x^2+1) - \frac{1}{2} \tan^{-1}x + C

Now, we use the initial condition

y=1

when

x=0

to find the constant

C

.

1 = \frac{1}{2} \log|0+1| + \frac{3}{4} \log(0^2+1) - \frac{1}{2} \tan^{-1}(0) + C

1 = \frac{1}{2} \log(1) + \frac{3}{4} \log(1) - \frac{1}{2}(0) + C

1 = \frac{1}{2}(0) + \frac{3}{4}(0) - 0 + C

C = 1

Substituting

C=1

back into the general solution gives the particular solution:

y = \frac{1}{2} \log|x+1| + \frac{3}{4} \log(x^2+1) - \frac{1}{2} \tan^{-1}x + 1

Final Answer: The particular solution of the given differential equation is

y = \frac{3}{4} \log(x^2+1) - \frac{1}{2} \tan^{-1}x + \frac{1}{2} \log|x+1| + 1

.

Q12EXERCISE 9.3

$x(x^{2}-1) \frac{d y}{d x}=1 ; y=0$ when $x=2$

Solution

Given: The differential equation is

x(x^{2}-1) \frac{d y}{d x}=1

, with the condition

y=0

when

x=2

.

To Find: The particular solution of the differential equation.

Solution: The given differential equation can be written as:

\frac{dy}{dx} = \frac{1}{x(x^2-1)}

\frac{dy}{dx} = \frac{1}{x(x-1)(x+1)}

This is a first-order differential equation. We can solve it by the method of separation of variables.

dy = \frac{1}{x(x-1)(x+1)} dx

To integrate the right-hand side, we use partial fractions. Let:

\frac{1}{x(x-1)(x+1)} = \frac{A}{x} + \frac{B}{x-1} + \frac{C}{x+1}

1 = A(x-1)(x+1) + Bx(x+1) + Cx(x-1)

To find the coefficients A, B, and C:

Put $x=0$ : $1 = A(-1)(1) \implies A = -1$ .
Put $x=1$ : $1 = B(1)(2) \implies B = \frac{1}{2}$ .
Put $x=-1$ : $1 = C(-1)(-2) \implies C = \frac{1}{2}$ .

So, the differential equation becomes:

dy = \left( -\frac{1}{x} + \frac{1}{2(x-1)} + \frac{1}{2(x+1)} \right) dx

Integrating both sides, we get:

\int dy = \int \left( -\frac{1}{x} + \frac{1}{2(x-1)} + \frac{1}{2(x+1)} \right) dx

y = -\ln|x| + \frac{1}{2}\ln|x-1| + \frac{1}{2}\ln|x+1| + C_1

Using properties of logarithms, we can simplify the expression:

y = \frac{1}{2}(\ln|x-1| + \ln|x+1|) - \ln|x| + C_1

y = \frac{1}{2}\ln|(x-1)(x+1)| - \ln|x| + C_1

y = \frac{1}{2}\ln|x^2-1| - \ln|x| + C_1

y = \ln(\sqrt{|x^2-1|}) - \ln|x| + C_1

y = \ln\left| \frac{\sqrt{x^2-1}}{x} \right| + C_1

Now, we use the given condition

y=0

when

x=2

to find the constant

C_1

.

0 = \ln\left| \frac{\sqrt{2^2-1}}{2} \right| + C_1

0 = \ln\left( \frac{\sqrt{3}}{2} \right) + C_1

C_1 = -\ln\left( \frac{\sqrt{3}}{2} \right) = \ln\left( \frac{2}{\sqrt{3}} \right)

Substituting the value of

C_1

back into the solution:

y = \ln\left| \frac{\sqrt{x^2-1}}{x} \right| + \ln\left( \frac{2}{\sqrt{3}} \right)

y = \ln\left| \frac{2\sqrt{x^2-1}}{\sqrt{3}x} \right|

Final Answer: The particular solution of the given differential equation is

y = \ln\left| \frac{2\sqrt{x^2-1}}{\sqrt{3}x} \right|

.

Q13EXERCISE 9.3

$\cos \left(\frac{d y}{d x}\right)=a(a \in \mathbf{R}) ; y=1$ when $x=0$

Solution

Given: The differential equation is

\cos \left(\frac{d y}{d x}\right)=a

, where

a \in \mathbf{R}

, with the condition

y=1

when

x=0

.

To Find: The particular solution of the differential equation.

Solution: The given differential equation is:

\cos \left(\frac{d y}{d x}\right)=a

To solve for

\frac{dy}{dx}

, we take the inverse cosine of both sides:

\frac{dy}{dx} = \cos^{-1}(a)

For

\cos^{-1}(a)

to be defined as a real number, we must have

-1 \leq a \leq 1

. Let's assume this condition holds.

Let

k = \cos^{-1}(a)

, which is a constant. The differential equation becomes:

\frac{dy}{dx} = k

This is a simple first-order differential equation that can be solved by separating the variables:

dy = k \, dx

Integrating both sides:

\int dy = \int k \, dx

y = kx + C

Substituting back

k = \cos^{-1}(a)

:

y = x \cos^{-1}(a) + C

This is the general solution of the differential equation.

Now, we use the given condition

y=1

when

x=0

to find the constant

C

.

1 = (0) \cos^{-1}(a) + C

C = 1

Substituting the value of

C

back into the general solution:

y = x \cos^{-1}(a) + 1

We can rearrange this equation to express it in a form similar to the original equation:

y - 1 = x \cos^{-1}(a)

\frac{y-1}{x} = \cos^{-1}(a)

Taking the cosine of both sides:

\cos\left(\frac{y-1}{x}\right) = a

Final Answer: The particular solution of the given differential equation is

\cos\left(\frac{y-1}{x}\right) = a

, which is valid for

x \neq 0

and

|a| \leq 1

.

Q14EXERCISE 9.3

$\frac{d y}{d x}=y \tan x ; y=1$ when $x=0$

Solution

Given: The differential equation is

\frac{d y}{d x}=y \tan x

, with the condition

y=1

when

x=0

.

To Find: The particular solution of the differential equation.

Solution: The given differential equation is:

\frac{dy}{dx} = y \tan x

This is a first-order differential equation. We can solve it by the method of separation of variables. Assuming

y \neq 0

(which is consistent with the initial condition):

\frac{1}{y} dy = \tan x \, dx

Integrating both sides:

\int \frac{1}{y} dy = \int \tan x \, dx

We know that

\int \frac{1}{y} dy = \ln|y|

and

\int \tan x \, dx = \ln|\sec x|

. So, the equation becomes:

\ln|y| = \ln|\sec x| + C_1

Let the constant of integration be

C_1 = \ln|C|

.

\ln|y| = \ln|\sec x| + \ln|C|

Using the property of logarithms,

\ln A + \ln B = \ln(AB)

:

\ln|y| = \ln|C \sec x|

Removing the logarithm from both sides:

|y| = |C \sec x|

y = \pm C \sec x

Let's denote

\pm C

by a new constant

C

. So, the general solution is:

y = C \sec x

Now, we use the given condition

y=1

when

x=0

to find the constant

C

.

1 = C \sec(0)

We know that

\sec(0) = \frac{1}{\cos(0)} = \frac{1}{1} = 1

.

1 = C \cdot 1

C = 1

Substituting the value of

C

back into the general solution:

y = 1 \cdot \sec x

y = \sec x

Final Answer: The particular solution of the given differential equation is

y = \sec x

.

Q15EXERCISE 9.3

Find the equation of a curve passing through the point $(0,0)$ and whose differential equation is $y^{\prime}=e^{x} \sin x$ .

Solution

Given: A curve passes through the point

(0,0)

and its differential equation is

y^{\prime}=e^{x} \sin x

.

To Find: The equation of the curve.

Solution: The differential equation is given by:

\frac{dy}{dx} = e^x \sin x

To find the equation of the curve, we need to integrate the differential equation:

dy = e^x \sin x \, dx

\int dy = \int e^x \sin x \, dx

y = \int e^x \sin x \, dx

Let

I = \int e^x \sin x \, dx

. We use integration by parts, with the formula

\int u \, dv = uv - \int v \, du

. Let

u = \sin x

and

dv = e^x dx

. Then

du = \cos x \, dx

and

v = e^x

.

I = (\sin x)(e^x) - \int e^x \cos x \, dx

I = e^x \sin x - \int e^x \cos x \, dx \quad ...(1)

Now, we need to evaluate

\int e^x \cos x \, dx

. We use integration by parts again. Let

u = \cos x

and

dv = e^x dx

. Then

du = -\sin x \, dx

and

v = e^x

.

\int e^x \cos x \, dx = (\cos x)(e^x) - \int e^x(-\sin x) \, dx

\int e^x \cos x \, dx = e^x \cos x + \int e^x \sin x \, dx

Since

I = \int e^x \sin x \, dx

, we have:

\int e^x \cos x \, dx = e^x \cos x + I

Substitute this back into equation (1):

I = e^x \sin x - (e^x \cos x + I)

I = e^x \sin x - e^x \cos x - I

2I = e^x (\sin x - \cos x)

I = \frac{e^x}{2}(\sin x - \cos x)

So, the general solution for

y

is:

y = \frac{e^x}{2}(\sin x - \cos x) + C

We are given that the curve passes through the point

(0,0)

. Substitute

x=0

and

y=0

to find

C

.

0 = \frac{e^0}{2}(\sin 0 - \cos 0) + C

0 = \frac{1}{2}(0 - 1) + C

0 = -\frac{1}{2} + C

C = \frac{1}{2}

Substitute the value of

C

back into the general solution to get the equation of the curve:

y = \frac{e^x}{2}(\sin x - \cos x) + \frac{1}{2}

y = \frac{1}{2} (e^x (\sin x - \cos x) + 1)

2y = e^x (\sin x - \cos x) + 1

Final Answer: The equation of the curve is

2y - 1 = e^x(\sin x - \cos x)

.

Q16EXERCISE 9.3

For the differential equation $x y \frac{d y}{d x}=(x+2)(y+2)$ , find the solution curve passing through the point $(1,-1)$ .

Solution

Given: The differential equation is

x y \frac{d y}{d x}=(x+2)(y+2)

, and the solution curve passes through the point

(1,-1)

.

To Find: The equation of the solution curve.

Solution: The given differential equation is:

x y \frac{d y}{d x}=(x+2)(y+2)

This is a first-order differential equation. We can solve it by the method of separation of variables. Rearranging the terms, we get:

\frac{y}{y+2} dy = \frac{x+2}{x} dx

This is valid for

x \neq 0

and

y \neq -2

. The given point

(1, -1)

satisfies these conditions.

To make integration easier, we can rewrite the fractions: LHS:

\frac{y}{y+2} = \frac{y+2-2}{y+2} = 1 - \frac{2}{y+2}

RHS:

\frac{x+2}{x} = 1 + \frac{2}{x}

So the differential equation becomes:

\left(1 - \frac{2}{y+2}\right) dy = \left(1 + \frac{2}{x}\right) dx

Now, we integrate both sides:

\int \left(1 - \frac{2}{y+2}\right) dy = \int \left(1 + \frac{2}{x}\right) dx

y - 2\ln|y+2| = x + 2\ln|x| + C

This is the general solution of the differential equation.

We are given that the curve passes through the point

(1, -1)

. Substitute

x=1

and

y=-1

into the general solution to find the constant

C

.

-1 - 2\ln|-1+2| = 1 + 2\ln|1| + C

-1 - 2\ln|1| = 1 + 2(0) + C

-1 - 2(0) = 1 + C

-1 = 1 + C

C = -2

Substitute the value of

C

back into the general solution:

y - 2\ln|y+2| = x + 2\ln|x| - 2

We can rearrange this equation to get the final form:

y - x + 2 = 2\ln|x| + 2\ln|y+2|

y - x + 2 = 2(\ln|x| + \ln|y+2|)

Using the property of logarithms,

\ln A + \ln B = \ln(AB)

:

y - x + 2 = 2\ln|x(y+2)|

Also,

2\ln A = \ln(A^2)

, so:

y - x + 2 = \ln((x(y+2))^2)

y - x + 2 = \ln(x^2(y+2)^2)

Final Answer: The equation of the solution curve is

y - x + 2 = 2\ln|x| + 2\ln|y+2|

, which can also be written as

y - x + 2 = \ln[x^2(y+2)^2]

.

Q17EXERCISE 9.3

Find the equation of a curve passing through the point $(0,-2)$ given that at any point $(x, y)$ on the curve, the product of the slope of its tangent and $y$ coordinate of the point is equal to the $x$ coordinate of the point.

Solution

Given: A curve passes through the point

(0, -2)

. At any point

(x, y)

on the curve, the product of the slope of its tangent and the

y

-coordinate of the point is equal to the

x

-coordinate of the point.

To Find: The equation of the curve.

Solution: Let

(x, y)

be any point on the curve. The slope of the tangent at this point is given by

\frac{dy}{dx}

. According to the problem statement, we have:

(\text{slope of tangent}) \times (y\text{-coordinate}) = (x\text{-coordinate})

\frac{dy}{dx} \cdot y = x

This is a differential equation. We can solve it using the method of separation of variables.

y \, dy = x \, dx

Integrating both sides, we get:

\int y \, dy = \int x \, dx

\frac{y^2}{2} = \frac{x^2}{2} + C

where

C

is the constant of integration.

To find the value of

C

, we use the fact that the curve passes through the point

(0, -2)

. Substituting

x=0

and

y=-2

into the equation:

\frac{(-2)^2}{2} = \frac{(0)^2}{2} + C

\frac{4}{2} = 0 + C

C = 2

Substituting the value of

C

back into the equation of the curve:

\frac{y^2}{2} = \frac{x^2}{2} + 2

Multiplying the entire equation by 2, we get:

y^2 = x^2 + 4

This can also be written as

y^2 - x^2 = 4

.

Final Answer: The equation of the curve is

y^2 - x^2 = 4

.

Q18EXERCISE 9.3

At any point $(x, y)$ of a curve, the slope of the tangent is twice the slope of the line segment joining the point of contact to the point $(-4,-3)$ . Find the equation of the curve given that it passes through $(-2,1)$ .

Solution

Given: At any point

(x, y)

on a curve, the slope of the tangent is twice the slope of the line segment joining the point of contact to the point

(-4, -3)

. The curve passes through the point

(-2, 1)

.

To Find: The equation of the curve.

Solution: Let

(x, y)

be the point of contact on the curve. The slope of the tangent at this point is

\frac{dy}{dx}

. The slope of the line segment joining the point of contact

(x, y)

and the point

(-4, -3)

is given by:

m = \frac{y - (-3)}{x - (-4)} = \frac{y+3}{x+4}

According to the problem statement, the slope of the tangent is twice this slope:

\frac{dy}{dx} = 2 \left( \frac{y+3}{x+4} \right)

This is a differential equation that can be solved by the method of separation of variables.

\frac{dy}{y+3} = \frac{2}{x+4} dx

Integrating both sides:

\int \frac{1}{y+3} dy = \int \frac{2}{x+4} dx

\ln|y+3| = 2 \ln|x+4| + C

where

C

is the constant of integration.

Using the properties of logarithms, we can write:

\ln|y+3| = \ln|(x+4)^2| + C

Let

C = \ln|C_1|

, where

C_1

is another constant.

\ln|y+3| = \ln|(x+4)^2| + \ln|C_1|

\ln|y+3| = \ln|C_1(x+4)^2|

Taking the antilogarithm on both sides:

y+3 = C_1(x+4)^2

To find the value of

C_1

, we use the fact that the curve passes through the point

(-2, 1)

. Substituting

x=-2

and

y=1

:

1+3 = C_1(-2+4)^2

4 = C_1(2)^2

4 = 4C_1

C_1 = 1

Substituting

C_1=1

back into the equation of the curve:

y+3 = 1 \cdot (x+4)^2

y+3 = (x+4)^2

Final Answer: The equation of the curve is

y+3 = (x+4)^2

.

Q19EXERCISE 9.3

The volume of spherical balloon being inflated changes at a constant rate. If initially its radius is 3 units and after 3 seconds it is 6 units. Find the radius of balloon after $t$ seconds.

Solution

Given: The volume of a spherical balloon being inflated changes at a constant rate. Initially (at

t=0

), its radius is

r=3

units. After 3 seconds (at

t=3

), its radius is

r=6

units.

To Find: The radius of the balloon after

t

seconds, i.e.,

r(t)

.

Solution: Let

V

be the volume and

r

be the radius of the spherical balloon at time

t

. The volume of a sphere is given by the formula

V = \frac{4}{3}\pi r^3

.

It is given that the volume changes at a constant rate. This can be expressed as a differential equation:

\frac{dV}{dt} = k

where

k

is a constant.

Integrating with respect to

t

:

\int dV = \int k \, dt

V = kt + C

where

C

is the constant of integration.

Substituting the formula for volume:

\frac{4}{3}\pi r^3 = kt + C

(Equation 1)

We use the given conditions to find the values of

k

and

C

.

Condition 1: At

t=0

,

r=3

. Substituting into Equation 1:

\frac{4}{3}\pi (3)^3 = k(0) + C

\frac{4}{3}\pi (27) = C

C = 36\pi

So, the equation becomes

\frac{4}{3}\pi r^3 = kt + 36\pi

.

Condition 2: At

t=3

,

r=6

. Substituting into the updated equation:

\frac{4}{3}\pi (6)^3 = k(3) + 36\pi

\frac{4}{3}\pi (216) = 3k + 36\pi

4\pi (72) = 3k + 36\pi

288\pi = 3k + 36\pi

3k = 288\pi - 36\pi

3k = 252\pi

k = 84\pi

Now, substitute the values of

k

and

C

back into Equation 1:

\frac{4}{3}\pi r^3 = 84\pi t + 36\pi

To find the radius

r

as a function of

t

, we solve for

r

. First, divide the entire equation by

\pi

:

\frac{4}{3} r^3 = 84t + 36

Now, divide by 4:

\frac{1}{3} r^3 = 21t + 9

Multiply by 3:

r^3 = 63t + 27

Finally, take the cube root:

r = (63t + 27)^{\frac{1}{3}}

Final Answer: The radius of the balloon after

t

seconds is

r = (63t + 27)^{\frac{1}{3}}

.

Q20EXERCISE 9.3

In a bank, principal increases continuously at the rate of $r\%$ per year. Find the value of $r$ if Rs 100 double itself in 10 years ( $\log _{e} 2=0.6931$ ).

Solution

Given: Principal increases continuously at the rate of

r\%

per year. Rs 100 doubles itself in 10 years.

\log_{e} 2 = 0.6931

.

To Find: The value of

r

.

Solution: Let

P

be the principal at any time

t

(in years). The principal increases continuously at a rate of

r\%

per year. This can be modelled by the differential equation:

\frac{dP}{dt} = \frac{r}{100} P

This is a separable differential equation:

\frac{dP}{P} = \frac{r}{100} dt

Integrating both sides:

\int \frac{dP}{P} = \int \frac{r}{100} dt

\ln P = \frac{r}{100} t + C

where

C

is the constant of integration.

Let

P_0

be the initial principal at

t=0

. Substituting this into the equation:

\ln P_0 = \frac{r}{100}(0) + C \implies C = \ln P_0

So the equation becomes:

\ln P = \frac{rt}{100} + \ln P_0

\ln P - \ln P_0 = \frac{rt}{100}

\ln \left( \frac{P}{P_0} \right) = \frac{rt}{100}

We are given that the initial principal

P_0 = \text{Rs } 100

. The amount doubles in 10 years, which means at

t=10

years, the principal

P = 2P_0 = \text{Rs } 200

.

Substituting these values into the equation:

\ln \left( \frac{200}{100} \right) = \frac{r(10)}{100}

\ln 2 = \frac{10r}{100}

\ln 2 = \frac{r}{10}

We are given that

\ln 2 = \log_{e} 2 = 0.6931

. Substituting this value:

0.6931 = \frac{r}{10}

Solving for

r

:

r = 10 \times 0.6931

r = 6.931

Final Answer: The value of

r

is

6.931

. This corresponds to a rate of

6.931\%

per year.

Q21EXERCISE 9.3

In a bank, principal increases continuously at the rate of $5\%$ per year. An amount of Rs 1000 is deposited with this bank, how much will it worth after 10 years $\left(e^{0.5}=1.648\right)$ .

Solution

Given: Principal increases continuously at the rate of

5\%

per year. An amount of Rs 1000 is deposited. Time period is 10 years.

e^{0.5} = 1.648

.

To Find: The worth of the amount after 10 years.

Solution: Let

P

be the principal at any time

t

(in years). The principal increases continuously at a rate of

5\%

per year. This is modeled by the differential equation:

\frac{dP}{dt} = \frac{5}{100} P = 0.05 P

This is a separable differential equation:

\frac{dP}{P} = 0.05 \, dt

Integrating both sides:

\int \frac{dP}{P} = \int 0.05 \, dt

\ln P = 0.05t + C

where

C

is the constant of integration.

Exponentiating both sides:

P = e^{0.05t + C} = e^{0.05t} \cdot e^C

Let

P_0 = e^C

, which is the principal at time

t=0

. So, the formula for the principal at time

t

is:

P(t) = P_0 e^{0.05t}

We are given the initial deposited amount,

P_0 = \text{Rs } 1000

. We need to find the amount after

t=10

years.

