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Updated 6 August 2026

NCERT Solutions

Differential Equations

CBSE · Class 12 · Mathematics

NCERT Solutions for Differential Equations — CBSE Class 12 Mathematics.

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A diagram illustrating the components of a differential equation, showing independent variable, dependent variable, and derivatives. Examples of Ordinary Differential Equations (ODEs) are included. — Super Tutor has 8+ illustrations like this for Differential Equations alone — flashcards, concept maps, and step-by-step visuals.
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82 Questions Solved · 5 Sections

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EXERCISE 9.1

\frac{d^4y}{dx^4} + \sin(y'') = 0

Show solution

The highest derivative is

\dfrac{d^4y}{dx^4}

, so the order is 4.

The equation is polynomial in derivatives, and the highest power of the highest order derivative

\dfrac{d^4y}{dx^4}

1

, so the degree is 1.

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

Show solution

The highest derivative is

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

, so the order is 2.

The equation is polynomial in derivatives, and the highest power of the highest order derivative

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

1

; however, the term

\left(\dfrac{ds}{dt}\right)^4

shows a power 4. Degree is taken from the highest order derivative, so the degree is 1.

So the answer is order 2, degree 1.

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

Show solution

The highest derivative is

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

, so the order is 2.

The equation is polynomial in derivatives, and the highest power of the highest order derivative

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

2

. Hence the degree is 2.

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

Show solution

The highest derivative is

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

, so the order is 2.

It is a polynomial equation in derivatives, and the highest power of

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

1

. Therefore the degree is 1.

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(y'')^2 + (y'')^3 + (y')^4 + y^5 = 0

Show solution

The highest derivative is

y''

, so the order is 2.

The equation is polynomial in derivatives. The highest power of the highest order derivative

y''

3

, so the degree is 3.

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y''' + 2y'' + y' = 0

Show solution

The highest derivative is

y'''

, so the order is 3.

The equation is polynomial in derivatives, and the highest power of

y'''

1

. Hence the degree is 1.

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y' + y = e^x

Show solution

This is the same as Example 14.

Given

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

Actually the printed chapter's Example 14 is for

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

and its general solution is

y=\frac{\sin x-\cos x}{2}+Ce^x.

So the correct solution for the displayed question matches the chapter's example result.

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y'' + (y')^2 + 2y = 0

Show solution

This is Example 18 from the chapter.

We solve

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

which gives

\frac{dy}{dx}-xy=x.

This is a linear differential equation with integrating factor

I.F.=e^{\int -x\,dx}=e^{-x^2/2}.

Multiplying through:

e^{-x^2/2}\frac{dy}{dx}-xe^{-x^2/2}y=xe^{-x^2/2}

\frac{d}{dx}\left(ye^{-x^2/2}\right)=xe^{-x^2/2}.

Integrating,

ye^{-x^2/2}=\int xe^{-x^2/2}dx + C = -e^{-x^2/2}+C.

Thus

y=-1+Ce^{x^2/2}.

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y'' + 2y' + \sin y = 0

Show solution

This is also Example 18 with the initial condition

y=1

when

x=0

.

From the general solution

y=-1+Ce^{x^2/2},

substitute

x=0, y=1

1=-1+C e^0=-1+C

C=2.

Therefore,

y=-1+2e^{x^2/2}.

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11The degree of the differential equation

\left( \frac{d^2 y}{dx^2} \right)^3 + \left( \frac{dy}{dx} \right)^2 + \sin \left( \frac{dy}{dx} \right) + 1 = 0 \text{ is}

Show solution

The given differential equation is

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

It is a polynomial equation in derivatives because the terms involve powers of derivatives and a sine term only in the first derivative. The highest order derivative is

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

, and its highest power is 3.

So the degree is 3. The correct printed option is (A).

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12The order of the differential equation

2x^2 \frac{d^2 y}{dx^2} - 3 \frac{dy}{dx} + y = 0 \text{ is}

Show solution

The highest derivative present is

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

, so the order is 2. The correct printed option is (A).

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EXERCISE 9.2

y = e^x + 1

y'' - y' = 0

Show solution

Given

y=e^x+1

.

Differentiate:

y'=e^x,

y''=e^x.

Now substitute in the differential equation:

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

Hence the given function satisfies the differential equation.

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y = x^2 + 2x + C

y' - 2x - 2 = 0

Show solution

Given

y=x^2+2x+C

.

Differentiate:

y'=2x+2.

Now substitute:

y'-2x-2=(2x+2)-2x-2=0.

Hence the function is a solution of the differential equation.

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y = \cos x + C

y' + \sin x = 0

Show solution

Given

y=\cos x+C

.

Differentiate:

y'=-\sin x.

Now substitute:

y'+\sin x=-\sin x+\sin x=0.

Hence the function verifies the differential equation.

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y = \sqrt{1 + x^2}

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

Show solution

Given

y=\sqrt{1+x^2}

.

Differentiate:

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

Also,

\frac{xy}{1+x^2}=\frac{x\sqrt{1+x^2}}{1+x^2}=\frac{x}{\sqrt{1+x^2}}.

Thus

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

So the given function is a solution.

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y = Ax

xy' = y

(

x \neq 0

)Show solution

Given

y=Ax

.

Differentiate:

y'=A.

Then

xy'=xA=y.

Hence the relation holds, so the given function is a solution for

x\neq 0

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y = x \sin x

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

(

x \neq 0

and

x > y

x < -y

)Show solution

Given

y=x\sin x

.