Substituting the values of

P_0

and

t

into the formula:

P(10) = 1000 \cdot e^{0.05 \times 10}

P(10) = 1000 \cdot e^{0.5}

We are given the value

e^{0.5} = 1.648

. Substituting this value:

P(10) = 1000 \times 1.648

P(10) = 1648

Final Answer: After 10 years, the amount will be worth Rs 1648.

Q22EXERCISE 9.3

In a culture, the bacteria count is $1,00,000$ . The number is increased by $10\%$ in 2 hours. In how many hours will the count reach $2,00,000$ , if the rate of growth of bacteria is proportional to the number present?

Solution

Given: Initial bacteria count,

P_0 = 1,00,000

. The number of bacteria increases by

10\%

in 2 hours.

To Find: The time required for the bacteria count to reach

2,00,000

.

Let: Let

P(t)

be the number of bacteria at any time

t

.

Equation: The rate of growth of bacteria is proportional to the number present. This can be modelled by the differential equation:

\frac{dP}{dt} = kP

where

k

is the constant of proportionality.

Solution: Separating the variables, we get:

\frac{dP}{P} = k \, dt

Integrating both sides:

\int \frac{dP}{P} = \int k \, dt

\ln|P| = kt + C

P = e^{kt+C} = e^C \cdot e^{kt}

Let

A = e^C

. The solution is of the form:

P(t) = A e^{kt}

At

t=0

,

P(0) = 1,00,000

. Substituting this into the equation:

1,00,000 = A e^{k \cdot 0} = A \cdot 1

So,

A = 1,00,000

. The equation becomes:

P(t) = 1,00,000 \, e^{kt}

After 2 hours, the count increases by

10\%

. So, at

t=2

, the count is:

P(2) = 1,00,000 + 10\% \text{ of } 1,00,000 = 1,00,000 + 10,000 = 1,10,000

Substituting

t=2

and

P(2)=1,10,000

:

1,10,000 = 1,00,000 \, e^{2k}

\frac{1,10,000}{1,00,000} = e^{2k}

1.1 = e^{2k}

Taking the natural logarithm on both sides:

\ln(1.1) = 2k

k = \frac{1}{2} \ln(1.1)

Now, we need to find the time

t

when the count reaches

2,00,000

. Let

P(t) = 2,00,000

.

2,00,000 = 1,00,000 \, e^{kt}

2 = e^{kt}

Taking the natural logarithm on both sides:

\ln(2) = kt

t = \frac{\ln(2)}{k}

Substituting the value of

k

we found:

t = \frac{\ln(2)}{\frac{1}{2} \ln(1.1)} = \frac{2 \ln(2)}{\ln(1.1)}

Final Answer: The time required for the count to reach

2,00,000

is

\frac{2 \ln(2)}{\ln(1.1)}

hours.

Q23EXERCISE 9.3

The general solution of the differential equation $\frac{d y}{d x}=e^{x+y}$ is

(A)

e^{x}+e^{-y}=\mathrm{C}

(B)

e^{x}+e^{y}=\mathrm{C}

(C)

e^{-x}+e^{y}=\mathrm{C}

(D)

e^{-x}+e^{-y}=\mathrm{C}

Solution

Given: The differential equation is

\frac{dy}{dx} = e^{x+y}

.

To Find: The general solution of the given differential equation.

Solution: We can rewrite the equation using the property of exponents

a^{m+n} = a^m a^n

:

\frac{dy}{dx} = e^x \cdot e^y

This is a separable differential equation. We can separate the variables by moving terms involving

y

to one side and terms involving

x

to the other side.

\frac{dy}{e^y} = e^x \, dx

e^{-y} \, dy = e^x \, dx

Now, we integrate both sides:

\int e^{-y} \, dy = \int e^x \, dx

Integrating with respect to their respective variables:

-e^{-y} = e^x + C'

where

C'

is the constant of integration.

To match the format of the given options, we can rearrange the terms:

e^x + e^{-y} = -C'

Let

C = -C'

, which is another arbitrary constant.

e^x + e^{-y} = C

Comparing this result with the given options: (A)

e^{x}+e^{-y}=\mathrm{C}

(B)

e^{x}+e^{y}=\mathrm{C}

(C)

e^{-x}+e^{y}=\mathrm{C}

(D)

e^{-x}+e^{-y}=\mathrm{C}

Our derived solution matches option (A).

Final Answer: The correct option is (A), which is

e^{x}+e^{-y}=\mathrm{C}

.

Q1EXERCISE 9.4

$(x^{2}+x y) d y=\left(x^{2}+y^{2}\right) d x$

Solution

Given: The differential equation is

(x^2 + xy) dy = (x^2 + y^2) dx

.

To Find: The general solution of the differential equation.

Solution: First, rearrange the equation into the form

\frac{dy}{dx} = F(x, y)

:

\frac{dy}{dx} = \frac{x^2 + y^2}{x^2 + xy}

This is a homogeneous differential equation because the numerator and denominator are homogeneous functions of the same degree (degree 2).

Let

y = vx

. Differentiating with respect to

x

, we get:

\frac{dy}{dx} = v + x \frac{dv}{dx}

Substituting

y=vx

and

\frac{dy}{dx}

into the differential equation:

v + x \frac{dv}{dx} = \frac{x^2 + (vx)^2}{x^2 + x(vx)}

v + x \frac{dv}{dx} = \frac{x^2(1 + v^2)}{x^2(1 + v)}

v + x \frac{dv}{dx} = \frac{1 + v^2}{1 + v}

Now, we separate the variables

v

and

x

:

x \frac{dv}{dx} = \frac{1 + v^2}{1 + v} - v

x \frac{dv}{dx} = \frac{1 + v^2 - v(1 + v)}{1 + v}

x \frac{dv}{dx} = \frac{1 + v^2 - v - v^2}{1 + v}

x \frac{dv}{dx} = \frac{1 - v}{1 + v}

\frac{1 + v}{1 - v} dv = \frac{dx}{x}

To integrate the left side, we can rewrite the fraction:

\frac{1 + v}{1 - v} = \frac{-(v-1-2)}{1-v} = \frac{-(v-1)+2}{-(v-1)} = -1 + \frac{2}{-(v-1)} = -1 - \frac{2}{v-1}

Alternatively,

\frac{1+v}{1-v} = \frac{2-(1-v)}{1-v} = \frac{2}{1-v} - 1

.

Integrating both sides:

\int \left( \frac{2}{1 - v} - 1 \right) dv = \int \frac{dx}{x}

-2 \ln|1 - v| - v = \ln|x| + C'

Now, substitute back

v = \frac{y}{x}

:

-2 \ln\left|1 - \frac{y}{x}\right| - \frac{y}{x} = \ln|x| + C'

-2 \ln\left|\frac{x - y}{x}\right| - \frac{y}{x} = \ln|x| + C'

Using the logarithm property

\ln(a/b) = \ln(a) - \ln(b)

:

-2 (\ln|x - y| - \ln|x|) - \frac{y}{x} = \ln|x| + C'

-2 \ln|x - y| + 2 \ln|x| - \frac{y}{x} = \ln|x| + C'

-2 \ln|x - y| + \ln|x| - \frac{y}{x} = C'

Let

C = -C'

.

2 \ln|x - y| - \ln|x| + \frac{y}{x} = C

Using logarithm properties:

\ln((x - y)^2) - \ln|x| + \frac{y}{x} = C

\ln\left|\frac{(x - y)^2}{x}\right| + \frac{y}{x} = C

Another way to write the solution is by rearranging from

-2 \ln|1 - v| - v = \ln|x| + C'

:

-v = \ln|x| + 2\ln|1-v| + C'

-v = \ln|x| + \ln((1-v)^2) + C'

-v = \ln|x(1-v)^2| + C'

Substituting

v = y/x

:

-\frac{y}{x} = \ln\left|x\left(1-\frac{y}{x}\right)^2\right| + C'

-\frac{y}{x} = \ln\left|x\left(\frac{x-y}{x}\right)^2\right| + C'

-\frac{y}{x} = \ln\left|\frac{(x-y)^2}{x}\right| + C'

Final Answer: The general solution is

-\frac{y}{x} = \ln\left|\frac{(x-y)^2}{x}\right| + C

, where C is an arbitrary constant.

Q2EXERCISE 9.4

$y'=\frac{x+y}{x}$

Solution

Given: The differential equation is

y' = \frac{x+y}{x}

.

To Find: The general solution of the differential equation.

Solution: The given equation can be written as:

\frac{dy}{dx} = \frac{x+y}{x}

This is a homogeneous differential equation as it can be expressed as a function of

\frac{y}{x}

:

\frac{dy}{dx} = 1 + \frac{y}{x}

Let

y = vx

. Differentiating with respect to

x

, we get:

\frac{dy}{dx} = v + x \frac{dv}{dx}

Substituting

y=vx

and

\frac{dy}{dx}

into the differential equation:

v + x \frac{dv}{dx} = 1 + v

Subtracting

v

from both sides:

x \frac{dv}{dx} = 1

Now, we separate the variables

v

and

x

:

dv = \frac{dx}{x}

Integrating both sides:

\int dv = \int \frac{dx}{x}

v = \ln|x| + C

where

C

is the constant of integration.

Now, substitute back

v = \frac{y}{x}

:

\frac{y}{x} = \ln|x| + C

Multiplying by

x

gives the explicit solution for

y

:

y = x(\ln|x| + C)

Final Answer: The general solution of the differential equation is

y = x(\ln|x| + C)

.

Q3EXERCISE 9.4

$(x-y) d y-(x+y) d x=0$

Solution

Given: The differential equation is

(x-y) dy - (x+y) dx = 0

.

To Find: The general solution of the differential equation.

Solution: First, rearrange the equation into the form

\frac{dy}{dx} = F(x, y)

:

(x-y) dy = (x+y) dx

\frac{dy}{dx} = \frac{x+y}{x-y}

This is a homogeneous differential equation because the numerator and denominator are homogeneous functions of the same degree (degree 1).

Let

y = vx

. Differentiating with respect to

x

, we get:

\frac{dy}{dx} = v + x \frac{dv}{dx}

Substituting

y=vx

and

\frac{dy}{dx}

into the differential equation:

v + x \frac{dv}{dx} = \frac{x+vx}{x-vx}

v + x \frac{dv}{dx} = \frac{x(1+v)}{x(1-v)}

v + x \frac{dv}{dx} = \frac{1+v}{1-v}

Now, we separate the variables

v

and

x

:

x \frac{dv}{dx} = \frac{1+v}{1-v} - v

x \frac{dv}{dx} = \frac{1+v - v(1-v)}{1-v}

x \frac{dv}{dx} = \frac{1+v - v + v^2}{1-v}

x \frac{dv}{dx} = \frac{1+v^2}{1-v}

\frac{1-v}{1+v^2} dv = \frac{dx}{x}

Split the fraction on the left side:

\left( \frac{1}{1+v^2} - \frac{v}{1+v^2} \right) dv = \frac{dx}{x}

Integrating both sides:

\int \left( \frac{1}{1+v^2} - \frac{v}{1+v^2} \right) dv = \int \frac{dx}{x}

\int \frac{1}{1+v^2} dv - \int \frac{v}{1+v^2} dv = \int \frac{dx}{x}

For the second integral on the left, let

u = 1+v^2

, so

du = 2v dv

, or

v dv = \frac{1}{2} du

.

\tan^{-1}(v) - \int \frac{1}{u} \frac{du}{2} = \ln|x| + C_1

\tan^{-1}(v) - \frac{1}{2} \ln|u| = \ln|x| + C_1

Substitute back

u = 1+v^2

:

\tan^{-1}(v) - \frac{1}{2} \ln|1+v^2| = \ln|x| + C_1

Now, substitute back

v = \frac{y}{x}

:

\tan^{-1}\left(\frac{y}{x}\right) - \frac{1}{2} \ln\left(1+\left(\frac{y}{x}\right)^2\right) = \ln|x| + C_1

\tan^{-1}\left(\frac{y}{x}\right) - \frac{1}{2} \ln\left(\frac{x^2+y^2}{x^2}\right) = \ln|x| + C_1

Using logarithm property

\ln(a/b) = \ln(a) - \ln(b)

:

\tan^{-1}\left(\frac{y}{x}\right) - \frac{1}{2} (\ln(x^2+y^2) - \ln(x^2)) = \ln|x| + C_1

\tan^{-1}\left(\frac{y}{x}\right) - \frac{1}{2} \ln(x^2+y^2) + \frac{1}{2} \ln(x^2) = \ln|x| + C_1

Since

\ln(x^2) = 2\ln|x|

:

\tan^{-1}\left(\frac{y}{x}\right) - \frac{1}{2} \ln(x^2+y^2) + \frac{1}{2} (2\ln|x|) = \ln|x| + C_1

\tan^{-1}\left(\frac{y}{x}\right) - \frac{1}{2} \ln(x^2+y^2) + \ln|x| = \ln|x| + C_1

\tan^{-1}\left(\frac{y}{x}\right) - \frac{1}{2} \ln(x^2+y^2) = C_1

Let

C = -2C_1

. Multiplying the equation by

-2

:

\ln(x^2+y^2) - 2\tan^{-1}\left(\frac{y}{x}\right) = C

Final Answer: The general solution of the differential equation is

\tan^{-1}\left(\frac{y}{x}\right) - \frac{1}{2} \ln(x^2+y^2) = C_1

, or equivalently,

\ln(x^2+y^2) - 2\tan^{-1}\left(\frac{y}{x}\right) = C

.

Q4EXERCISE 9.4

$\left(x^{2}-y^{2}\right) d x+2 x y d y=0$

Solution

Given: The differential equation is

(x^2 - y^2)dx + 2xydy = 0

.

Solution: The given differential equation can be rewritten as:

2xydy = -(x^2 - y^2)dx

\frac{dy}{dx} = \frac{-(x^2 - y^2)}{2xy} = \frac{y^2 - x^2}{2xy} \quad ...(1)

Let

F(x, y) = \frac{y^2 - x^2}{2xy}

. Now, let's check for homogeneity:

F(\lambda x, \lambda y) = \frac{(\lambda y)^2 - (\lambda x)^2}{2(\lambda x)(\lambda y)} = \frac{\lambda^2(y^2 - x^2)}{\lambda^2(2xy)} = \frac{y^2 - x^2}{2xy} = \lambda^0 F(x, y)

Since the function is a homogeneous function of degree zero, the differential equation is a homogeneous differential equation.

To solve it, we make the substitution

y = vx

. Differentiating with respect to

x

, we get:

\frac{dy}{dx} = v + x\frac{dv}{dx}

Substituting the expressions for

y

and

\frac{dy}{dx}

in equation (1):

v + x\frac{dv}{dx} = \frac{(vx)^2 - x^2}{2x(vx)}

v + x\frac{dv}{dx} = \frac{x^2(v^2 - 1)}{2vx^2}

v + x\frac{dv}{dx} = \frac{v^2 - 1}{2v}

x\frac{dv}{dx} = \frac{v^2 - 1}{2v} - v

x\frac{dv}{dx} = \frac{v^2 - 1 - 2v^2}{2v}

x\frac{dv}{dx} = \frac{-v^2 - 1}{2v} = -\frac{v^2 + 1}{2v}

Now, separating the variables:

\frac{2v}{v^2 + 1}dv = -\frac{1}{x}dx

Integrating both sides:

\int \frac{2v}{v^2 + 1}dv = -\int \frac{1}{x}dx

Let

u = v^2 + 1

, then

du = 2vdv

. The integral becomes

\int \frac{du}{u} = \ln|u|

.

\ln|v^2 + 1| = -\ln|x| + C

Since

v^2+1

is always positive, we can write:

\ln(v^2 + 1) = -\ln|x| + \ln|C_1|

(where

C = \ln|C_1|

)

\ln(v^2 + 1) = \ln\left|\frac{C_1}{x}\right|

v^2 + 1 = \frac{C_1}{x}

Substituting back

v = \frac{y}{x}

:

\left(\frac{y}{x}\right)^2 + 1 = \frac{C_1}{x}

\frac{y^2}{x^2} + 1 = \frac{C_1}{x}

\frac{y^2 + x^2}{x^2} = \frac{C_1}{x}

x^2 + y^2 = C_1x

Let

C_1 = C

.

x^2 + y^2 = Cx

Final Answer: The general solution of the given differential equation is

x^2 + y^2 = Cx

.

Q5EXERCISE 9.4

$x^{2} \frac{d y}{d x}=x^{2}-2 y^{2}+x y$

Solution

Given: The differential equation is

x^2 \frac{dy}{dx} = x^2 - 2y^2 + xy

.

Solution: The given differential equation can be rewritten as:

\frac{dy}{dx} = \frac{x^2 - 2y^2 + xy}{x^2}

\frac{dy}{dx} = 1 - 2\left(\frac{y}{x}\right)^2 + \frac{y}{x} \quad ...(1)

This is a homogeneous differential equation of the form

\frac{dy}{dx} = F\left(\frac{y}{x}\right)

.

To solve it, we make the substitution

y = vx

. Differentiating with respect to

x

, we get:

\frac{dy}{dx} = v + x\frac{dv}{dx}

Substituting the expressions for

\frac{y}{x}

and

\frac{dy}{dx}

in equation (1):

v + x\frac{dv}{dx} = 1 - 2v^2 + v

x\frac{dv}{dx} = 1 - 2v^2

Now, separating the variables:

\frac{dv}{1 - 2v^2} = \frac{dx}{x}

Integrating both sides:

\int \frac{dv}{1 - 2v^2} = \int \frac{dx}{x}

\int \frac{dv}{1 - (\sqrt{2}v)^2} = \int \frac{dx}{x}

We use the standard integral formula

\int \frac{dx}{a^2 - x^2} = \frac{1}{2a}\ln\left|\frac{a+x}{a-x}\right|

. Here

a=1

and

x

is replaced by

\sqrt{2}v

.

\frac{1}{\sqrt{2}} \int \frac{\sqrt{2}dv}{1^2 - (\sqrt{2}v)^2} = \ln|x| + C

\frac{1}{\sqrt{2}} \cdot \frac{1}{2 \cdot 1} \ln\left|\frac{1+\sqrt{2}v}{1-\sqrt{2}v}\right| = \ln|x| + C

\frac{1}{2\sqrt{2}} \ln\left|\frac{1+\sqrt{2}v}{1-\sqrt{2}v}\right| = \ln|x| + C

Substituting back

v = \frac{y}{x}

:

\frac{1}{2\sqrt{2}} \ln\left|\frac{1+\sqrt{2}\frac{y}{x}}{1-\sqrt{2}\frac{y}{x}}\right| = \ln|x| + C

\frac{1}{2\sqrt{2}} \ln\left|\frac{\frac{x+\sqrt{2}y}{x}}{\frac{x-\sqrt{2}y}{x}}\right| = \ln|x| + C

\frac{1}{2\sqrt{2}} \ln\left|\frac{x+\sqrt{2}y}{x-\sqrt{2}y}\right| = \ln|x| + C

Final Answer: The general solution of the given differential equation is

\frac{1}{2\sqrt{2}} \ln\left|\frac{x+\sqrt{2}y}{x-\sqrt{2}y}\right| = \ln|x| + C

.

Q6EXERCISE 9.4

$x d y-y d x=\sqrt{x^{2}+y^{2}} d x$

Solution

Given: The differential equation is

x dy - y dx = \sqrt{x^2 + y^2} dx

.

Solution: The given differential equation can be rewritten as:

x dy = (y + \sqrt{x^2 + y^2}) dx

\frac{dy}{dx} = \frac{y + \sqrt{x^2 + y^2}}{x}

\frac{dy}{dx} = \frac{y}{x} + \frac{\sqrt{x^2 + y^2}}{x}

Assuming

x > 0

, we have:

\frac{dy}{dx} = \frac{y}{x} + \sqrt{\frac{x^2 + y^2}{x^2}} = \frac{y}{x} + \sqrt{1 + \left(\frac{y}{x}\right)^2} \quad ...(1)

This is a homogeneous differential equation of the form

\frac{dy}{dx} = F\left(\frac{y}{x}\right)

.

To solve it, we make the substitution

y = vx

. Differentiating with respect to

x

, we get:

\frac{dy}{dx} = v + x\frac{dv}{dx}

Substituting the expressions for

\frac{y}{x}

and

\frac{dy}{dx}

in equation (1):

v + x\frac{dv}{dx} = v + \sqrt{1 + v^2}

x\frac{dv}{dx} = \sqrt{1 + v^2}

Now, separating the variables:

\frac{dv}{\sqrt{1 + v^2}} = \frac{dx}{x}

Integrating both sides:

\int \frac{dv}{\sqrt{1 + v^2}} = \int \frac{dx}{x}

We use the standard integral formula

\int \frac{dx}{\sqrt{x^2 + a^2}} = \ln|x + \sqrt{x^2 + a^2}|

.

\ln|v + \sqrt{1 + v^2}| = \ln|x| + C

Let

C = \ln|C_1|

for some constant

C_1

.

\ln|v + \sqrt{1 + v^2}| = \ln|x| + \ln|C_1|

\ln|v + \sqrt{1 + v^2}| = \ln|C_1x|

v + \sqrt{1 + v^2} = C_1x

Substituting back

v = \frac{y}{x}

:

\frac{y}{x} + \sqrt{1 + \left(\frac{y}{x}\right)^2} = C_1x

\frac{y}{x} + \sqrt{\frac{x^2 + y^2}{x^2}} = C_1x

\frac{y}{x} + \frac{\sqrt{x^2 + y^2}}{|x|} = C_1x

Assuming

x>0

,

|x|=x

.

\frac{y + \sqrt{x^2 + y^2}}{x} = C_1x

y + \sqrt{x^2 + y^2} = C_1x^2

Let

C_1 = C

.

y + \sqrt{x^2 + y^2} = Cx^2

Final Answer: The general solution of the given differential equation is

y + \sqrt{x^2 + y^2} = Cx^2

.

Q7EXERCISE 9.4

$\left\{x \cos \left(\frac{y}{x}\right)+y \sin \left(\frac{y}{x}\right)\right\} y d x=\left\{y \sin \left(\frac{y}{x}\right)-x \cos \left(\frac{y}{x}\right)\right\} x d y$

Solution

Given: The differential equation is

\left\{x \cos\left(\frac{y}{x}\right) + y \sin\left(\frac{y}{x}\right)\right\} y dx = \left\{y \sin\left(\frac{y}{x}\right) - x \cos\left(\frac{y}{x}\right)\right\} x dy

.

Solution: The given differential equation can be rewritten to find

\frac{dy}{dx}

:

\frac{dy}{dx} = \frac{y\left\{x \cos\left(\frac{y}{x}\right) + y \sin\left(\frac{y}{x}\right)\right\}}{x\left\{y \sin\left(\frac{y}{x}\right) - x \cos\left(\frac{y}{x}\right)\right\}}

Divide the numerator and denominator of the fraction on the right by

x^2

:

\frac{dy}{dx} = \frac{\frac{y}{x}\left\{\cos\left(\frac{y}{x}\right) + \frac{y}{x} \sin\left(\frac{y}{x}\right)\right\}}{\left\{\frac{y}{x} \sin\left(\frac{y}{x}\right) - \cos\left(\frac{y}{x}\right)\right\}} \quad ...(1)

This is a homogeneous differential equation of the form

\frac{dy}{dx} = F\left(\frac{y}{x}\right)

.

To solve it, we make the substitution

y = vx

. Then

\frac{dy}{dx} = v + x\frac{dv}{dx}

. Substituting into equation (1):

v + x\frac{dv}{dx} = \frac{v(\cos v + v \sin v)}{v \sin v - \cos v}

x\frac{dv}{dx} = \frac{v(\cos v + v \sin v)}{v \sin v - \cos v} - v

x\frac{dv}{dx} = \frac{v\cos v + v^2\sin v - v(v\sin v - \cos v)}{v \sin v - \cos v}

x\frac{dv}{dx} = \frac{v\cos v + v^2\sin v - v^2\sin v + v\cos v}{v \sin v - \cos v}

x\frac{dv}{dx} = \frac{2v\cos v}{v \sin v - \cos v}

Now, separating the variables:

\frac{v \sin v - \cos v}{v \cos v} dv = \frac{2}{x} dx

\left(\frac{v \sin v}{v \cos v} - \frac{\cos v}{v \cos v}\right) dv = \frac{2}{x} dx

\left(\tan v - \frac{1}{v}\right) dv = \frac{2}{x} dx

Integrating both sides:

\int \left(\tan v - \frac{1}{v}\right) dv = \int \frac{2}{x} dx

\int \tan v dv - \int \frac{1}{v} dv = 2 \int \frac{1}{x} dx

\ln|\sec v| - \ln|v| = 2\ln|x| + C

\ln\left|\frac{\sec v}{v}\right| = \ln(x^2) + \ln|C_1|

(where

C = \ln|C_1|

)

\ln\left|\frac{\sec v}{v}\right| = \ln|C_1x^2|

\frac{\sec v}{v} = C_1x^2

Substituting back

v = \frac{y}{x}

:

\frac{\sec(y/x)}{y/x} = C_1x^2

\frac{x}{y} \sec\left(\frac{y}{x}\right) = C_1x^2

\sec\left(\frac{y}{x}\right) = C_1xy

\frac{1}{\cos(y/x)} = C_1xy

1 = C_1xy \cos\left(\frac{y}{x}\right)

Let

C = \frac{1}{C_1}

. Then

C = xy \cos\left(\frac{y}{x}\right)

.

Final Answer: The general solution of the given differential equation is

xy \cos\left(\frac{y}{x}\right) = C

.

Q8EXERCISE 9.4

$x \frac{d y}{d x}-y+x \sin \left(\frac{y}{x}\right)=0$

Solution

Given: The differential equation is

x \frac{dy}{dx} - y + x \sin\left(\frac{y}{x}\right) = 0

.

Solution: The given differential equation can be rewritten as:

x \frac{dy}{dx} = y - x \sin\left(\frac{y}{x}\right)

\frac{dy}{dx} = \frac{y}{x} - \sin\left(\frac{y}{x}\right) \quad ...(1)

This is a homogeneous differential equation of the form

\frac{dy}{dx} = F\left(\frac{y}{x}\right)

.

To solve it, we make the substitution

y = vx

. Then

\frac{dy}{dx} = v + x\frac{dv}{dx}

. Substituting into equation (1):

v + x\frac{dv}{dx} = v - \sin v

x\frac{dv}{dx} = -\sin v

Now, separating the variables:

\frac{dv}{\sin v} = -\frac{dx}{x}

\csc v dv = -\frac{dx}{x}

Integrating both sides:

\int \csc v dv = -\int \frac{dx}{x}

The integral of

\csc v

is

\ln|\csc v - \cot v|

.

\ln|\csc v - \cot v| = -\ln|x| + C

Let

C = \ln|C_1|

.

\ln|\csc v - \cot v| = -\ln|x| + \ln|C_1|

\ln|\csc v - \cot v| = \ln\left|\frac{C_1}{x}\right|

\csc v - \cot v = \frac{C_1}{x}

x(\csc v - \cot v) = C_1

Substituting back

v = \frac{y}{x}

:

x\left(\csc\left(\frac{y}{x}\right) - \cot\left(\frac{y}{x}\right)\right) = C_1

Let

C_1 = C

. We can also simplify the term

\csc v - \cot v

:

\csc v - \cot v = \frac{1}{\sin v} - \frac{\cos v}{\sin v} = \frac{1 - \cos v}{\sin v} = \frac{2\sin^2(v/2)}{2\sin(v/2)\cos(v/2)} = \tan(v/2)

So, the solution is also

\tan(v/2) = C_1/x

. Substituting

v=y/x

, we get

\tan(y/(2x)) = C_1/x

. This is an alternate form.

Final Answer: The general solution of the given differential equation is

x\left(\csc\left(\frac{y}{x}\right) - \cot\left(\frac{y}{x}\right)\right) = C

.

Q9EXERCISE 9.4

$y d x+x \log \left(\frac{y}{x}\right) d y-2 x d y=0$

Solution

Given: The differential equation is

y d x+x \log \left(\frac{y}{x}\right) d y-2 x d y=0

.

Solution: Rearranging the given differential equation:

y d x = (2x - x \log\left(\frac{y}{x}\right)) dy

\frac{dx}{dy} = \frac{2x - x \log\left(\frac{y}{x}\right)}{y}

\frac{dx}{dy} = 2\frac{x}{y} - \frac{x}{y} \log\left(\frac{y}{x}\right)

\frac{dx}{dy} = 2\frac{x}{y} - \frac{x}{y} \log\left(\left(\frac{x}{y}\right)^{-1}\right)

\frac{dx}{dy} = 2\frac{x}{y} + \frac{x}{y} \log\left(\frac{x}{y}\right)

This is a homogeneous differential equation. Let

x = vy

. Then, differentiating with respect to

y

, we get:

\frac{dx}{dy} = v + y \frac{dv}{dy}

Substituting these into the equation:

v + y \frac{dv}{dy} = 2v + v \log(v)

y \frac{dv}{dy} = v + v \log(v)

y \frac{dv}{dy} = v(1 + \log(v))

Separating the variables:

\frac{dv}{v(1 + \log(v))} = \frac{dy}{y}

Integrating both sides:

\int \frac{dv}{v(1 + \log(v))} = \int \frac{dy}{y}

For the left-hand side integral, let

t = 1 + \log(v)

. Then

dt = \frac{1}{v} dv

.

\int \frac{dt}{t} = \log|t| = \log|1 + \log(v)|

So, the equation becomes:

\log|1 + \log(v)| = \log|y| + \log|C|

where C is the integration constant.

\log|1 + \log(v)| = \log|Cy|

1 + \log(v) = Cy

Substitute back

v = \frac{x}{y}

:

1 + \log\left(\frac{x}{y}\right) = Cy

1 + \log(x) - \log(y) = Cy

\log(x) = \log(y) + Cy - 1

Final Answer: The general solution of the differential equation is

\log(x) = \log(y) + Cy - 1

.

Q10EXERCISE 9.4

$\left(1+e^{\frac{x}{y}}\right) d x+e^{\frac{x}{y}}\left(1-\frac{x}{y}\right) d y=0$

Solution

Given: The differential equation is

\left(1+e^{\frac{x}{y}}\right) d x+e^{\frac{x}{y}}\left(1-\frac{x}{y}\right) d y=0

.

Solution: Rearranging the given differential equation:

\left(1+e^{\frac{x}{y}}\right) d x = -e^{\frac{x}{y}}\left(1-\frac{x}{y}\right) d y

\frac{dx}{dy} = -\frac{e^{\frac{x}{y}}\left(1-\frac{x}{y}\right)}{1+e^{\frac{x}{y}}}

This is a homogeneous differential equation. Let

x = vy

. Then, differentiating with respect to

y

, we get:

\frac{dx}{dy} = v + y \frac{dv}{dy}

Substituting these into the equation:

v + y \frac{dv}{dy} = -\frac{e^v(1-v)}{1+e^v}

y \frac{dv}{dy} = -\frac{e^v(1-v)}{1+e^v} - v

y \frac{dv}{dy} = \frac{-e^v + ve^v - v(1+e^v)}{1+e^v}

y \frac{dv}{dy} = \frac{-e^v + ve^v - v - ve^v}{1+e^v}

y \frac{dv}{dy} = \frac{-(v+e^v)}{1+e^v}

Separating the variables:

\frac{1+e^v}{v+e^v} dv = -\frac{dy}{y}

Integrating both sides:

\int \frac{1+e^v}{v+e^v} dv = -\int \frac{dy}{y}

For the left-hand side integral, let

t = v + e^v

. Then

dt = (1 + e^v) dv

.

\int \frac{dt}{t} = \log|t| = \log|v + e^v|

So, the equation becomes:

\log|v + e^v| = -\log|y| + C

\log|v + e^v| + \log|y| = C

\log|y(v + e^v)| = C

y(v + e^v) = e^C = k

where

k

is an arbitrary constant. Substitute back

v = \frac{x}{y}

:

y\left(\frac{x}{y} + e^{\frac{x}{y}}\right) = k

x + y e^{\frac{x}{y}} = k

Final Answer: The general solution of the differential equation is

x + y e^{\frac{x}{y}} = k

.

Q11EXERCISE 9.4

$(x+y) d y+(x-y) d x=0 ; y=1 \text { when } x=1$

Solution

Given: The differential equation is

(x+y) d y+(x-y) d x=0

, with the initial condition

y=1

when

x=1

.

Solution: Rearranging the given differential equation:

(x+y) dy = -(x-y) dx

\frac{dy}{dx} = -\frac{x-y}{x+y} = \frac{y-x}{y+x}

This is a homogeneous differential equation. Let

y = vx

. Then, differentiating with respect to

x

, we get:

\frac{dy}{dx} = v + x \frac{dv}{dx}

Substituting these into the equation:

v + x \frac{dv}{dx} = \frac{vx-x}{vx+x} = \frac{v-1}{v+1}

x \frac{dv}{dx} = \frac{v-1}{v+1} - v

x \frac{dv}{dx} = \frac{v-1 - v(v+1)}{v+1}

x \frac{dv}{dx} = \frac{v-1 - v^2 - v}{v+1}

x \frac{dv}{dx} = \frac{-v^2-1}{v+1} = -\frac{v^2+1}{v+1}

Separating the variables:

\frac{v+1}{v^2+1} dv = -\frac{dx}{x}

\left(\frac{v}{v^2+1} + \frac{1}{v^2+1}\right) dv = -\frac{dx}{x}

Integrating both sides:

\int \frac{v}{v^2+1} dv + \int \frac{1}{v^2+1} dv = -\int \frac{dx}{x}

\frac{1}{2} \log(v^2+1) + \tan^{-1}(v) = -\log|x| + C

Substitute back

v = \frac{y}{x}

:

\frac{1}{2} \log\left(\left(\frac{y}{x}\right)^2+1\right) + \tan^{-1}\left(\frac{y}{x}\right) = -\log|x| + C

\frac{1}{2} \log\left(\frac{y^2+x^2}{x^2}\right) + \tan^{-1}\left(\frac{y}{x}\right) = -\log|x| + C

\frac{1}{2} (\log(y^2+x^2) - \log(x^2)) + \tan^{-1}\left(\frac{y}{x}\right) = -\log|x| + C

\frac{1}{2} \log(y^2+x^2) - \log|x| + \tan^{-1}\left(\frac{y}{x}\right) = -\log|x| + C

\frac{1}{2} \log(x^2+y^2) + \tan^{-1}\left(\frac{y}{x}\right) = C

This is the general solution. Now, we apply the initial condition

y=1

when

x=1

.

\frac{1}{2} \log(1^2+1^2) + \tan^{-1}\left(\frac{1}{1}\right) = C

\frac{1}{2} \log(2) + \frac{\pi}{4} = C

Substituting the value of C back into the general solution:

\frac{1}{2} \log(x^2+y^2) + \tan^{-1}\left(\frac{y}{x}\right) = \frac{1}{2} \log(2) + \frac{\pi}{4}

Multiplying by 2:

\log(x^2+y^2) + 2\tan^{-1}\left(\frac{y}{x}\right) = \log(2) + \frac{\pi}{2}

Final Answer: The particular solution is

\log(x^2+y^2) + 2\tan^{-1}\left(\frac{y}{x}\right) = \log(2) + \frac{\pi}{2}

.

Q12EXERCISE 9.4

$x^{2} d y+\left(x y+y^{2}\right) d x=0 ; y=1 \text { when } x=1$

Solution

Given: The differential equation is

x^{2} d y+\left(x y+y^{2}\right) d x=0

, with the initial condition

y=1

when

x=1

.

Solution: Rearranging the given differential equation:

x^2 dy = -(xy+y^2) dx

\frac{dy}{dx} = -\frac{xy+y^2}{x^2} = -\frac{y}{x} - \left(\frac{y}{x}\right)^2

This is a homogeneous differential equation. Let

y = vx

. Then, differentiating with respect to

x

, we get:

\frac{dy}{dx} = v + x \frac{dv}{dx}

Substituting these into the equation:

v + x \frac{dv}{dx} = -v - v^2

x \frac{dv}{dx} = -2v - v^2 = -v(v+2)

Separating the variables:

\frac{dv}{v(v+2)} = -\frac{dx}{x}

Using partial fractions for the left side:

\frac{1}{v(v+2)} = \frac{A}{v} + \frac{B}{v+2}

1 = A(v+2) + Bv

Setting

v=0

, we get

1 = 2A \Rightarrow A = 1/2

. Setting

v=-2

, we get

1 = -2B \Rightarrow B = -1/2

. So, the equation becomes:

\frac{1}{2}\left(\frac{1}{v} - \frac{1}{v+2}\right)dv = -\frac{dx}{x}

Integrating both sides:

\frac{1}{2} \int \left(\frac{1}{v} - \frac{1}{v+2}\right)dv = -\int \frac{dx}{x}

\frac{1}{2} (\log|v| - \log|v+2|) = -\log|x| + C

\log\left|\frac{v}{v+2}\right| = -2\log|x| + 2C

\log\left|\frac{v}{v+2}\right| = \log\left(\frac{1}{x^2}\right) + C_1

where

C_1 = 2C

.

\frac{v}{v+2} = e^{\log(1/x^2)+C_1} = e^{C_1} \cdot \frac{1}{x^2} = \frac{K}{x^2}

Substitute back

v = \frac{y}{x}

:

\frac{y/x}{y/x + 2} = \frac{K}{x^2}

\frac{y/x}{(y+2x)/x} = \frac{K}{x^2}

\frac{y}{y+2x} = \frac{K}{x^2}

x^2 y = K(y+2x)

This is the general solution. Now, apply the initial condition

y=1

when

x=1

.

1^2(1) = K(1+2(1))

1 = 3K \Rightarrow K = \frac{1}{3}

Substituting the value of K back into the general solution:

x^2 y = \frac{1}{3}(y+2x)

3x^2 y = y+2x

3x^2 y - y = 2x

y(3x^2 - 1) = 2x

Final Answer: The particular solution is

y(3x^2 - 1) = 2x

or

y = \frac{2x}{3x^2-1}

.

Q13EXERCISE 9.4

$\left[x \sin ^{2}\left(\frac{y}{x}\right)-y\right] d x+x d y=0 ; y=\frac{\pi}{4} \text { when } x=1$

Solution

Given: The differential equation is

\left[x \sin ^{2}\left(\frac{y}{x}\right)-y\right] d x+x d y=0

, with the initial condition

y=\frac{\pi}{4}

when

x=1

.

Solution: Rearranging the given differential equation:

x dy = -\left[x \sin ^{2}\left(\frac{y}{x}\right)-y\right] dx

\frac{dy}{dx} = \frac{-\left[x \sin ^{2}\left(\frac{y}{x}\right)-y\right]}{x}

\frac{dy}{dx} = -\sin^2\left(\frac{y}{x}\right) + \frac{y}{x}

This is a homogeneous differential equation. Let

y = vx

. Then, differentiating with respect to

x

, we get:

\frac{dy}{dx} = v + x \frac{dv}{dx}

Substituting these into the equation:

v + x \frac{dv}{dx} = -\sin^2(v) + v

x \frac{dv}{dx} = -\sin^2(v)

Separating the variables:

\frac{dv}{\sin^2(v)} = -\frac{dx}{x}

\csc^2(v) dv = -\frac{dx}{x}

Integrating both sides:

\int \csc^2(v) dv = -\int \frac{dx}{x}

-\cot(v) = -\log|x| + C

\cot(v) = \log|x| - C

Let

K = -C

, so

\cot(v) = \log|x| + K

. Substitute back

v = \frac{y}{x}

:

\cot\left(\frac{y}{x}\right) = \log|x| + K

This is the general solution. Now, we apply the initial condition

y=\frac{\pi}{4}

when

x=1

.

\cot\left(\frac{\pi/4}{1}\right) = \log|1| + K

\cot\left(\frac{\pi}{4}\right) = 0 + K

1 = K

Substituting the value of K back into the general solution:

\cot\left(\frac{y}{x}\right) = \log|x| + 1

Final Answer: The particular solution is

\cot\left(\frac{y}{x}\right) = \log|x| + 1

.

Q14EXERCISE 9.4

$\frac{d y}{d x}-\frac{y}{x}+\operatorname{cosec}\left(\frac{y}{x}\right)=0 ; y=0 \text { when } x=1$

Solution

Given: The differential equation is

\frac{d y}{d x}-\frac{y}{x}+\operatorname{cosec}\left(\frac{y}{x}\right)=0

with the condition

y=0

when

x=1

.

To Find: The particular solution of the differential equation.

Solution: The given differential equation can be written as:

\frac{dy}{dx} = \frac{y}{x} - \operatorname{cosec}\left(\frac{y}{x}\right) \quad ...(1)

This is a homogeneous differential equation.

Let

y = vx

. Then, differentiating with respect to

x

, we get:

\frac{dy}{dx} = v + x\frac{dv}{dx}

Substituting these values in equation (1):

v + x\frac{dv}{dx} = v - \operatorname{cosec}(v)

x\frac{dv}{dx} = -\operatorname{cosec}(v)

Separating the variables, we get:

\frac{dv}{\operatorname{cosec}(v)} = -\frac{dx}{x}

\sin(v) dv = -\frac{1}{x} dx

Integrating both sides:

\int \sin(v) dv = -\int \frac{1}{x} dx

-\cos(v) = -\ln|x| + C'

\cos(v) = \ln|x| - C'

Let

C = -C'

.

\cos(v) = \ln|x| + C

Now, substitute back

v = \frac{y}{x}

:

\cos\left(\frac{y}{x}\right) = \ln|x| + C \quad ...(2)

We are given the condition that

y=0

when

x=1

. Substituting these values into equation (2):

\cos\left(\frac{0}{1}\right) = \ln|1| + C

\cos(0) = 0 + C

1 = C

Substituting

C=1

back into equation (2), we get the particular solution:

\cos\left(\frac{y}{x}\right) = \ln|x| + 1

Final Answer: The particular solution to the given differential equation is

\cos\left(\frac{y}{x}\right) = \ln|x| + 1

.

Q15EXERCISE 9.4

$2 x y+y^{2}-2 x^{2} \frac{d y}{d x}=0 ; y=2 \text { when } x=1$

Solution

Given: The differential equation is

2 x y+y^{2}-2 x^{2} \frac{d y}{d x}=0

with the condition

y=2

when

x=1

.

To Find: The particular solution of the differential equation.

Solution: The given differential equation can be written as:

2x^2 \frac{dy}{dx} = 2xy + y^2

\frac{dy}{dx} = \frac{2xy + y^2}{2x^2} \quad ...(1)

This is a homogeneous differential equation.

Let

y = vx

. Then, differentiating with respect to

x

, we get:

\frac{dy}{dx} = v + x\frac{dv}{dx}

Substituting these values in equation (1):

v + x\frac{dv}{dx} = \frac{2x(vx) + (vx)^2}{2x^2}

v + x\frac{dv}{dx} = \frac{2vx^2 + v^2x^2}{2x^2}

v + x\frac{dv}{dx} = \frac{x^2(2v + v^2)}{2x^2}

v + x\frac{dv}{dx} = \frac{2v + v^2}{2}

x\frac{dv}{dx} = \frac{2v + v^2}{2} - v

x\frac{dv}{dx} = \frac{2v + v^2 - 2v}{2}

x\frac{dv}{dx} = \frac{v^2}{2}

Separating the variables, we get:

\frac{2}{v^2} dv = \frac{1}{x} dx

Integrating both sides:

\int \frac{2}{v^2} dv = \int \frac{1}{x} dx

2 \int v^{-2} dv = \int \frac{1}{x} dx

2 \frac{v^{-1}}{-1} = \ln|x| + C

-\frac{2}{v} = \ln|x| + C

Now, substitute back

v = \frac{y}{x}

:

-\frac{2}{y/x} = \ln|x| + C

-\frac{2x}{y} = \ln|x| + C \quad ...(2)

We are given the condition that

y=2

when

x=1

. Substituting these values into equation (2):

-\frac{2(1)}{2} = \ln|1| + C

-1 = 0 + C

C = -1

Substituting

C=-1

back into equation (2), we get the particular solution:

-\frac{2x}{y} = \ln|x| - 1

\frac{2x}{y} = 1 - \ln|x|

y(1 - \ln|x|) = 2x

y = \frac{2x}{1 - \ln|x|}

Final Answer: The particular solution to the given differential equation is

y = \frac{2x}{1 - \ln|x|}

.

Q16EXERCISE 9.4

A homogeneous differential equation of the from $\frac{d x}{d y}=h\left(\frac{x}{y}\right)$ can be solved by making the substitution.

(A)

y=v x

(B)

v=y x

(C)

x=v y

(D)

x=v

Solution

Given: A homogeneous differential equation of the form

\frac{d x}{d y}=h\left(\frac{x}{y}\right)

.

To Find: The substitution to solve this equation.

Solution: The given differential equation is

\frac{dx}{dy} = h\left(\frac{x}{y}\right)

. This equation is homogeneous in

x

and

y

. To solve such an equation, we make a substitution that simplifies the expression

h\left(\frac{x}{y}\right)

.

Let

v = \frac{x}{y}

. This implies

x = vy

.

Now, we differentiate

x = vy

with respect to

y

using the product rule:

\frac{dx}{dy} = v \cdot 1 + y \cdot \frac{dv}{dy}

\frac{dx}{dy} = v + y\frac{dv}{dy}

Substitute this expression for

\frac{dx}{dy}

and

v

for

\frac{x}{y}

back into the original differential equation:

v + y\frac{dv}{dy} = h(v)

This can be rearranged to separate the variables:

y\frac{dv}{dy} = h(v) - v

\frac{dv}{h(v) - v} = \frac{dy}{y}

This is a variables separable form, which can be integrated to find the solution.

Thus, the correct substitution is

x = vy

.

Comparing with the given options: (A)

y=v x

(B)

v=y x

(C)

x=v y

(D)

x=v

The correct option is (C).

Final Answer: The correct option is (C) $x=v y$ .

Q17EXERCISE 9.4

Which of the following is a homogeneous differential equation?

(A)

(4 x+6 y+5) d y-(3 y+2 x+4) d x=0

(B)

(x y) d x-\left(x^{3}+y^{3}\right) d y=0

(C)

\left(x^{3}+2 y^{2}\right) d x+2 x y d y=0

(D)

y^{2} d x+\left(x^{2}-x y-y^{2}\right) d y=0

Solution

To Find: Which of the given options is a homogeneous differential equation.

Concept: A differential equation of the form

\frac{dy}{dx} = F(x, y)

is homogeneous if the function

F(x, y)

can be expressed as a function of

\frac{y}{x}

or

\frac{x}{y}

. This is equivalent to checking if

F(\lambda x, \lambda y) = F(x, y)

for any non-zero constant

\lambda

. In other words, the degree of each term in the numerator and denominator of

F(x, y)

must be the same.

Analysis of Options:

(A) $(4 x+6 y+5) d y-(3 y+2 x+4) d x=0$

\frac{dy}{dx} = \frac{3y+2x+4}{4x+6y+5}

Let

F(x, y) = \frac{3y+2x+4}{4x+6y+5}

.

F(\lambda x, \lambda y) = \frac{3(\lambda y)+2(\lambda x)+4}{4(\lambda x)+6(\lambda y)+5} = \frac{\lambda(3y+2x)+4}{\lambda(4x+6y)+5} \neq F(x, y)

The presence of the constant terms 4 and 5 prevents it from being homogeneous.

(B) $(x y) d x-\left(x^{3}+y^{3}\right) d y=0$

\frac{dy}{dx} = \frac{xy}{x^3+y^3}

Let

F(x, y) = \frac{xy}{x^3+y^3}

. The degree of the numerator is

1+1=2

. The degree of the denominator is 3. Since the degrees are not the same, it is not a homogeneous equation.

F(\lambda x, \lambda y) = \frac{(\lambda x)(\lambda y)}{(\lambda x)^3+(\lambda y)^3} = \frac{\lambda^2 xy}{\lambda^3(x^3+y^3)} = \frac{1}{\lambda} F(x, y) \neq F(x, y)

(C) $\left(x^{3}+2 y^{2}\right) d x+2 x y d y=0$

\frac{dy}{dx} = -\frac{x^3+2y^2}{2xy}

Let

F(x, y) = -\frac{x^3+2y^2}{2xy}

. The numerator has terms of degree 3 (

x^3

) and degree 2 (

y^2

). Since the terms in the numerator do not have the same degree, it is not a homogeneous equation.

F(\lambda x, \lambda y) = -\frac{(\lambda x)^3+2(\lambda y)^2}{2(\lambda x)(\lambda y)} = -\frac{\lambda^3 x^3+2\lambda^2 y^2}{2\lambda^2 xy} \neq F(x, y)

(D) $y^{2} d x+\left(x^{2}-x y-y^{2}\right) d y=0$

\frac{dy}{dx} = -\frac{y^2}{x^2-xy-y^2} = \frac{y^2}{y^2+xy-x^2}

Let

F(x, y) = \frac{y^2}{y^2+xy-x^2}

. The degree of the numerator (

y^2

) is 2. Each term in the denominator (

y^2, xy, x^2

) also has a degree of 2. So, this is a homogeneous equation. Let's verify:

F(\lambda x, \lambda y) = \frac{(\lambda y)^2}{(\lambda y)^2+(\lambda x)(\lambda y)-(\lambda x)^2} = \frac{\lambda^2 y^2}{\lambda^2 y^2+\lambda^2 xy-\lambda^2 x^2} = \frac{\lambda^2(y^2)}{\lambda^2(y^2+xy-x^2)} = F(x, y)

Since

F(\lambda x, \lambda y) = F(x, y)

, the equation is homogeneous.

Final Answer: The correct option is (D) $y^{2} d x+\left(x^{2}-x y-y^{2}\right) d y=0$ .

Q1EXERCISE 9.5

$\frac{d y}{d x}+2 y=\sin x$

Solution

Given: The differential equation is

\frac{d y}{d x}+2 y=\sin x

.

To Find: The general solution of the differential equation.

Solution: The given differential equation is of the form of a linear differential equation:

\frac{dy}{dx} + P(x)y = Q(x)

Comparing the given equation with the standard form, we have:

P(x) = 2

Q(x) = \sin x

First, we find the Integrating Factor (I.F.):

I.F. = e^{\int P(x) dx}

I.F. = e^{\int 2 dx} = e^{2x}

The general solution of a linear differential equation is given by:

y \times (I.F.) = \int (Q(x) \times I.F.) dx + C

y \cdot e^{2x} = \int (\sin x \cdot e^{2x}) dx + C \quad ...(1)

Let's evaluate the integral

I = \int e^{2x} \sin x \, dx

. We use integration by parts, with the formula

\int u \, dv = uv - \int v \, du

. Let

u = \sin x

and

dv = e^{2x} dx

. Then

du = \cos x \, dx

and

v = \frac{e^{2x}}{2}

.

I = \sin x \cdot \frac{e^{2x}}{2} - \int \frac{e^{2x}}{2} \cos x \, dx

I = \frac{1}{2}e^{2x}\sin x - \frac{1}{2} \int e^{2x} \cos x \, dx

Now we apply integration by parts again for

\int e^{2x} \cos x \, dx

. Let

u = \cos x

and

dv = e^{2x} dx

. Then

du = -\sin x \, dx

and

v = \frac{e^{2x}}{2}

.

\int e^{2x} \cos x \, dx = \cos x \cdot \frac{e^{2x}}{2} - \int \frac{e^{2x}}{2} (-\sin x) \, dx

\int e^{2x} \cos x \, dx = \frac{1}{2}e^{2x}\cos x + \frac{1}{2} \int e^{2x} \sin x \, dx

\int e^{2x} \cos x \, dx = \frac{1}{2}e^{2x}\cos x + \frac{1}{2}I

Substitute this back into the expression for

I

:

I = \frac{1}{2}e^{2x}\sin x - \frac{1}{2} \left( \frac{1}{2}e^{2x}\cos x + \frac{1}{2}I \right)

I = \frac{1}{2}e^{2x}\sin x - \frac{1}{4}e^{2x}\cos x - \frac{1}{4}I

I + \frac{1}{4}I = \frac{1}{2}e^{2x}\sin x - \frac{1}{4}e^{2x}\cos x

\frac{5}{4}I = e^{2x} \left( \frac{1}{2}\sin x - \frac{1}{4}\cos x \right)

I = \frac{4}{5} e^{2x} \left( \frac{2\sin x - \cos x}{4} \right)

I = \frac{e^{2x}}{5} (2\sin x - \cos x)

Alternatively, we can use the standard formula:

\int e^{ax} \sin(bx) dx = \frac{e^{ax}}{a^2+b^2}(a\sin(bx) - b\cos(bx))

Here,

a=2

and

b=1

. So,

I = \frac{e^{2x}}{2^2+1^2}(2\sin x - 1\cos x) = \frac{e^{2x}}{5}(2\sin x - \cos x)

Now, substitute the value of the integral back into equation (1):

y \cdot e^{2x} = \frac{e^{2x}}{5}(2\sin x - \cos x) + C

Divide the entire equation by

e^{2x}

to find

y

:

y = \frac{1}{5}(2\sin x - \cos x) + C e^{-2x}

Final Answer: The general solution of the given differential equation is

y = \frac{1}{5}(2\sin x - \cos x) + C e^{-2x}

.

Q2EXERCISE 9.5

$\frac{d y}{d x}+3 y=e^{-2 x}$

Solution

Given: The differential equation is

\frac{dy}{dx}+3 y=e^{-2 x}

.

To Find: The general solution of the differential equation.

Solution: The given equation is a linear differential equation of the form

\frac{dy}{dx} + Py = Q

, where

P = 3

and

Q = e^{-2x}

.

First, we find the Integrating Factor (I.F.). The formula for the I.F. is

e^{\int P dx}

.

I.F. = e^{\int 3 dx} = e^{3x}

The general solution of a linear differential equation is given by the formula:

y \times (I.F.) = \int (Q \times I.F.) dx + C

Substituting the values of I.F. and Q:

y \cdot e^{3x} = \int (e^{-2x} \cdot e^{3x}) dx + C

y \cdot e^{3x} = \int e^{-2x+3x} dx + C

y \cdot e^{3x} = \int e^x dx + C

Integrating

e^x

with respect to

x

gives

e^x

.

y \cdot e^{3x} = e^x + C

To get the explicit solution for

y

, we divide the entire equation by

e^{3x}

:

y = \frac{e^x}{e^{3x}} + \frac{C}{e^{3x}}

y = e^{x-3x} + C e^{-3x}

y = e^{-2x} + C e^{-3x}

Final Answer: The general solution of the given differential equation is

y = e^{-2x} + C e^{-3x}

.

Q3EXERCISE 9.5

$\frac{d y}{d x}+\frac{y}{x}=x^{2}$

Solution

Given: The differential equation is

\frac{d y}{d x}+\frac{y}{x}=x^{2}

.

To Find: The general solution of the differential equation.

Solution: The given equation is a linear differential equation of the form

\frac{dy}{dx} + Py = Q

, where

P = \frac{1}{x}

and

Q = x^2

. The presence of

x

in the denominator implies

x \neq 0

.

First, we find the Integrating Factor (I.F.). The formula for the I.F. is

e^{\int P dx}

.

I.F. = e^{\int \frac{1}{x} dx} = e^{\ln|x|} = |x|

Assuming

x > 0

, the I.F. is

x

. If

x < 0

, the I.F. is

-x

. The solution form remains the same. Let's proceed with I.F. =

x

.

The general solution of a linear differential equation is given by the formula:

y \times (I.F.) = \int (Q \times I.F.) dx + C

Substituting the values of I.F. and Q:

y \cdot x = \int (x^2 \cdot x) dx + C

y \cdot x = \int x^3 dx + C

Integrating

x^3

with respect to

x

gives

\frac{x^4}{4}

.

y \cdot x = \frac{x^4}{4} + C

To get the explicit solution for

y

, we divide the entire equation by

x

:

y = \frac{x^4}{4x} + \frac{C}{x}

y = \frac{x^3}{4} + \frac{C}{x}

Final Answer: The general solution of the given differential equation is

y = \frac{x^3}{4} + \frac{C}{x}

.

Q4EXERCISE 9.5

$\frac{d y}{d x}+(\sec x) y=\tan x\left(0 \leq x<\frac{\pi}{2}\right)$

Solution

Given: The differential equation is

\frac{d y}{d x}+(\sec x) y=\tan x

for

0 \leq x<\frac{\pi}{2}

.

To Find: The general solution of the differential equation.

Solution: The given equation is a linear differential equation of the form

\frac{dy}{dx} + Py = Q

, where

P = \sec x

and

Q = \tan x

. The condition

0 \leq x < \frac{\pi}{2}

ensures that

\sec x

and

\tan x

are well-defined and non-negative.

First, we find the Integrating Factor (I.F.).

I.F. = e^{\int P dx} = e^{\int \sec x dx}

Since

\int \sec x dx = \ln|\sec x + \tan x|

,

I.F. = e^{\ln|\sec x + \tan x|} = |\sec x + \tan x|

For

0 \leq x < \frac{\pi}{2}

,

\sec x > 0

and

\tan x \geq 0

, so

\sec x + \tan x > 0

. Thus, I.F. =

\sec x + \tan x

.

The general solution is given by:

y \times (I.F.) = \int (Q \times I.F.) dx + C

y(\sec x + \tan x) = \int \tan x (\sec x + \tan x) dx + C

y(\sec x + \tan x) = \int (\sec x \tan x + \tan^2 x) dx + C

We know that

\tan^2 x = \sec^2 x - 1

. Substituting this into the integral:

y(\sec x + \tan x) = \int (\sec x \tan x + \sec^2 x - 1) dx + C

Now, we integrate term by term:

\int \sec x \tan x dx = \sec x

\int \sec^2 x dx = \tan x

\int 1 dx = x

So, the equation becomes:

y(\sec x + \tan x) = \sec x + \tan x - x + C

To get the explicit solution for

y

, we can divide by

(\sec x + \tan x)

:

y = \frac{\sec x + \tan x - x + C}{\sec x + \tan x}

y = 1 - \frac{x - C}{\sec x + \tan x}

Final Answer: The general solution of the given differential equation is

y(\sec x + \tan x) = \sec x + \tan x - x + C

.

Q5EXERCISE 9.5

$\cos ^{2} x \frac{d y}{d x}+y=\tan x\left(0 \leq x<\frac{\pi}{2}\right)$

Solution

Given: The differential equation is

\cos ^{2} x \frac{d y}{d x}+y=\tan x

for

0 \leq x<\frac{\pi}{2}

.

To Find: The general solution of the differential equation.

Solution: First, we rewrite the equation in the standard linear form

\frac{dy}{dx} + Py = Q

. To do this, we divide the entire equation by

\cos^2 x

.

\frac{dy}{dx} + \frac{1}{\cos^2 x} y = \frac{\tan x}{\cos^2 x}

Using the identities

\sec x = \frac{1}{\cos x}

and

\tan x = \frac{\sin x}{\cos x}

, we get:

\frac{dy}{dx} + (\sec^2 x) y = \tan x \sec^2 x

This is a linear differential equation with

P = \sec^2 x

and

Q = \tan x \sec^2 x

.

Next, we find the Integrating Factor (I.F.).

I.F. = e^{\int P dx} = e^{\int \sec^2 x dx}

Since

\int \sec^2 x dx = \tan x

,

I.F. = e^{\tan x}

The general solution is given by:

y \times (I.F.) = \int (Q \times I.F.) dx + C

y \cdot e^{\tan x} = \int (\tan x \sec^2 x \cdot e^{\tan x}) dx + C

To evaluate the integral on the right side, we use the substitution method. Let

t = \tan x

. Then

dt = \sec^2 x dx

.

\int (\tan x \sec^2 x \cdot e^{\tan x}) dx = \int t e^t dt

We solve

\int t e^t dt

using integration by parts,

\int u dv = uv - \int v du

. Let

u=t

and

dv=e^t dt

. Then

du=dt

and

v=e^t

.

\int t e^t dt = t e^t - \int e^t dt = t e^t - e^t = e^t(t-1)

Substitute back

t = \tan x

:

\int (\tan x \sec^2 x \cdot e^{\tan x}) dx = e^{\tan x}(\tan x - 1)

Now, substitute this result back into the general solution equation:

y \cdot e^{\tan x} = e^{\tan x}(\tan x - 1) + C

To find the explicit solution for

y

, divide by

e^{\tan x}

:

y = \frac{e^{\tan x}(\tan x - 1)}{e^{\tan x}} + \frac{C}{e^{\tan x}}

y = \tan x - 1 + C e^{-\tan x}

Final Answer: The general solution of the given differential equation is

y = \tan x - 1 + C e^{-\tan x}

.

Q6EXERCISE 9.5

$x \frac{d y}{d x}+2 y=x^{2} \log x$

Solution

Given: The differential equation is

x \frac{d y}{d x}+2 y=x^{2} \log x

.

To Find: The general solution of the differential equation.

Solution: First, we rewrite the equation in the standard linear form

\frac{dy}{dx} + Py = Q

. Since

\log x

is present, we assume

x > 0

. We divide the entire equation by

x

.

\frac{dy}{dx} + \frac{2}{x} y = x \log x

This is a linear differential equation with

P = \frac{2}{x}

and

Q = x \log x

.

Next, we find the Integrating Factor (I.F.).

I.F. = e^{\int P dx} = e^{\int \frac{2}{x} dx} = e^{2 \ln x} = e^{\ln(x^2)} = x^2

The general solution is given by:

y \times (I.F.) = \int (Q \times I.F.) dx + C

y \cdot x^2 = \int (x \log x \cdot x^2) dx + C

y \cdot x^2 = \int x^3 \log x dx + C

To evaluate the integral

\int x^3 \log x dx

, we use integration by parts,

\int u dv = uv - \int v du

. Let

u = \log x

and

dv = x^3 dx

. Then

du = \frac{1}{x} dx

and

v = \int x^3 dx = \frac{x^4}{4}

.

\int x^3 \log x dx = (\log x) \left(\frac{x^4}{4}\right) - \int \left(\frac{x^4}{4}\right) \left(\frac{1}{x}\right) dx

= \frac{x^4 \log x}{4} - \frac{1}{4} \int x^3 dx

= \frac{x^4 \log x}{4} - \frac{1}{4} \left(\frac{x^4}{4}\right)

= \frac{x^4 \log x}{4} - \frac{x^4}{16}

Now, substitute this result back into the general solution equation:

y \cdot x^2 = \frac{x^4 \log x}{4} - \frac{x^4}{16} + C

To find the explicit solution for

y

, divide by

x^2

:

y = \frac{x^2 \log x}{4} - \frac{x^2}{16} + \frac{C}{x^2}

This can also be written as:

y = \frac{x^2}{16}(4 \log x - 1) + C x^{-2}

Final Answer: The general solution of the given differential equation is

y = \frac{x^2}{16}(4 \log x - 1) + C x^{-2}

.

Q7EXERCISE 9.5

$x \log x \frac{d y}{d x}+y=\frac{2}{x} \log x$

Solution

Given: The differential equation is

x \log x \frac{d y}{d x}+y=\frac{2}{x} \log x

.

To Find: The general solution of the differential equation.

Solution: The given differential equation is:

x \log x \frac{d y}{d x}+y=\frac{2}{x} \log x

To write it in the standard linear form

\frac{dy}{dx} + Py = Q

, we divide the entire equation by

x \log x

:

\frac{dy}{dx} + \frac{1}{x \log x} y = \frac{\frac{2}{x} \log x}{x \log x}

\frac{dy}{dx} + \left(\frac{1}{x \log x}\right) y = \frac{2}{x^2}

This is a linear differential equation where:

P = \frac{1}{x \log x}

and

Q = \frac{2}{x^2}

.

First, we find the Integrating Factor (I.F.):

I.F. = e^{\int P dx} = e^{\int \frac{1}{x \log x} dx}

To evaluate the integral, let

t = \log x

. Then

dt = \frac{1}{x} dx

.

\int \frac{1}{x \log x} dx = \int \frac{1}{t} dt = \log|t| = \log|\log x|

Assuming

\log x > 0

, we have:

I.F. = e^{\log(\log x)} = \log x

The general solution of a linear differential equation is given by:

y \times (I.F.) = \int (Q \times I.F.) dx + C

y \log x = \int \left(\frac{2}{x^2} \cdot \log x\right) dx + C

Now, we evaluate the integral on the right-hand side using integration by parts,

\int u dv = uv - \int v du

. Let

u = \log x

and

dv = \frac{2}{x^2} dx

. Then

du = \frac{1}{x} dx

and

v = \int \frac{2}{x^2} dx = 2 \int x^{-2} dx = 2 \frac{x^{-1}}{-1} = -\frac{2}{x}

.

\int \frac{2}{x^2} \log x dx = (\log x) \left(-\frac{2}{x}\right) - \int \left(-\frac{2}{x}\right) \left(\frac{1}{x}\right) dx

= -\frac{2 \log x}{x} + \int \frac{2}{x^2} dx

= -\frac{2 \log x}{x} - \frac{2}{x}

Substituting this back into the solution equation:

y \log x = -\frac{2 \log x}{x} - \frac{2}{x} + C

Final Answer: The general solution is

y \log x = -\frac{2}{x}(1 + \log x) + C

, or explicitly for y:

y = -\frac{2(1 + \log x)}{x \log x} + \frac{C}{\log x}

Q8EXERCISE 9.5

$\left(1+x^{2}\right) d y+2 x y d x=\cot x d x(x \neq 0)$

Solution

Given: The differential equation is

\left(1+x^{2}\right) d y+2 x y d x=\cot x d x(x \neq 0)

.

To Find: The general solution of the differential equation.

Solution: The given differential equation can be rewritten as:

\left(1+x^{2}\right) d y = (\cot x - 2xy) dx

\frac{dy}{dx} = \frac{\cot x - 2xy}{1+x^2}

\frac{dy}{dx} = \frac{\cot x}{1+x^2} - \frac{2xy}{1+x^2}

Rearranging to the standard linear form

\frac{dy}{dx} + Py = Q

:

\frac{dy}{dx} + \frac{2x}{1+x^2} y = \frac{\cot x}{1+x^2}

This is a linear differential equation where:

P = \frac{2x}{1+x^2}

and

Q = \frac{\cot x}{1+x^2}

.

First, we find the Integrating Factor (I.F.):

I.F. = e^{\int P dx} = e^{\int \frac{2x}{1+x^2} dx}

To evaluate the integral, let

t = 1+x^2

. Then

dt = 2x dx

.

\int \frac{2x}{1+x^2} dx = \int \frac{1}{t} dt = \log|t| = \log(1+x^2)

(since

1+x^2 > 0

for all real

x

)

I.F. = e^{\log(1+x^2)} = 1+x^2

The general solution of a linear differential equation is given by:

y \times (I.F.) = \int (Q \times I.F.) dx + C

y (1+x^2) = \int \left(\frac{\cot x}{1+x^2} \cdot (1+x^2)\right) dx + C

y (1+x^2) = \int \cot x dx + C

The integral of

\cot x

is

\log|\sin x|

.

y (1+x^2) = \log|\sin x| + C

Final Answer: The general solution is

y (1+x^2) = \log|\sin x| + C

.

Q9EXERCISE 9.5

$x \frac{d y}{d x}+y-x+x y \cot x=0(x \neq 0)$

Solution

Given: The differential equation is

x \frac{d y}{d x}+y-x+x y \cot x=0(x \neq 0)

.

To Find: The general solution of the differential equation.

Solution: The given differential equation is:

x \frac{d y}{d x}+y-x+x y \cot x=0

First, we group the terms involving

y

and move the term with

x

to the other side:

x \frac{dy}{dx} + y(1 + x \cot x) = x

To write it in the standard linear form

\frac{dy}{dx} + Py = Q

, we divide the entire equation by

x

:

\frac{dy}{dx} + \frac{y(1 + x \cot x)}{x} = \frac{x}{x}

\frac{dy}{dx} + \left(\frac{1}{x} + \cot x\right) y = 1

This is a linear differential equation where:

P = \frac{1}{x} + \cot x

and

Q = 1

.

Next, we find the Integrating Factor (I.F.):

I.F. = e^{\int P dx} = e^{\int (\frac{1}{x} + \cot x) dx}

I.F. = e^{\int \frac{1}{x} dx + \int \cot x dx} = e^{\log|x| + \log|\sin x|}

Using the property

\log a + \log b = \log(ab)

:

I.F. = e^{\log|x \sin x|} = x \sin x

(assuming

x \sin x > 0

) The general solution of a linear differential equation is given by:

y \times (I.F.) = \int (Q \times I.F.) dx + C

y (x \sin x) = \int (1 \cdot x \sin x) dx + C

y (x \sin x) = \int x \sin x dx + C

Now, we evaluate the integral on the right-hand side using integration by parts,

\int u dv = uv - \int v du

. Let

u = x

and

dv = \sin x dx

. Then

du = dx

and

v = \int \sin x dx = -\cos x

.

\int x \sin x dx = x(-\cos x) - \int (-\cos x) dx

= -x \cos x + \int \cos x dx

= -x \cos x + \sin x

Substituting this back into the solution equation:

y (x \sin x) = -x \cos x + \sin x + C

Final Answer: The general solution is

y (x \sin x) = -x \cos x + \sin x + C

.

Q10EXERCISE 9.5

$(x+y) \frac{d y}{d x}=1$

Solution

Given: The differential equation is

(x+y) \frac{d y}{d x}=1

.

To Find: The general solution of the differential equation.

Solution: The given equation is:

(x+y) \frac{d y}{d x}=1

This equation can be written as

\frac{dy}{dx} = \frac{1}{x+y}

. This form is not linear in

y

. Let's try to express

x

as a function of

y

. We can rewrite the equation as:

\frac{dx}{dy} = x+y

Rearranging this to the standard linear form for

x

, which is

\frac{dx}{dy} + Px = Q

(where P and Q are functions of y):

\frac{dx}{dy} - x = y

This is a linear differential equation in

x

with the independent variable

y

. Here:

P = -1

and

Q = y

.

First, we find the Integrating Factor (I.F.):

I.F. = e^{\int P dy} = e^{\int (-1) dy} = e^{-y}

The general solution of this type of linear differential equation is given by:

x \times (I.F.) = \int (Q \times I.F.) dy + C

x e^{-y} = \int y e^{-y} dy + C

Now, we evaluate the integral on the right-hand side using integration by parts,

\int u dv = uv - \int v du

. Let

u = y

and

dv = e^{-y} dy

. Then

du = dy

and

v = \int e^{-y} dy = -e^{-y}

.

\int y e^{-y} dy = y(-e^{-y}) - \int (-e^{-y}) dy

= -y e^{-y} + \int e^{-y} dy

= -y e^{-y} - e^{-y}

Substituting this back into the solution equation:

x e^{-y} = -y e^{-y} - e^{-y} + C

To simplify, we can multiply the entire equation by

e^y

:

x = -y - 1 + C e^y

Rearranging the terms, we get:

x + y + 1 = C e^y

Final Answer: The general solution is

x + y + 1 = C e^y

.

Q11EXERCISE 9.5

$y d x+\left(x-y^{2}\right) d y=0$

Solution

Given: The differential equation is

y d x+\left(x-y^{2}\right) d y=0

.

To Find: The general solution of the differential equation.

Solution: The given differential equation is:

y dx + (x - y^2) dy = 0

If we try to write it in the form

\frac{dy}{dx}

, we get:

(x-y^2)dy = -y dx \implies \frac{dy}{dx} = \frac{-y}{x-y^2} = \frac{y}{y^2-x}

This is not a standard linear form in

y

. Let's try to express

x

as a function of

y

. We can rewrite the original equation as:

y \frac{dx}{dy} + x - y^2 = 0

y \frac{dx}{dy} + x = y^2

To get the standard linear form

\frac{dx}{dy} + Px = Q

(where P and Q are functions of y), we divide by

y

(assuming

y \neq 0

):

\frac{dx}{dy} + \frac{1}{y} x = y

This is a linear differential equation in

x

with the independent variable

y

. Here:

P = \frac{1}{y}

and

Q = y

.

First, we find the Integrating Factor (I.F.):

I.F. = e^{\int P dy} = e^{\int \frac{1}{y} dy} = e^{\log|y|} = |y|

Let's consider the case where

y>0

, so

I.F. = y

. (The solution for

y<0

would have a similar form).

I.F. = y

The general solution of this type of linear differential equation is given by:

x \times (I.F.) = \int (Q \times I.F.) dy + C

x \cdot y = \int (y \cdot y) dy + C

xy = \int y^2 dy + C

xy = \frac{y^3}{3} + C

To get an explicit solution for

x

, we can divide by

y

:

x = \frac{y^2}{3} + \frac{C}{y}

Final Answer: The general solution is

xy = \frac{y^3}{3} + C

.

Q12EXERCISE 9.5

$(x+3 y^{2}) \frac{d y}{d x}=y(y>0)$ .

Solution

Given: The differential equation

(x+3 y^{2}) \frac{d y}{d x}=y(y>0)

.

To Find: The general solution of the differential equation.

Solution: The given differential equation is:

(x+3 y^{2}) \frac{d y}{d x}=y

This can be rewritten as:

\frac{dx}{dy} = \frac{x+3y^2}{y}

\frac{dx}{dy} = \frac{x}{y} + 3y

\frac{dx}{dy} - \frac{1}{y}x = 3y

This is a linear differential equation of the form

\frac{dx}{dy} + P(y)x = Q(y)

, where

P(y) = -\frac{1}{y}

and

Q(y) = 3y

.

The integrating factor (I.F.) is given by:

\text{I.F.} = e^{\int P(y) dy} = e^{\int -\frac{1}{y} dy} = e^{-\ln y} = e^{\ln(y^{-1})} = y^{-1} = \frac{1}{y}

The solution of the linear differential equation is given by:

x \cdot (\text{I.F.}) = \int Q(y) \cdot (\text{I.F.}) dy + C

x \cdot \frac{1}{y} = \int (3y) \cdot \left(\frac{1}{y}\right) dy + C

\frac{x}{y} = \int 3 dy + C

\frac{x}{y} = 3y + C

x = 3y^2 + Cy

Final Answer: The general solution of the given differential equation is

x = 3y^2 + Cy

.

Q13EXERCISE 9.5

$\frac{d y}{d x}+2 y \tan x=\sin x ; y=0$ when $x=\frac{\pi}{3}$

Solution

Given: The differential equation

\frac{d y}{d x}+2 y \tan x=\sin x

, with the condition

y=0

when

x=\frac{\pi}{3}

.

To Find: The particular solution of the differential equation.

Solution: The given differential equation is in the linear form

\frac{dy}{dx} + P(x)y = Q(x)

. Here,

P(x) = 2 \tan x

and

Q(x) = \sin x

.

The integrating factor (I.F.) is given by:

\text{I.F.} = e^{\int P(x) dx} = e^{\int 2 \tan x dx} = e^{2 \ln|\sec x|} = e^{\ln(\sec^2 x)} = \sec^2 x

The solution of the linear differential equation is given by:

y \cdot (\text{I.F.}) = \int Q(x) \cdot (\text{I.F.}) dx + C

y \cdot \sec^2 x = \int \sin x \cdot \sec^2 x dx + C

y \sec^2 x = \int \frac{\sin x}{\cos^2 x} dx + C

y \sec^2 x = \int \frac{\sin x}{\cos x} \cdot \frac{1}{\cos x} dx + C

y \sec^2 x = \int \tan x \sec x dx + C

y \sec^2 x = \sec x + C

This is the general solution.

Now, we apply the given condition:

y=0

when

x=\frac{\pi}{3}

.

0 \cdot \sec^2\left(\frac{\pi}{3}\right) = \sec\left(\frac{\pi}{3}\right) + C

0 = 2 + C

C = -2

Substituting the value of

C

back into the general solution:

y \sec^2 x = \sec x - 2

y = \frac{\sec x}{\sec^2 x} - \frac{2}{\sec^2 x}

y = \cos x - 2\cos^2 x

Final Answer: The particular solution of the given differential equation is

y = \cos x - 2\cos^2 x

.

Q14EXERCISE 9.5

$(1+x^{2}) \frac{d y}{d x}+2 x y=\frac{1}{1+x^{2}} ; y=0$ when $x=1$

Solution

Given: The differential equation

(1+x^{2}) \frac{d y}{d x}+2 x y=\frac{1}{1+x^{2}}

, with the condition

y=0

when

x=1

.

To Find: The particular solution of the differential equation.

Solution: First, we write the given equation in the standard linear form

\frac{dy}{dx} + P(x)y = Q(x)

. Divide the entire equation by

(1+x^2)

:

\frac{d y}{d x}+\frac{2 x}{1+x^{2}} y=\frac{1}{(1+x^{2})^2}

This is a linear differential equation where

P(x) = \frac{2x}{1+x^2}

and

Q(x) = \frac{1}{(1+x^2)^2}

.

The integrating factor (I.F.) is given by:

\text{I.F.} = e^{\int P(x) dx} = e^{\int \frac{2x}{1+x^2} dx}

Let

t = 1+x^2

, so

dt = 2x dx

. The integral becomes

\int \frac{1}{t} dt = \ln|t| = \ln(1+x^2)

.

\text{I.F.} = e^{\ln(1+x^2)} = 1+x^2

The solution of the linear differential equation is given by:

y \cdot (\text{I.F.}) = \int Q(x) \cdot (\text{I.F.}) dx + C

y(1+x^2) = \int \frac{1}{(1+x^2)^2} \cdot (1+x^2) dx + C

y(1+x^2) = \int \frac{1}{1+x^2} dx + C

y(1+x^2) = \tan^{-1}x + C

This is the general solution.

Now, we apply the given condition:

y=0

when

x=1

.

0(1+1^2) = \tan^{-1}(1) + C

0 = \frac{\pi}{4} + C

C = -\frac{\pi}{4}

Substituting the value of

C

back into the general solution:

y(1+x^2) = \tan^{-1}x - \frac{\pi}{4}

Final Answer: The particular solution of the given differential equation is

y(1+x^2) = \tan^{-1}x - \frac{\pi}{4}

.

Q15EXERCISE 9.5

$\frac{d y}{d x}-3 y \cot x=\sin 2 x ; y=2$ when $x=\frac{\pi}{2}$

Solution

Given: The differential equation

\frac{d y}{d x}-3 y \cot x=\sin 2 x

, with the condition

y=2

when

x=\frac{\pi}{2}

.

To Find: The particular solution of the differential equation.

Solution: The given differential equation is in the linear form

\frac{dy}{dx} + P(x)y = Q(x)

. Here,

P(x) = -3 \cot x

and

Q(x) = \sin 2x

.

The integrating factor (I.F.) is given by:

\text{I.F.} = e^{\int P(x) dx} = e^{\int -3 \cot x dx} = e^{-3 \ln|\sin x|} = e^{\ln((\sin x)^{-3})} = (\sin x)^{-3} = \frac{1}{\sin^3 x}

The solution of the linear differential equation is given by:

y \cdot (\text{I.F.}) = \int Q(x) \cdot (\text{I.F.}) dx + C

y \cdot \frac{1}{\sin^3 x} = \int \sin 2x \cdot \frac{1}{\sin^3 x} dx + C

\frac{y}{\sin^3 x} = \int \frac{2 \sin x \cos x}{\sin^3 x} dx + C

\frac{y}{\sin^3 x} = \int \frac{2 \cos x}{\sin^2 x} dx + C

To evaluate the integral, let

u = \sin x

, so

du = \cos x dx

. The integral becomes:

\int \frac{2}{u^2} du = 2 \int u^{-2} du = 2 \frac{u^{-1}}{-1} = -\frac{2}{u} = -\frac{2}{\sin x}

So, the solution is:

\frac{y}{\sin^3 x} = -\frac{2}{\sin x} + C

y = -2\sin^2 x + C \sin^3 x

This is the general solution.

Now, we apply the given condition:

y=2

when

x=\frac{\pi}{2}

.

2 = -2\sin^2\left(\frac{\pi}{2}\right) + C \sin^3\left(\frac{\pi}{2}\right)

2 = -2(1)^2 + C(1)^3

2 = -2 + C

C = 4

Substituting the value of

C

back into the general solution:

y = -2\sin^2 x + 4\sin^3 x

Final Answer: The particular solution of the given differential equation is

y = 4\sin^3 x - 2\sin^2 x

.

Q16EXERCISE 9.5

Find the equation of a curve passing through the origin given that the slope of the tangent to the curve at any point $(x, y)$ is equal to the sum of the coordinates of the point.

Solution

Given: A curve passes through the origin

(0,0)

. The slope of the tangent to the curve at any point

(x, y)

is equal to the sum of the coordinates of the point.

To Find: The equation of the curve.

Solution: Let the equation of the curve be

y = f(x)

. The slope of the tangent at any point

(x,y)

is given by

\frac{dy}{dx}

. According to the problem statement, the slope is equal to the sum of the coordinates, so:

\frac{dy}{dx} = x + y

We can rearrange this equation into the standard linear form:

\frac{dy}{dx} - y = x

This is a linear differential equation of the form

\frac{dy}{dx} + P(x)y = Q(x)

, where

P(x) = -1

and

Q(x) = x

.

The integrating factor (I.F.) is given by:

\text{I.F.} = e^{\int P(x) dx} = e^{\int -1 dx} = e^{-x}

The solution of the linear differential equation is given by:

y \cdot (\text{I.F.}) = \int Q(x) \cdot (\text{I.F.}) dx + C

y e^{-x} = \int x e^{-x} dx + C

We evaluate the integral

\int x e^{-x} dx

using integration by parts,

\int u dv = uv - \int v du

. Let

u = x

and

dv = e^{-x} dx

. Then

du = dx

and

v = -e^{-x}

.

\int x e^{-x} dx = x(-e^{-x}) - \int (-e^{-x}) dx

= -xe^{-x} + \int e^{-x} dx

= -xe^{-x} - e^{-x} = -(x+1)e^{-x}

Substituting this back into the solution equation:

y e^{-x} = -(x+1)e^{-x} + C

Multiplying the entire equation by

e^x

to solve for

y

:

y = -(x+1) + C e^x

This is the general solution.

The curve passes through the origin

(0,0)

. So, we substitute

x=0

and

y=0

to find

C

.

0 = -(0+1) + C e^0

0 = -1 + C(1)

C = 1

Substituting

C=1

back into the general solution gives the particular solution:

y = -(x+1) + e^x

y = e^x - x - 1

This can also be written as

x + y + 1 = e^x

.

Final Answer: The equation of the curve is

x+y+1=e^x

.

Q17EXERCISE 9.5

Find the equation of a curve passing through the point $(0,2)$ given that the sum of the coordinates of any point on the curve exceeds the magnitude of the slope of the tangent to the curve at that point by 5.

Solution

Given: A curve passes through the point

(0, 2)

. The sum of the coordinates of any point

(x, y)

on the curve exceeds the magnitude of the slope of the tangent to the curve at that point by 5.

Interpretation of the problem statement: Let

(x, y)

be any point on the curve. The slope of the tangent at this point is

\frac{dy}{dx}

. The sum of the coordinates is

x+y

. The magnitude of the slope is

|\frac{dy}{dx}|

.

The given condition is:

(x+y) - |\frac{dy}{dx}| = 5

, which means

x+y-5 = |\frac{dy}{dx}|

.

Let's check this condition at the given point

(0, 2)

:

0+2-5 = |\frac{dy}{dx}|_{(0,2)}

-3 = |\frac{dy}{dx}|_{(0,2)}

This is a contradiction, as the magnitude of a real number cannot be negative. This suggests a possible misstatement in the question as presented in the textbook.

A plausible interpretation is that the word "magnitude" was not intended. Let's assume the problem meant: "the sum of the coordinates of any point on the curve exceeds the slope of the tangent to the curve at that point by 5".

Formulating the differential equation: Under this assumption, the condition becomes:

(x+y) - \frac{dy}{dx} = 5

Rearranging the terms, we get a linear differential equation:

\frac{dy}{dx} = x+y-5

\frac{dy}{dx} - y = x-5

This is a linear differential equation of the form

\frac{dy}{dx} + P(x)y = Q(x)

, where

P(x) = -1

and

Q(x) = x-5

.

Solving the differential equation: First, we find the Integrating Factor (I.F.):

\text{I.F.} = e^{\int P(x) dx} = e^{\int -1 dx} = e^{-x}

The general solution is given by:

y \cdot (\text{I.F.}) = \int Q(x) \cdot (\text{I.F.}) dx + C

y e^{-x} = \int (x-5) e^{-x} dx + C

We solve the integral on the right-hand side using integration by parts,

\int u dv = uv - \int v du

. Let

u = x-5

and

dv = e^{-x} dx

. Then

du = dx

and

v = -e^{-x}

.

\int (x-5) e^{-x} dx = (x-5)(-e^{-x}) - \int (-e^{-x}) dx

= -(x-5)e^{-x} + \int e^{-x} dx

= -(x-5)e^{-x} - e^{-x}

= (-x+5-1)e^{-x} = (4-x)e^{-x}

Substituting this back into the solution equation:

y e^{-x} = (4-x)e^{-x} + C

To get the explicit solution for

y

, we multiply the entire equation by

e^x

:

y = 4-x+Ce^x

Finding the particular solution: The curve passes through the point

(0, 2)

. We substitute

x=0

and

y=2

into the general solution to find the constant

C

.

2 = 4-0+Ce^0

2 = 4+C

C = 2 - 4 = -2

Substituting

C=-2

back into the general solution gives the equation of the required curve.

Final Answer: The equation of the curve is

y = 4-x-2e^x

.

Q18EXERCISE 9.5

The Integrating Factor of the differential equation $x \frac{d y}{d x}-y=2 x^{2}$ is

(A)

e^{-x}

(B)

e^{-y}

(C)

\frac{1}{x}

(D)

x

Solution

Given: The differential equation is

x \frac{dy}{dx} - y = 2x^2

.

To Find: The Integrating Factor (I.F.) of the differential equation.

Solution: First, we need to write the given differential equation in the standard form of a linear differential equation, which is

\frac{dy}{dx} + P(x)y = Q(x)

.

To do this, we divide the entire equation by

x

(assuming

x \neq 0

):

\frac{x \frac{dy}{dx}}{x} - \frac{y}{x} = \frac{2x^2}{x}

\frac{dy}{dx} - \frac{1}{x} y = 2x

Now, the equation is in the standard form. By comparing it with

\frac{dy}{dx} + P(x)y = Q(x)

, we can identify

P(x)

:

P(x) = -\frac{1}{x}

Formula: The Integrating Factor (I.F.) is given by the formula:

\text{I.F.} = e^{\int P(x) dx}

Calculation: Substitute the expression for

P(x)

into the formula:

\text{I.F.} = e^{\int -\frac{1}{x} dx}

\text{I.F.} = e^{-\int \frac{1}{x} dx}

\text{I.F.} = e^{-\log|x|}

Using the property of logarithms,

m \log a = \log a^m

:

\text{I.F.} = e^{\log|x|^{-1}}

\text{I.F.} = e^{\log\frac{1}{|x|}}

Using the property

e^{\log z} = z

:

\text{I.F.} = \frac{1}{|x|}

Since the options do not involve absolute values, we consider the case for

x>0

, so

|x|=x

.

\text{I.F.} = \frac{1}{x}

Conclusion: The Integrating Factor is

\frac{1}{x}

. This matches option (C).

Final Answer: The correct option is (C)

\frac{1}{x}

.

Q19EXERCISE 9.5

The Integrating Factor of the differential equation $(1-y^{2}) \frac{d x}{d y}+y x=a y(-1<y<1)$ is

(A)

\frac{1}{y^{2}-1}

(B)

\frac{1}{\sqrt{y^{2}-1}}

(C)

\frac{1}{1-y^{2}}

(D)

\frac{1}{\sqrt{1-y^{2}}}

Solution

Given: The differential equation is

(1-y^2) \frac{dx}{dy} + yx = ay

, where

-1 < y < 1

.

To Find: The Integrating Factor (I.F.) of the differential equation.

Solution: The given equation is a linear differential equation in

x

. To find the Integrating Factor, we first need to write it in the standard form

\frac{dx}{dy} + P(y)x = Q(y)

.

Divide the entire equation by

(1-y^2)

:

\frac{(1-y^2) \frac{dx}{dy}}{1-y^2} + \frac{yx}{1-y^2} = \frac{ay}{1-y^2}

\frac{dx}{dy} + \frac{y}{1-y^2} x = \frac{ay}{1-y^2}

Now the equation is in the standard form. By comparing it with

\frac{dx}{dy} + P(y)x = Q(y)

, we can identify

P(y)

:

P(y) = \frac{y}{1-y^2}

Formula: The Integrating Factor (I.F.) is given by the formula:

\text{I.F.} = e^{\int P(y) dy}

Calculation: Substitute the expression for

P(y)

into the formula:

\text{I.F.} = e^{\int \frac{y}{1-y^2} dy}

To evaluate the integral, we use the substitution method. Let

t = 1-y^2

. Then,

dt = -2y dy

, which implies

y dy = -\frac{1}{2} dt

.

\int \frac{y}{1-y^2} dy = \int \frac{1}{t} \left(-\frac{1}{2} dt\right) = -\frac{1}{2} \int \frac{1}{t} dt

= -\frac{1}{2} \log|t|

Substitute back

t = 1-y^2

:

= -\frac{1}{2} \log|1-y^2|

Since it is given that

-1 < y < 1

, we have

y^2 < 1

, which means

1-y^2 > 0

. So,

|1-y^2| = 1-y^2

.

= -\frac{1}{2} \log(1-y^2)

Using the property of logarithms,

m \log a = \log a^m

:

= \log((1-y^2)^{-1/2}) = \log\left(\frac{1}{\sqrt{1-y^2}}\right)

Now, substitute this back into the I.F. formula:

\text{I.F.} = e^{\log\left(\frac{1}{\sqrt{1-y^2}}\right)}

Using the property

e^{\log z} = z

:

\text{I.F.} = \frac{1}{\sqrt{1-y^2}}

Conclusion: The Integrating Factor is

\frac{1}{\sqrt{1-y^2}}

. This matches option (D).

Final Answer: The correct option is (D)

\frac{1}{\sqrt{1-y^2}}

.

Q1Miscellaneous Exercise on Chapter 9

For each of the differential equations given below, indicate its order and degree (if defined).

(i)

$\frac{d^{2} y}{d x^{2}}+5 x\left(\frac{d y}{d x}\right)^{2}-6 y=\log x$

(ii)

$\left(\frac{d y}{d x}\right)^{3}-4\left(\frac{d y}{d x}\right)^{2}+7 y=\sin x$

(iii)

$\frac{d^{4} y}{d x^{4}}-\sin \left(\frac{d^{3} y}{d x^{3}}\right)=0$

Solution

To Find: For each differential equation, determine its order and degree (if defined).

Definitions:

Order: The order of a differential equation is the order of the highest derivative appearing in the equation.
Degree: The degree of a differential equation is the power of the highest order derivative, after the equation has been made a polynomial in derivatives (i.e., free from radicals and fractions involving derivatives).

Solution:

(i) $\frac{d^{2} y}{d x^{2}}+5 x\left(\frac{d y}{d x}\right)^{2}-6 y=\log x$

Order: The derivatives present are $\frac{d^2y}{dx^2}$ (second order) and $\frac{dy}{dx}$ (first order). The highest order derivative is $\frac{d^2y}{dx^2}$ , which is of order 2. Therefore, the order of the equation is 2.
Degree: The equation is a polynomial in its derivatives. The highest order derivative is $\frac{d^2y}{dx^2}$ , and its power is 1. Therefore, the degree of the equation is 1.

Answer (i): Order = 2, Degree = 1.

(ii) $\left(\frac{d y}{d x}\right)^{3}-4\left(\frac{d y}{d x}\right)^{2}+7 y=\sin x$

Order: The only derivative present is $\frac{dy}{dx}$ , which is of order 1. Therefore, the order of the equation is 1.
Degree: The equation is a polynomial in its derivative. The highest order derivative is $\frac{dy}{dx}$ , and its highest power in the equation is 3. Therefore, the degree of the equation is 3.

Answer (ii): Order = 1, Degree = 3.

(iii) $\frac{d^{4} y}{d x^{4}}-\sin \left(\frac{d^{3} y}{d x^{3}}\right)=0$

Order: The derivatives present are $\frac{d^4y}{dx^4}$ (fourth order) and $\frac{d^3y}{dx^3}$ (third order). The highest order derivative is $\frac{d^4y}{dx^4}$ , which is of order 4. Therefore, the order of the equation is 4.
Degree: The equation contains the term $\sin(\frac{d^3y}{dx^3})$ . Because the derivative $\frac{d^3y}{dx^3}$ is an argument of the trigonometric function $\sin$ , the differential equation cannot be expressed as a polynomial in its derivatives. Therefore, the degree of the equation is not defined.

Answer (iii): Order = 4, Degree is not defined.

Final Answer:

(i)

Order: 2, Degree: 1

(ii)

Order: 1, Degree: 3

(iii)

Order: 4, Degree: not defined

Q2Miscellaneous Exercise on Chapter 9

For each of the exercises given below, verify that the given function (implicit or explicit) is a solution of the corresponding differential equation.

(i)

$x y=a e^{x}+b e^{-x}+x^{2} \quad: x \frac{d^{2} y}{d x^{2}}+2 \frac{d y}{d x}-x y+x^{2}-2=0$

(ii)

$y=e^{x}(a \cos x+b \sin x): \frac{d^{2} y}{d x^{2}}-2 \frac{d y}{d x}+2 y=0$

(iii)

$y=x \sin 3 x \quad:\frac{d^{2} y}{d x^{2}}+9 y-6 \cos 3 x=0$

(iv)

$x^{2}=2 y^{2} \log y \quad:\left(x^{2}+y^{2}\right) \frac{d y}{d x}-x y=0$

Solution

To Verify: For each case, verify that the given function is a solution of the corresponding differential equation.

Solution:

(i) Function: $x y=a e^{x}+b e^{-x}+x^{2}$ Differential Equation: $x \frac{d^{2} y}{d x^{2}}+2 \frac{d y}{d x}-x y+x^{2}-2=0$

Verification: Given function:

xy = ae^x + be^{-x} + x^2

. First, let's express

y

explicitly:

y = \frac{ae^x + be^{-x} + x^2}{x}

. Now, we differentiate with respect to

x

using the product rule on

xy = ...

:

1 \cdot y + x \frac{dy}{dx} = a e^x - b e^{-x} + 2x \quad ... (1)

Differentiate again with respect to

x

:

\frac{dy}{dx} + (1 \cdot \frac{dy}{dx} + x \frac{d^2y}{dx^2}) = a e^x + b e^{-x} + 2

2 \frac{dy}{dx} + x \frac{d^2y}{dx^2} = a e^x + b e^{-x} + 2

From the given function, we have

ae^x + be^{-x} = xy - x^2

. Substitute this into the second derivative equation:

2 \frac{dy}{dx} + x \frac{d^2y}{dx^2} = (xy - x^2) + 2

Rearranging the terms to match the differential equation:

x \frac{d^{2} y}{d x^{2}}+2 \frac{d y}{d x}-x y+x^{2}-2=0

This is the same as the given differential equation. Hence, the given function is a solution of the differential equation.

(ii) Function: $y=e^{x}(a \cos x+b \sin x)$ Differential Equation: $\frac{d^{2} y}{d x^{2}}-2 \frac{d y}{d x}+2 y=0$

Verification: Given function:

y = e^x(a \cos x + b \sin x)

. First derivative,

\frac{dy}{dx}

(using product rule):

\frac{dy}{dx} = e^x(a \cos x + b \sin x) + e^x(-a \sin x + b \cos x)

\frac{dy}{dx} = y + e^x(-a \sin x + b \cos x) \quad ... (1)

Second derivative,

\frac{d^2y}{dx^2}

:

\frac{d^2y}{dx^2} = \frac{dy}{dx} + [e^x(-a \sin x + b \cos x) + e^x(-a \cos x - b \sin x)]

From (1), we have

e^x(-a \sin x + b \cos x) = \frac{dy}{dx} - y

. Substitute this:

\frac{d^2y}{dx^2} = \frac{dy}{dx} + (\frac{dy}{dx} - y) + e^x(-a \cos x - b \sin x)

\frac{d^2y}{dx^2} = 2\frac{dy}{dx} - y - e^x(a \cos x + b \sin x)

From the given function,

e^x(a \cos x + b \sin x) = y

. Substitute this:

\frac{d^2y}{dx^2} = 2\frac{dy}{dx} - y - y

\frac{d^2y}{dx^2} = 2\frac{dy}{dx} - 2y

Rearranging the terms:

\frac{d^{2} y}{d x^{2}}-2 \frac{d y}{d x}+2 y=0

This is the same as the given differential equation. Hence, the given function is a solution of the differential equation.

(iii) Function: $y=x \sin 3 x$ Differential Equation: $\frac{d^{2} y}{d x^{2}}+9 y-6 \cos 3 x=0$

Verification: Given function:

y = x \sin 3x

. First derivative,

\frac{dy}{dx}

(using product rule):

\frac{dy}{dx} = 1 \cdot \sin 3x + x \cdot (\cos 3x \cdot 3)

\frac{dy}{dx} = \sin 3x + 3x \cos 3x

Second derivative,

\frac{d^2y}{dx^2}

:

\frac{d^2y}{dx^2} = \cos 3x \cdot 3 + [3 \cdot \cos 3x + 3x \cdot (-\sin 3x \cdot 3)]

\frac{d^2y}{dx^2} = 3\cos 3x + 3\cos 3x - 9x \sin 3x

\frac{d^2y}{dx^2} = 6\cos 3x - 9(x \sin 3x)

From the given function,

y = x \sin 3x

. Substitute this:

\frac{d^2y}{dx^2} = 6\cos 3x - 9y

Rearranging the terms:

\frac{d^{2} y}{d x^{2}} + 9y - 6\cos 3x = 0

This is the same as the given differential equation. Hence, the given function is a solution of the differential equation.

(iv) Function: $x^{2}=2 y^{2} \log y$ Differential Equation: $\left(x^{2}+y^{2}\right) \frac{d y}{d x}-x y=0$

Verification: Given function:

x^2 = 2y^2 \log y

. Differentiate implicitly with respect to

x

:

2x = 2 \left[ (2y \frac{dy}{dx}) \log y + y^2 (\frac{1}{y} \frac{dy}{dx}) \right]

2x = 2 \left[ 2y \log y \frac{dy}{dx} + y \frac{dy}{dx} \right]

x = (2y \log y + y) \frac{dy}{dx}

x = y(2 \log y + 1) \frac{dy}{dx}

Now we solve for

\frac{dy}{dx}

:

\frac{dy}{dx} = \frac{x}{y(2 \log y + 1)}

From the given function,

x^2 = 2y^2 \log y

, we can write

\log y = \frac{x^2}{2y^2}

. Substitute this into the expression for

\frac{dy}{dx}

:

\frac{dy}{dx} = \frac{x}{y(2(\frac{x^2}{2y^2}) + 1)}

\frac{dy}{dx} = \frac{x}{y(\frac{x^2}{y^2} + 1)}

\frac{dy}{dx} = \frac{x}{y(\frac{x^2+y^2}{y^2})}

\frac{dy}{dx} = \frac{x}{\frac{x^2+y^2}{y}}

\frac{dy}{dx} = \frac{xy}{x^2+y^2}

Now, rearrange this to match the given differential equation:

(x^2+y^2) \frac{dy}{dx} = xy

(x^2+y^2) \frac{dy}{dx} - xy = 0

This is the same as the given differential equation. Hence, the given function is a solution of the differential equation.

Final Answer: In all four cases, the given function is verified to be a solution of the corresponding differential equation.

Q3Miscellaneous Exercise on Chapter 9

Prove that $x^{2}-y^{2}=c (x^{2}+y^{2})^{2}$ is the general solution of differential equation $(x^{3}-3 x y^{2}) d x= (y^{3}-3 x^{2} y) d y$ , where $c$ is a parameter.

Solution

To Prove:

x^{2}-y^{2}=c(x^{2}+y^{2})^{2}

is the general solution of the differential equation

(x^{3}-3 x y^{2}) d x=(y^{3}-3 x^{2} y) d y

.

Proof: The given potential solution is:

x^{2}-y^{2}=c(x^{2}+y^{2})^{2} \quad (*)

We will differentiate this equation with respect to

x

and then eliminate the parameter

c

to see if we arrive at the given differential equation.

Differentiating both sides of

(*)

with respect to

x

:

\frac{d}{dx}(x^2 - y^2) = \frac{d}{dx}[c(x^2+y^2)^2]

2x - 2y \frac{dy}{dx} = c \cdot 2(x^2+y^2) \cdot \frac{d}{dx}(x^2+y^2)

2(x - y \frac{dy}{dx}) = 2c(x^2+y^2) (2x + 2y \frac{dy}{dx})

x - y \frac{dy}{dx} = 2c(x^2+y^2) (x + y \frac{dy}{dx})

From the given solution

(*)

, we can express the parameter

c

as:

c = \frac{x^2-y^2}{(x^2+y^2)^2}

Substitute this expression for

c

back into the differentiated equation:

x - y \frac{dy}{dx} = 2 \left( \frac{x^2-y^2}{(x^2+y^2)^2} \right) (x^2+y^2) (x + y \frac{dy}{dx})

x - y \frac{dy}{dx} = \frac{2(x^2-y^2)}{x^2+y^2} (x + y \frac{dy}{dx})

Multiply both sides by

(x^2+y^2)

to eliminate the fraction:

(x^2+y^2)(x - y \frac{dy}{dx}) = 2(x^2-y^2)(x + y \frac{dy}{dx})

Expand both sides:

x(x^2+y^2) - y(x^2+y^2)\frac{dy}{dx} = 2x(x^2-y^2) + 2y(x^2-y^2)\frac{dy}{dx}

x^3+xy^2 - (x^2y+y^3)\frac{dy}{dx} = 2x^3-2xy^2 + (2x^2y-2y^3)\frac{dy}{dx}

Now, group the terms with

\frac{dy}{dx}

on one side and the other terms on the other side:

x^3+xy^2 - (2x^3-2xy^2) = (x^2y+y^3)\frac{dy}{dx} + (2x^2y-2y^3)\frac{dy}{dx}

-x^3+3xy^2 = (x^2y+y^3 + 2x^2y-2y^3)\frac{dy}{dx}

-(x^3-3xy^2) = (3x^2y-y^3)\frac{dy}{dx}

-(x^3-3xy^2) = -(y^3-3x^2y)\frac{dy}{dx}

(x^3-3xy^2) = (y^3-3x^2y)\frac{dy}{dx}

Rearranging into the differential form:

(x^3-3xy^2)dx = (y^3-3x^2y)dy

This is the given differential equation. Thus, the given relation is indeed the general solution.

Hence Proved.

Q4Miscellaneous Exercise on Chapter 9

Find the general solution of the differential equation $\frac{d y}{d x}+\sqrt{\frac{1-y^{2}}{1-x^{2}}}=0$ .

Solution

Given: The differential equation

\frac{d y}{d x}+\sqrt{\frac{1-y^{2}}{1-x^{2}}}=0

.

To Find: The general solution of the differential equation.

Solution: The given differential equation can be written as:

\frac{dy}{dx} = -\sqrt{\frac{1-y^2}{1-x^2}}

\frac{dy}{dx} = -\frac{\sqrt{1-y^2}}{\sqrt{1-x^2}}

This is a variable-separable differential equation. We can separate the variables

x

and

y

as follows:

\frac{dy}{\sqrt{1-y^2}} = -\frac{dx}{\sqrt{1-x^2}}

Now, we integrate both sides of the equation:

\int \frac{1}{\sqrt{1-y^2}} dy = -\int \frac{1}{\sqrt{1-x^2}} dx

We use the standard integral formula

\int \frac{1}{\sqrt{1-u^2}} du = \sin^{-1}(u)

. Applying this formula to both sides, we get:

\sin^{-1}(y) = -\sin^{-1}(x) + C

where

C

is the constant of integration.

Rearranging the terms, we get the general solution:

\sin^{-1}(x) + \sin^{-1}(y) = C

Final Answer: The general solution of the given differential equation is

\sin^{-1}(x) + \sin^{-1}(y) = C

.

Q5Miscellaneous Exercise on Chapter 9

Show that the general solution of the differential equation $\frac{d y}{d x}+\frac{y^{2}+y+1}{x^{2}+x+1}=0$ is given by $(x+y+1)=\mathrm{A}(1-x-y-2 x y)$ , where A is parameter.

Solution

To Show: The general solution of the differential equation

\frac{d y}{d x}+\frac{y^{2}+y+1}{x^{2}+x+1}=0

is given by

(x+y+1)=\mathrm{A}(1-x-y-2 x y)

.

Proof: We will solve the given differential equation and show that its solution can be expressed in the required form.

The differential equation is:

\frac{dy}{dx} = -\frac{y^2+y+1}{x^2+x+1}

This is a variable-separable equation. Separating the variables, we get:

\frac{dy}{y^2+y+1} = -\frac{dx}{x^2+x+1}

Integrate both sides:

\int \frac{dy}{y^2+y+1} = -\int \frac{dx}{x^2+x+1} + C_0

where

C_0

is the constant of integration.

Let's evaluate the integral

\int \frac{du}{u^2+u+1}

. We complete the square in the denominator:

u^2+u+1 = (u^2+u+\frac{1}{4}) - \frac{1}{4} + 1 = (u+\frac{1}{2})^2 + \frac{3}{4} = (u+\frac{1}{2})^2 + (\frac{\sqrt{3}}{2})^2

So, the integral becomes:

\int \frac{du}{(u+\frac{1}{2})^2 + (\frac{\sqrt{3}}{2})^2}

Using the formula

\int \frac{1}{t^2+a^2}dt = \frac{1}{a}\tan^{-1}(\frac{t}{a})

with

t=u+\frac{1}{2}

and

a=\frac{\sqrt{3}}{2}

:

\int \frac{du}{u^2+u+1} = \frac{1}{\sqrt{3}/2} \tan^{-1}\left(\frac{u+1/2}{\sqrt{3}/2}\right) = \frac{2}{\sqrt{3}} \tan^{-1}\left(\frac{2u+1}{\sqrt{3}}\right)

Applying this result to our integrated differential equation:

\frac{2}{\sqrt{3}} \tan^{-1}\left(\frac{2y+1}{\sqrt{3}}\right) = -\frac{2}{\sqrt{3}} \tan^{-1}\left(\frac{2x+1}{\sqrt{3}}\right) + C_0

Multiplying by

\frac{\sqrt{3}}{2}

:

\tan^{-1}\left(\frac{2y+1}{\sqrt{3}}\right) = -\tan^{-1}\left(\frac{2x+1}{\sqrt{3}}\right) + C_1

where

C_1 = \frac{\sqrt{3}}{2}C_0

.

\tan^{-1}\left(\frac{2y+1}{\sqrt{3}}\right) + \tan^{-1}\left(\frac{2x+1}{\sqrt{3}}\right) = C_1

Using the identity

\tan^{-1}(a) + \tan^{-1}(b) = \tan^{-1}\left(\frac{a+b}{1-ab}\right)

:

\tan^{-1}\left( \frac{\frac{2y+1}{\sqrt{3}} + \frac{2x+1}{\sqrt{3}}}{1 - \frac{(2y+1)(2x+1)}{(\sqrt{3})^2}} \right) = C_1

\tan^{-1}\left( \frac{\frac{2(x+y+1)}{\sqrt{3}}}{1 - \frac{4xy+2x+2y+1}{3}} \right) = C_1

\tan^{-1}\left( \frac{\frac{2(x+y+1)}{\sqrt{3}}}{\frac{3 - (4xy+2x+2y+1)}{3}} \right) = C_1

\tan^{-1}\left( \frac{2(x+y+1)}{\sqrt{3}} \cdot \frac{3}{2-2x-2y-4xy} \right) = C_1

\tan^{-1}\left( \frac{\sqrt{3}(x+y+1)}{1-x-y-2xy} \right) = C_1

Taking the tangent of both sides:

\frac{\sqrt{3}(x+y+1)}{1-x-y-2xy} = \tan(C_1)

Let

\tan(C_1) = K

, where

K

is a new arbitrary constant.

\sqrt{3}(x+y+1) = K(1-x-y-2xy)

(x+y+1) = \frac{K}{\sqrt{3}}(1-x-y-2xy)

Let

\mathrm{A} = \frac{K}{\sqrt{3}}

. Since

K

is an arbitrary constant, so is

\mathrm{A}

.

(x+y+1) = \mathrm{A}(1-x-y-2xy)

This is the required form of the solution.

Hence Shown.

Q6Miscellaneous Exercise on Chapter 9

Find the equation of the curve passing through the point $\left(0, \frac{\pi}{4}\right)$ whose differential equation is $\sin x \cos y d x+\cos x \sin y d y=0$ .

Solution

Given: The differential equation of a curve is

\sin x \cos y d x+\cos x \sin y d y=0

. The curve passes through the point

(0, \frac{\pi}{4})

.

To Find: The equation of the curve.

Solution: The given differential equation is:

\sin x \cos y dx + \cos x \sin y dy = 0

This equation can be solved by separating the variables. We can rearrange the equation to group terms with

x

and terms with

y

. Divide the entire equation by

\cos x \cos y

(assuming

\cos x \neq 0

and

\cos y \neq 0

).

\frac{\sin x \cos y}{\cos x \cos y} dx + \frac{\cos x \sin y}{\cos x \cos y} dy = 0

\frac{\sin x}{\cos x} dx + \frac{\sin y}{\cos y} dy = 0

\tan x dx + \tan y dy = 0

Now, integrate both sides:

\int \tan x dx + \int \tan y dy = C

where

C

is the constant of integration.

We know that

\int \tan u du = \ln|\sec u|

. So, the integrated equation is:

\ln|\sec x| + \ln|\sec y| = C

Using the property of logarithms,

\ln A + \ln B = \ln(AB)

:

\ln|\sec x \sec y| = C

To find the constant

C

, we use the given condition that the curve passes through the point

(x, y) = (0, \frac{\pi}{4})

. Substitute these values into the equation:

\ln|\sec(0) \sec(\frac{\pi}{4})| = C

We have

\sec(0) = 1

and

\sec(\frac{\pi}{4}) = \sqrt{2}

.

C = \ln|1 \cdot \sqrt{2}| = \ln(\sqrt{2})

Now substitute the value of

C

back into the general solution:

\ln|\sec x \sec y| = \ln(\sqrt{2})

Taking the exponent of both sides (or equating the arguments of the logarithm):

|\sec x \sec y| = \sqrt{2}

Since

\sec(0)=1>0

and

\sec(\pi/4)=\sqrt{2}>0

, we can consider the positive case in the neighborhood of the given point.

\sec x \sec y = \sqrt{2}

This can also be written in terms of cosine:

\frac{1}{\cos x \cos y} = \sqrt{2}

\sqrt{2} \cos x \cos y = 1

Final Answer: The equation of the curve is

\sqrt{2} \cos x \cos y = 1

.

Q7Miscellaneous Exercise on Chapter 9

Find the particular solution of the differential equation $\left(1+e^{2 x}\right) d y+\left(1+y^{2}\right) e^{x} d x=0$ , given that $y=1$ when $x=0$ .

Solution

Given: The differential equation

(1+e^{2 x}) d y+\left(1+y^{2}\right) e^{x} d x=0

, with the condition that

y=1

when

x=0

.

To Find: The particular solution of the differential equation.

Solution: The given differential equation is:

(1+e^{2x}) dy + (1+y^2)e^x dx = 0

This is a variable-separable differential equation. We can rearrange the terms to separate variables

x

and

y

.

(1+e^{2x}) dy = -(1+y^2)e^x dx

\frac{dy}{1+y^2} = -\frac{e^x}{1+e^{2x}} dx

Now, we integrate both sides:

\int \frac{1}{1+y^2} dy = -\int \frac{e^x}{1+(e^x)^2} dx

For the left-hand side integral:

\int \frac{1}{1+y^2} dy = \tan^{-1}(y)

For the right-hand side integral, let

t = e^x

. Then

dt = e^x dx

.

-\int \frac{e^x}{1+(e^x)^2} dx = -\int \frac{1}{1+t^2} dt = -\tan^{-1}(t)

Substituting back

t = e^x

, we get:

-\tan^{-1}(e^x)

Combining the results, the general solution is:

\tan^{-1}(y) = -\tan^{-1}(e^x) + C

where

C

is the constant of integration.

\tan^{-1}(y) + \tan^{-1}(e^x) = C

Now, we use the given condition

y=1

when

x=0

to find the value of

C

. Substitute

x=0

and

y=1

into the general solution:

\tan^{-1}(1) + \tan^{-1}(e^0) = C

\tan^{-1}(1) + \tan^{-1}(1) = C

\frac{\pi}{4} + \frac{\pi}{4} = C

C = \frac{\pi}{2}

Substitute the value of

C

back into the general solution to get the particular solution:

\tan^{-1}(y) + \tan^{-1}(e^x) = \frac{\pi}{2}

We can simplify this further. We know that

\tan^{-1}(A) + \tan^{-1}(B) = \frac{\pi}{2}

implies

AB=1

. Or, using the identity

\tan^{-1}(A) + \cot^{-1}(A) = \frac{\pi}{2}

, we can write

\frac{\pi}{2} - \tan^{-1}(e^x) = \cot^{-1}(e^x)

. So,

\tan^{-1}(y) = \cot^{-1}(e^x)

. Let

\theta = \cot^{-1}(e^x)

, then

\cot(\theta) = e^x

. This means

\tan(\theta) = \frac{1}{e^x}

. So,

\tan^{-1}(y) = \tan^{-1}(\frac{1}{e^x})

. Taking the tangent of both sides:

y = \frac{1}{e^x} = e^{-x}

Alternatively, using the formula

\tan^{-1}(A) + \tan^{-1}(B) = \tan^{-1}\left(\frac{A+B}{1-AB}\right)

on the solution

\tan^{-1}(y) + \tan^{-1}(e^x) = \frac{\pi}{2}

:

\tan^{-1}\left(\frac{y+e^x}{1-ye^x}\right) = \frac{\pi}{2}

This implies the denominator must be zero, as

\tan(\frac{\pi}{2})

is undefined.

1 - ye^x = 0

ye^x = 1

y = \frac{1}{e^x} = e^{-x}

Final Answer: The particular solution of the differential equation is

y=e^{-x}

.

Q8Miscellaneous Exercise on Chapter 9

Solve the differential equation $y e^{\frac{x}{y}} d x=\left(x e^{\frac{x}{y}}+y^{2}\right) d y(y \neq 0)$ .

Solution

Given: The differential equation is

y e^{\frac{x}{y}} d x=\left(x e^{\frac{x}{y}}+y^{2}\right) d y

, with

y \neq 0

.

To Find: The general solution of the differential equation.

Solution: The given differential equation is:

y e^{\frac{x}{y}} d x=\left(x e^{\frac{x}{y}}+y^{2}\right) d y

Rearranging the terms, we get:

y e^{\frac{x}{y}} d x - x e^{\frac{x}{y}} d y = y^{2} d y

e^{\frac{x}{y}} (y dx - x dy) = y^2 dy

Since

y \neq 0

, we can divide the entire equation by

y^2

:

e^{\frac{x}{y}} \frac{y dx - x dy}{y^2} = dy

We know that the differential of the quotient

\frac{x}{y}

is given by:

d\left(\frac{x}{y}\right) = \frac{y dx - x d(y)}{y^2} = \frac{y dx - x dy}{y^2}

Substituting this into our equation, we get:

e^{\frac{x}{y}} d\left(\frac{x}{y}\right) = dy

Let

v = \frac{x}{y}

. The equation becomes:

e^v dv = dy

This is a variable separable form. Integrating both sides:

\int e^v dv = \int dy

e^v = y + C

Now, substitute back the value of

v = \frac{x}{y}

:

e^{\frac{x}{y}} = y + C

Final Answer: The general solution of the given differential equation is

e^{\frac{x}{y}} = y + C

.

Q9Miscellaneous Exercise on Chapter 9

Find a particular solution of the differential equation $(x-y)(d x+d y)=d x-d y$ , given that $y=-1$ , when $x=0$ . (Hint: put $x-y=t$ )

Solution

Given: The differential equation is

(x-y)(d x+d y)=d x-d y

, with the initial condition

y=-1

when

x=0

.

To Find: The particular solution of the differential equation.

Solution: The given differential equation is:

(x-y)(d x+d y)=d x-d y

We are given the hint to substitute

t = x-y

. Differentiating this substitution, we get:

dt = dx - dy

Also, from

t=x-y

, we have

y = x-t

, so

dy = dx - dt

. Then,

dx+dy = dx + (dx - dt) = 2dx - dt

.

Now, substitute

x-y=t

,

dx-dy=dt

, and

dx+dy=2dx-dt

into the original equation:

t(2dx - dt) = dt

2t dx - t dt = dt

2t dx = dt + t dt

2t dx = (1+t) dt

This is a variable separable equation. Rearranging to separate the variables:

dx = \frac{1+t}{2t} dt

dx = \left(\frac{1}{2t} + \frac{t}{2t}\right) dt

dx = \left(\frac{1}{2t} + \frac{1}{2}\right) dt

Integrating both sides:

\int dx = \int \left(\frac{1}{2t} + \frac{1}{2}\right) dt

x = \frac{1}{2} \ln|t| + \frac{1}{2} t + C

Substitute back

t = x-y

:

x = \frac{1}{2} \ln|x-y| + \frac{1}{2} (x-y) + C

Multiplying by 2 to simplify:

2x = \ln|x-y| + x - y + 2C

x + y = \ln|x-y| + C_1

where

C_1 = 2C

is another constant. This is the general solution.

To find the particular solution, we use the given condition:

y=-1

when

x=0

. Substitute these values into the general solution:

0 + (-1) = \ln|0 - (-1)| + C_1

-1 = \ln|1| + C_1

-1 = 0 + C_1

C_1 = -1

Substitute

C_1 = -1

back into the general solution to get the particular solution:

x+y = \ln|x-y| - 1

This can also be written as

x+y+1 = \ln|x-y|

.

Final Answer: The particular solution of the given differential equation is

x+y+1 = \ln|x-y|

.

Q10Miscellaneous Exercise on Chapter 9

Solve the differential equation $\left[\frac{e^{-2 \sqrt{x}}}{\sqrt{x}}-\frac{y}{\sqrt{x}}\right] \frac{d x}{d y}=1(x \neq 0)$ .

Solution

Given: The differential equation is

\left[\frac{e^{-2 \sqrt{x}}}{\sqrt{x}}-\frac{y}{\sqrt{x}}\right] \frac{d x}{d y}=1

, with

x \neq 0

.

To Find: The general solution of the differential equation.

Solution: The given differential equation is:

\left[\frac{e^{-2 \sqrt{x}}}{\sqrt{x}}-\frac{y}{\sqrt{x}}\right] \frac{d x}{d y}=1

We can rewrite this equation as:

\frac{dy}{dx} = \frac{e^{-2 \sqrt{x}}}{\sqrt{x}}-\frac{y}{\sqrt{x}}

Rearranging the terms to match the standard form of a linear differential equation,

\frac{dy}{dx} + P(x)y = Q(x)

:

\frac{dy}{dx} + \frac{1}{\sqrt{x}} y = \frac{e^{-2 \sqrt{x}}}{\sqrt{x}}

This is a linear differential equation where:

P(x) = \frac{1}{\sqrt{x}}

Q(x) = \frac{e^{-2 \sqrt{x}}}{\sqrt{x}}

First, we calculate the integrating factor (I.F.):

\text{I.F.} = e^{\int P(x) dx} = e^{\int \frac{1}{\sqrt{x}} dx}

\int \frac{1}{\sqrt{x}} dx = \int x^{-1/2} dx = \frac{x^{1/2}}{1/2} = 2\sqrt{x}

So, the integrating factor is:

\text{I.F.} = e^{2\sqrt{x}}

The general solution of a linear differential equation is given by the formula:

y \times (\text{I.F.}) = \int Q(x) \times (\text{I.F.}) dx + C

Substituting the values of I.F. and

Q(x)

:

y \cdot e^{2\sqrt{x}} = \int \left(\frac{e^{-2 \sqrt{x}}}{\sqrt{x}}\right) \cdot e^{2\sqrt{x}} dx + C

y \cdot e^{2\sqrt{x}} = \int \frac{e^{-2\sqrt{x} + 2\sqrt{x}}}{\sqrt{x}} dx + C

y \cdot e^{2\sqrt{x}} = \int \frac{e^0}{\sqrt{x}} dx + C

y \cdot e^{2\sqrt{x}} = \int \frac{1}{\sqrt{x}} dx + C

y \cdot e^{2\sqrt{x}} = 2\sqrt{x} + C

This is the general solution of the differential equation.

Final Answer: The general solution of the given differential equation is

y e^{2\sqrt{x}} = 2\sqrt{x} + C

.

Q11Miscellaneous Exercise on Chapter 9

Find a particular solution of the differential equation $\frac{d y}{d x}+y \cot x=4 x \operatorname{cosec} x (x \neq 0)$ , given that $y=0$ when $x=\frac{\pi}{2}$ .

Solution

Given: The differential equation is

\frac{d y}{d x}+y \cot x=4 x \operatorname{cosec} x

, with

x \neq 0

, and the initial condition

y=0

when

x=\frac{\pi}{2}

.

To Find: The particular solution of the differential equation.

Solution: The given differential equation is:

\frac{d y}{d x}+y \cot x=4 x \csc x

This equation is in the standard linear differential equation form

\frac{dy}{dx} + P(x)y = Q(x)

, where:

P(x) = \cot x

Q(x) = 4x \csc x

First, we find the integrating factor (I.F.):

\text{I.F.} = e^{\int P(x) dx} = e^{\int \cot x dx}

Since

\int \cot x dx = \ln|\sin x|

, the integrating factor is:

\text{I.F.} = e^{\ln|\sin x|} = |\sin x|

Since the initial condition is given at

x = \frac{\pi}{2}

, which is in the interval

(0, \pi)

where

\sin x > 0

, we can take the integrating factor as

\sin x

.

The general solution is given by:

y \times (\text{I.F.}) = \int Q(x) \times (\text{I.F.}) dx + C

y \sin x = \int (4x \csc x) (\sin x) dx + C

y \sin x = \int 4x \left(\frac{1}{\sin x}\right) (\sin x) dx + C

y \sin x = \int 4x dx + C

y \sin x = 4 \frac{x^2}{2} + C

y \sin x = 2x^2 + C

This is the general solution.

To find the particular solution, we use the given condition:

y=0

when

x=\frac{\pi}{2}

. Substitute these values into the general solution:

0 \cdot \sin\left(\frac{\pi}{2}\right) = 2\left(\frac{\pi}{2}\right)^2 + C

0 \cdot 1 = 2\left(\frac{\pi^2}{4}\right) + C

0 = \frac{\pi^2}{2} + C

C = -\frac{\pi^2}{2}

Now, substitute the value of

C

back into the general solution:

y \sin x = 2x^2 - \frac{\pi^2}{2}

Final Answer: The particular solution of the given differential equation is

y \sin x = 2x^2 - \frac{\pi^2}{2}

.

Q12Miscellaneous Exercise on Chapter 9

Find a particular solution of the differential equation $(x+1) \frac{d y}{d x}=2 e^{-y}-1$ , given that $y=0$ when $x=0$ .

Solution

Given: The differential equation is

(x+1) \frac{d y}{d x}=2 e^{-y}-1

, with the initial condition

y=0

when

x=0

.

To Find: The particular solution of the differential equation.

Solution: The given differential equation is:

(x+1) \frac{dy}{dx} = 2e^{-y} - 1

This equation can be solved using the method of separation of variables. First, rewrite the equation to group terms with

y

and terms with

x

on opposite sides.

(x+1) \frac{dy}{dx} = \frac{2}{e^y} - 1 = \frac{2-e^y}{e^y}

Now, separate the variables by moving all

y

terms to the left side and all

x

terms to the right side:

\frac{e^y}{2-e^y} dy = \frac{1}{x+1} dx

To integrate the left side, we can use a substitution. Let

u = 2-e^y

. Then

du = -e^y dy

, which means

e^y dy = -du

. The integral becomes:

\int \frac{e^y}{2-e^y} dy = \int \frac{-du}{u} = -\ln|u| = -\ln|2-e^y|

So, integrating both sides of the separated equation:

\int \frac{e^y}{2-e^y} dy = \int \frac{1}{x+1} dx

-\ln|2-e^y| = \ln|x+1| + C

where

C

is the constant of integration.

To find the particular solution, we use the given condition:

y=0

when

x=0

. Substitute these values into the general solution:

-\ln|2-e^0| = \ln|0+1| + C

-\ln|2-1| = \ln|1| + C

-\ln(1) = 0 + C

0 = C

So, the constant

C

is 0.

Substitute

C=0

back into the general solution:

-\ln|2-e^y| = \ln|x+1|

\ln\left(\frac{1}{|2-e^y|}\right) = \ln|x+1|

Equating the arguments of the logarithm:

\frac{1}{|2-e^y|} = |x+1|

|x+1| \cdot |2-e^y| = 1

|(x+1)(2-e^y)| = 1

This implies either

(x+1)(2-e^y) = 1

or

(x+1)(2-e^y) = -1

. Let's check the condition

x=0, y=0

:

(0+1)(2-e^0) = 1(2-1)=1

. This satisfies the first case. So, the particular solution is

(x+1)(2-e^y) = 1

. We can also express

y

in terms of

x

:

2-e^y = \frac{1}{x+1}

e^y = 2 - \frac{1}{x+1} = \frac{2(x+1)-1}{x+1} = \frac{2x+1}{x+1}

y = \ln\left|\frac{2x+1}{x+1}\right|

Final Answer: The particular solution of the given differential equation is

y = \ln\left|\frac{2x+1}{x+1}\right|

.

Q13Miscellaneous Exercise on Chapter 9

The general solution of the differential equation $\frac{y d x-x d y}{y}=0$ is

(A)

x y=\mathrm{C}

(B)

x=\mathrm{C} y^{2}

(C)

y=\mathrm{C} x

(D)

y=\mathrm{C} x^{2}

Solution

Given: The differential equation is

\frac{y d x-x d y}{y}=0

.

To Find: The general solution of the given differential equation.

Solution: The given differential equation is:

\frac{y d x-x d y}{y}=0

Assuming

y \neq 0

, we can simplify the equation to:

y dx - x dy = 0

y dx = x dy

This is a differential equation with separable variables. Separating the variables, we get:

\frac{dx}{x} = \frac{dy}{y}

Now, we integrate both sides:

\int \frac{1}{x} dx = \int \frac{1}{y} dy

\ln|x| = \ln|y| + C_1

where

C_1

is the constant of integration. We can write

C_1

as

\ln|C|

for convenience.

\ln|x| = \ln|y| + \ln|C|

Using the property of logarithms,

\ln a + \ln b = \ln(ab)

:

\ln|x| = \ln|Cy|

Removing the logarithm from both sides:

|x| = |Cy|

x = \pm Cy

Let a new constant

k = \pm C

. Then:

x = ky

This can be rewritten as

y = \frac{1}{k} x

. Let's denote the constant

\frac{1}{k}

as

C_{new}

.

y = C_{new}x

This is of the form

y = Cx

.

Comparing this with the given options: (A)

xy=\mathrm{C}

(B)

x=\mathrm{C} y^{2}

(C)

y=\mathrm{C} x

(D)

y=\mathrm{C} x^{2}

The derived solution matches option (C).

Final Answer: The correct option is (C), which is

y=\mathrm{C} x

.

Q14Miscellaneous Exercise on Chapter 9

The general solution of a differential equation of the type $\frac{d x}{d y}+P_{1} x=Q_{1}$ is

(A)

y e^{\int P_{1} d y}=\int\left(Q_{1} e^{\int P_{1} d y}\right) d y+C

(B)

y \cdot e^{\int P_{1} d x}=\int\left(Q_{1} e^{\int P_{1} d x}\right) d x+C

(C)

x e^{\int P_{1} d y}=\int\left(Q_{1} e^{\int P_{1} d y}\right) d y+C

(D)

x e^{\int P_{1} d x}=\int\left(Q_{1} e^{\int P_{1} d x}\right) d x+C

Solution

Given: A differential equation of the type

\frac{d x}{d y}+P_{1} x=Q_{1}

.

To Find: The general solution for this type of differential equation.

Solution: The given equation is a linear differential equation of the first order, with

x

as the dependent variable and

y

as the independent variable. The standard form is:

\frac{dx}{dy} + P(y)x = Q(y)

In this case,

P_1

and

Q_1

are functions of

y

(or constants).

To solve this type of equation, we first find the integrating factor (I.F.). The formula for the integrating factor is:

\text{I.F.} = e^{\int P_1 dy}

Next, the general solution of the linear differential equation is given by the formula:

x \times (\text{I.F.}) = \int (Q_1 \times (\text{I.F.})) dy + C

Substituting the expression for the integrating factor, we get:

x e^{\int P_{1} d y}=\int\left(Q_{1} e^{\int P_{1} d y}\right) d y+C

Now, let's compare this result with the given options: (A)

y e^{\int P_{1} d y}=\int\left(Q_{1} e^{\int P_{1} d y}\right) d y+C

(Incorrect, the term outside the integral should be

x \times \text{I.F.}

) (B)

y \cdot e^{\int P_{1} d x}=\int\left(Q_{1} e^{\int P_{1} d x}\right) d x+C

(Incorrect, the integration should be with respect to

dy

) (C)

x e^{\int P_{1} d y}=\int\left(Q_{1} e^{\int P_{1} d y}\right) d y+C

(Correct) (D)

x e^{\int P_{1} d x}=\int\left(Q_{1} e^{\int P_{1} d x}\right) d x+C

(Incorrect, the integration should be with respect to

dy

)

The derived formula for the general solution matches option (C).

Final Answer: The correct option is (C), which is

x e^{\int P_{1} d y}=\int\left(Q_{1} e^{\int P_{1} d y}\right) d y+C

.

Q15Miscellaneous Exercise on Chapter 9

The general solution of the differential equation $e^{x} d y+\left(y e^{x}+2 x\right) d x=0$ is

(A)

x e^{y}+x^{2}=\mathrm{C}

(B)

x e^{y}+y^{2}=\mathrm{C}

(C)

y e^{x}+x^{2}=\mathrm{C}

(D)

y e^{y}+x^{2}=\mathrm{C}

Solution

Given: The differential equation

e^{x} d y+\left(y e^{x}+2 x\right) d x=0

.

To Find: The general solution of the given differential equation.

Solution: The given differential equation is:

e^{x} d y+\left(y e^{x}+2 x\right) d x=0

We can rearrange this equation to identify its form. Dividing by

dx

and rearranging gives:

e^{x} \frac{dy}{dx} + y e^{x} + 2x = 0

Dividing the entire equation by

e^x

(which is never zero):

\frac{dy}{dx} + y + \frac{2x}{e^x} = 0

\frac{dy}{dx} + y = -2x e^{-x}

This is a linear differential equation of the first order in the form

\frac{dy}{dx} + Py = Q

, where:

P = 1

Q = -2x e^{-x}

The first step is to find the integrating factor (I.F.):

\text{I.F.} = e^{\int P dx} = e^{\int 1 dx} = e^x

The general solution is given by the formula:

y \times (\text{I.F.}) = \int (Q \times (\text{I.F.})) dx + C

Substituting the values of

Q

and I.F.:

y \cdot e^x = \int (-2x e^{-x} \cdot e^x) dx + C

y e^x = \int -2x dx + C

Now, we perform the integration:

y e^x = -2 \int x dx + C

y e^x = -2 \left(\frac{x^2}{2}\right) + C

y e^x = -x^2 + C

Rearranging the terms to match the options:

y e^x + x^2 = C

Comparing this with the given options: (A)

x e^{y}+x^{2}=\mathrm{C}

(B)

x e^{y}+y^{2}=\mathrm{C}

(C)

y e^{x}+x^{2}=\mathrm{C}

(D)

y e^{y}+x^{2}=\mathrm{C}

The derived solution matches option (C).

Final Answer: The correct option is (C), which is

y e^{x}+x^{2}=\mathrm{C}

.

Differential EquationsClass 12 Mathematics NCERT Solutions

rac{d^{4} y}{d x^{4}}+\sin \left(y^{\prime \prime \prime}\right)=0

Solution

y′+5y=0y^{\prime}+5 y=0y′+5y=0

Solution

\left( rac{d s}{d t}\right)^{4}+3 s \frac{d^{2} s}{d t^{2}}=0

Solution

\left( rac{d^{2} y}{d x^{2}}\right)^{2}+\cos \left( rac{d y}{d x}\right)=0

Solution

d2ydx2=cos⁡3x+sin⁡3x\frac{d^{2} y}{d x^{2}}=\cos 3 x+\sin 3 xdx2d2y​=cos3x+sin3x

Solution

(y′′′)2+(y′′)3+(y′)4+y5=0\left(y^{\prime \prime \prime}\right)^{2}+\left(y^{\prime \prime}\right)^{3}+\left(y^{\prime}\right)^{4}+y^{5}=0(y′′′)2+(y′′)3+(y′)4+y5=0

Solution

y′′′+2y′′+y′=0y^{\prime \prime \prime}+2 y^{\prime \prime}+y^{\prime}=0y′′′+2y′′+y′=0

Solution

y′+y=exy^{\prime}+y=e^{x}y′+y=ex

Solution

y′′+(y′)2+2y=0y^{\prime \prime}+\left(y^{\prime}\right)^{2}+2 y=0y′′+(y′)2+2y=0

Solution

y′′+2y′+sin⁡y=0y^{\prime \prime}+2 y^{\prime}+\sin y=0y′′+2y′+siny=0

Solution

The degree of the differential equation \left( rac{d^{2} y}{d x^{2}}\right)^{3}+\left( rac{d y}{d x}\right)^{2}+\sin \left( rac{d y}{d x}\right)+1=0 is

Solution

The order of the differential equation 2x2d2ydx2−3dydx+y=02 x^{2} \frac{d^{2} y}{d x^{2}}-3 \frac{d y}{d x}+y=02x2dx2d2y​−3dxdy​+y=0 is

Solution

y=ex+1y=e^{x}+1y=ex+1 : y′′−y′=0y''-y'=0y′′−y′=0

Solution

y=x2+2x+Cy=x^{2}+2 x+\mathrm{C}y=x2+2x+C : y′−2x−2=0y'-2 x-2=0y′−2x−2=0

Solution

y=cos⁡x+Cy=\cos x+\mathrm{C}y=cosx+C : y′+sin⁡x=0\quad y'+\sin x=0y′+sinx=0

Solution

y=1+x2y=\sqrt{1+x^{2}}y=1+x2​ : y′=xy1+x2y'=\frac{x y}{1+x^{2}}y′=1+x2xy​

Solution

y=Axy=\mathrm{A} xy=Ax : xy′=y(x≠0)\quad x y'=y(x \neq 0)xy′=y(x=0)

Solution

y=xsin⁡x:xy′=y+xx2−y2(x≠0y=x \sin x : \quad x y'=y+x \sqrt{x^{2}-y^{2}}(x \neq 0y=xsinx:xy′=y+xx2−y2​(x=0 and x>yx>yx>y or x<−y)x<-y)x<−y)

Solution

: y′=y21−xy(xy≠1)y'=\frac{y^{2}}{1-x y}(x y \neq 1)y′=1−xyy2​(xy=1)

Solution

:(ysin⁡y+cos⁡y+x)y′=y:(y \sin y+\cos y+x) y'=y:(ysiny+cosy+x)y′=y

Solution

x+y=tan⁡−1yx+y=\tan ^{-1} yx+y=tan−1y : y2y′+y2+1=0y^{2} y'+y^{2}+1=0y2y′+y2+1=0

Solution

y=a2−x2x∈(−a,a):x+ydydx=0(y≠0)y=\sqrt{a^{2}-x^{2}} x \in(-a, a): \quad x+y \frac{d y}{d x}=0(y \neq 0)y=a2−x2​x∈(−a,a):x+ydxdy​=0(y=0)

Solution

The number of arbitrary constants in the general solution of a differential equation of fourth order are:

Solution

The number of arbitrary constants in the particular solution of a differential equation of third order are:

Solution

dydx=1−cos⁡x1+cos⁡x\frac{d y}{d x}=\frac{1-\cos x}{1+\cos x}dxdy​=1+cosx1−cosx​

Solution

dydx=4−y2(−2<y<2)\frac{d y}{d x}=\sqrt{4-y^{2}}\quad(-2<y<2)dxdy​=4−y2​(−2<y<2)

Solution

dydx+y=1(y≠1)\frac{d y}{d x}+y=1(y \neq 1)dxdy​+y=1(y=1)

Solution

sec⁡2xtan⁡ydx+sec⁡2ytan⁡xdy=0\sec ^{2} x \tan y d x+\sec ^{2} y \tan x d y=0sec2xtanydx+sec2ytanxdy=0

Solution

(ex+e−x)dy−(ex−e−x)dx=0(e^{x}+e^{-x}) d y-(e^{x}-e^{-x}) d x=0(ex+e−x)dy−(ex−e−x)dx=0

Solution

dydx=(1+x2)(1+y2)\frac{d y}{d x}=\left(1+x^{2}\right)\left(1+y^{2}\right)dxdy​=(1+x2)(1+y2)

Solution

ylog⁡ydx−xdy=0y \log y d x-x d y=0ylogydx−xdy=0

Solution

x5dydx=−y5x^{5} \frac{d y}{d x}=-y^{5}x5dxdy​=−y5

Solution

dydx=sin⁡−1x\frac{d y}{d x}=\sin ^{-1} xdxdy​=sin−1x

Solution

extan⁡ydx+(1−ex)sec⁡2ydy=0e^{x} \tan y d x+\left(1-e^{x}\right) \sec ^{2} y d y=0extanydx+(1−ex)sec2ydy=0

Solution

(x3+x2+x+1)dydx=2x2+x;y=1(x^{3}+x^{2}+x+1) \frac{d y}{d x}=2 x^{2}+x ; y=1(x3+x2+x+1)dxdy​=2x2+x;y=1 when x=0x=0x=0

Solution

x(x2−1)dydx=1;y=0x(x^{2}-1) \frac{d y}{d x}=1 ; y=0x(x2−1)dxdy​=1;y=0 when x=2x=2x=2

Solution

cos⁡(dydx)=a(a∈R);y=1\cos \left(\frac{d y}{d x}\right)=a(a \in \mathbf{R}) ; y=1cos(dxdy​)=a(a∈R);y=1 when x=0x=0x=0

Solution

dydx=ytan⁡x;y=1\frac{d y}{d x}=y \tan x ; y=1dxdy​=ytanx;y=1 when x=0x=0x=0

Solution

Find the equation of a curve passing through the point (0,0)(0,0)(0,0) and whose differential equation is y′=exsin⁡xy^{\prime}=e^{x} \sin xy′=exsinx.

Solution

For the differential equation xydydx=(x+2)(y+2)x y \frac{d y}{d x}=(x+2)(y+2)xydxdy​=(x+2)(y+2), find the solution curve passing through the point (1,−1)(1,-1)(1,−1).

$rac{d^{4} y}{d x^{4}}+\sin \left(y^{\prime \prime \prime}\right)=0$

$y^{\prime}+5 y=0$

$\left( rac{d s}{d t}\right)^{4}+3 s \frac{d^{2} s}{d t^{2}}=0$

$\left( rac{d^{2} y}{d x^{2}}\right)^{2}+\cos \left( rac{d y}{d x}\right)=0$

$\frac{d^{2} y}{d x^{2}}=\cos 3 x+\sin 3 x$

$\left(y^{\prime \prime \prime}\right)^{2}+\left(y^{\prime \prime}\right)^{3}+\left(y^{\prime}\right)^{4}+y^{5}=0$

$y^{\prime \prime \prime}+2 y^{\prime \prime}+y^{\prime}=0$

$y^{\prime}+y=e^{x}$

$y^{\prime \prime}+\left(y^{\prime}\right)^{2}+2 y=0$

$y^{\prime \prime}+2 y^{\prime}+\sin y=0$

The degree of the differential equation $\left( rac{d^{2} y}{d x^{2}}\right)^{3}+\left( rac{d y}{d x}\right)^{2}+\sin \left( rac{d y}{d x}\right)+1=0$ is

The order of the differential equation $2 x^{2} \frac{d^{2} y}{d x^{2}}-3 \frac{d y}{d x}+y=0$ is

$y=e^{x}+1$ : $y''-y'=0$

$y=x^{2}+2 x+\mathrm{C}$ : $y'-2 x-2=0$

$y=\cos x+\mathrm{C}$ : $\quad y'+\sin x=0$

$y=\sqrt{1+x^{2}}$ : $y'=\frac{x y}{1+x^{2}}$

$y=\mathrm{A} x$ : $\quad x y'=y(x \neq 0)$

$y=x \sin x : \quad x y'=y+x \sqrt{x^{2}-y^{2}}(x \neq 0$ and $x>y$ or $x<-y)$

: $y'=\frac{y^{2}}{1-x y}(x y \neq 1)$

$:(y \sin y+\cos y+x) y'=y$

$x+y=\tan ^{-1} y$ : $y^{2} y'+y^{2}+1=0$

$y=\sqrt{a^{2}-x^{2}} x \in(-a, a): \quad x+y \frac{d y}{d x}=0(y \neq 0)$

$\frac{d y}{d x}=\frac{1-\cos x}{1+\cos x}$

$\frac{d y}{d x}=\sqrt{4-y^{2}}\quad(-2<y<2)$

$\frac{d y}{d x}+y=1(y \neq 1)$

$\sec ^{2} x \tan y d x+\sec ^{2} y \tan x d y=0$

$(e^{x}+e^{-x}) d y-(e^{x}-e^{-x}) d x=0$

$\frac{d y}{d x}=\left(1+x^{2}\right)\left(1+y^{2}\right)$

$y \log y d x-x d y=0$

$x^{5} \frac{d y}{d x}=-y^{5}$

$\frac{d y}{d x}=\sin ^{-1} x$

$e^{x} \tan y d x+\left(1-e^{x}\right) \sec ^{2} y d y=0$

$(x^{3}+x^{2}+x+1) \frac{d y}{d x}=2 x^{2}+x ; y=1$ when $x=0$

$x(x^{2}-1) \frac{d y}{d x}=1 ; y=0$ when $x=2$

$\cos \left(\frac{d y}{d x}\right)=a(a \in \mathbf{R}) ; y=1$ when $x=0$

$\frac{d y}{d x}=y \tan x ; y=1$ when $x=0$

Find the equation of a curve passing through the point $(0,0)$ and whose differential equation is $y^{\prime}=e^{x} \sin x$ .

For the differential equation $x y \frac{d y}{d x}=(x+2)(y+2)$ , find the solution curve passing through the point $(1,-1)$ .

Find the equation of a curve passing through the point $(0,-2)$ given that at any point $(x, y)$ on the curve, the product of the slope of its tangent and $y$ coordinate of the point is equal to the $x$ coordinate of the point.

At any point $(x, y)$ of a curve, the slope of the tangent is twice the slope of the line segment joining the point of contact to the point $(-4,-3)$ . Find the equation of the curve given that it passes through $(-2,1)$ .

The volume of spherical balloon being inflated changes at a constant rate. If initially its radius is 3 units and after 3 seconds it is 6 units. Find the radius of balloon after $t$ seconds.

In a bank, principal increases continuously at the rate of $r\%$ per year. Find the value of $r$ if Rs 100 double itself in 10 years ( $\log _{e} 2=0.6931$ ).

In a bank, principal increases continuously at the rate of $5\%$ per year. An amount of Rs 1000 is deposited with this bank, how much will it worth after 10 years $\left(e^{0.5}=1.648\right)$ .

In a culture, the bacteria count is $1,00,000$ . The number is increased by $10\%$ in 2 hours. In how many hours will the count reach $2,00,000$ , if the rate of growth of bacteria is proportional to the number present?

The general solution of the differential equation $\frac{d y}{d x}=e^{x+y}$ is

$(x^{2}+x y) d y=\left(x^{2}+y^{2}\right) d x$

$y'=\frac{x+y}{x}$

$(x-y) d y-(x+y) d x=0$

$\left(x^{2}-y^{2}\right) d x+2 x y d y=0$

$x^{2} \frac{d y}{d x}=x^{2}-2 y^{2}+x y$

$x d y-y d x=\sqrt{x^{2}+y^{2}} d x$

$\left\{x \cos \left(\frac{y}{x}\right)+y \sin \left(\frac{y}{x}\right)\right\} y d x=\left\{y \sin \left(\frac{y}{x}\right)-x \cos \left(\frac{y}{x}\right)\right\} x d y$

$x \frac{d y}{d x}-y+x \sin \left(\frac{y}{x}\right)=0$

$y d x+x \log \left(\frac{y}{x}\right) d y-2 x d y=0$

$\left(1+e^{\frac{x}{y}}\right) d x+e^{\frac{x}{y}}\left(1-\frac{x}{y}\right) d y=0$

$(x+y) d y+(x-y) d x=0 ; y=1 \text { when } x=1$

$x^{2} d y+\left(x y+y^{2}\right) d x=0 ; y=1 \text { when } x=1$

$\left[x \sin ^{2}\left(\frac{y}{x}\right)-y\right] d x+x d y=0 ; y=\frac{\pi}{4} \text { when } x=1$

$\frac{d y}{d x}-\frac{y}{x}+\operatorname{cosec}\left(\frac{y}{x}\right)=0 ; y=0 \text { when } x=1$

$2 x y+y^{2}-2 x^{2} \frac{d y}{d x}=0 ; y=2 \text { when } x=1$

A homogeneous differential equation of the from $\frac{d x}{d y}=h\left(\frac{x}{y}\right)$ can be solved by making the substitution.

$\frac{d y}{d x}+2 y=\sin x$

$\frac{d y}{d x}+3 y=e^{-2 x}$

$\frac{d y}{d x}+\frac{y}{x}=x^{2}$

$\frac{d y}{d x}+(\sec x) y=\tan x\left(0 \leq x<\frac{\pi}{2}\right)$

$\cos ^{2} x \frac{d y}{d x}+y=\tan x\left(0 \leq x<\frac{\pi}{2}\right)$

$x \frac{d y}{d x}+2 y=x^{2} \log x$

$x \log x \frac{d y}{d x}+y=\frac{2}{x} \log x$

$\left(1+x^{2}\right) d y+2 x y d x=\cot x d x(x \neq 0)$

$x \frac{d y}{d x}+y-x+x y \cot x=0(x \neq 0)$

$(x+y) \frac{d y}{d x}=1$

$y d x+\left(x-y^{2}\right) d y=0$