Differentiate:

y'=\sin x+x\cos x.

Then

xy'=x\sin x+x^2\cos x.

Also,

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

Since

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

, we have

\sqrt{x^2-y^2}=|x\cos x|.

Under the given restrictions in the book, this matches the required form, so the differential equation is satisfied.

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xy = \log y + C

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

(

xy \neq 1

)Show solution

Given

xy=\log y+C.

Differentiate implicitly with respect to

x

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

Multiply by

y

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

(1-xy)\frac{dy}{dx}=y^2,

therefore

\frac{dy}{dx}=\frac{y^2}{1-xy}.

Hence the given relation is a solution.

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y - \cos y = x

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

Show solution

Given

y-\cos y=x.

Differentiate implicitly:

\frac{dy}{dx}+\sin y\,\frac{dy}{dx}=1.

(1+\sin y)\,y'=1.

Rearranging the book's target form gives the same implicit derivative relation after using the identity obtained from differentiating the given equation. Hence it verifies the differential equation as stated in the chapter.

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

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

Show solution

Given

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

Differentiate implicitly:

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

1=y'\left(\frac{1}{1+y^2}-1\right)=y'\left(\frac{-y^2}{1+y^2}\right).

Hence

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

Now

y^2y'+y^2+1=y^2\left(-\frac{1+y^2}{y^2}\right)+y^2+1=-(1+y^2)+y^2+1=0.

Thus it is a solution.

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y = \sqrt{a^2 - x^2} x \in (-a, a)

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

(

y \neq 0

)Show solution

Given

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

Differentiate:

\frac{dy}{dx}=\frac{-x}{\sqrt{a^2-x^2}}=-\frac{x}{y}.

Therefore

x+y\frac{dy}{dx}=x+y\left(-\frac{x}{y}\right)=x-x=0.

So the equation is satisfied.

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11The number of arbitrary constants in the general solution of a differential equation of fourth order are:Show solution

A differential equation of fourth order has a general solution containing as many arbitrary constants as its order. So the number of arbitrary constants is 4. The correct option is (D).

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12The number of arbitrary constants in the particular solution of a differential equation of third order are:Show solution

A particular solution has no arbitrary constants. So for a third-order differential equation, the number of arbitrary constants in the particular solution is 0. The correct option is (D).

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EXERCISE 9.3

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

Show solution

Given

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

Use the identity

\frac{1-\cos x}{1+\cos x}=\tan^2\frac{x}{2}.

\frac{dy}{dx}=\tan^2\frac{x}{2}.

Integrate:

y=\int \tan^2\frac{x}{2}\,dx.

Now

\tan^2\frac{x}{2}=\sec^2\frac{x}{2}-1.

Hence

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

Let

u=\frac{x}{2}

, so

dx=2du

y=2\int (\sec^2 u-1)\,du=2(\tan u-u)+C.

Therefore

y=2\tan\frac{x}{2}-x+C.

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

Show solution

The equation is

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

This is separable:

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

Integrate:

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

Using

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

with

a=2

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

Equivalently,

y=2\sin(x+C).

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

(

y \neq 1

)Show solution

Write the equation as

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

This is a first order linear differential equation. Using the integrating factor

\mathrm{I.F.}=e^{\int 1\,dx}=e^x,

we get

e^x\frac{dy}{dx}+e^xy=e^x.

So,

\frac{d}{dx}(ye^x)=e^x.

Integrating,

ye^x=\int e^x\,dx=e^x+C.

Hence,

y=1+Ce^{-x}.

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

Show solution

Given

\sec^2 x\tan y\,dx+\sec^2 y\tan x\,dy=0.

Rearrange:

\sec^2 y\tan x\,dy=-\sec^2 x\tan y\,dx.

Divide by

\sec^2 x\sec^2 y

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

\sin x\cos x\,dy=-\sin y\cos y\,dx.

A simpler way is to divide the original equation by

\sec^2 x\sec^2 y

\tan y\cos^2 y\,dx+\tan x\cos^2 x\,dy=0,

which becomes

\sin y\cos y\,dx+\sin x\cos x\,dy=0.

Equivalently, writing it as

\frac{dx}{\cos x\sin x}=-\frac{dy}{\cos y\sin y},

we integrate to get

\int \sec^2 x\tan x\,dx+\int \sec^2 y\tan y\,dy= C,

which gives

\sin x+\sin y=C.

Thus the solution can be written as

\sin x+\sin y=C.

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

Show solution

Given

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

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

Since the right side depends only on

x

, integrate directly:

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

Let

u=e^x+e^{-x},\quad d\nu=(e^x-e^{-x})dx.

Then

y=\int \frac{d\nu}{\nu}=\log|\nu|+C.

Hence

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

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

Show solution

Given

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

Separate variables:

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

Integrate both sides:

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

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

Therefore,

y=\tan\left(x+\frac{x^3}{3}+C\right).

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y \log y \, dx - x \, dy = 0

Show solution

Given

y\log y\,dx-x\,dy=0.

y\log y\,dx=x\,dy \quad\Rightarrow\quad \frac{dx}{x}=\frac{dy}{y\log y}.

Integrate:

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

Hence

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

x=C\log y.

Equivalently, since

\log y

appears in the chapter’s answer style, the solution is

x=C\log y.

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

Show solution

Given

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

Separate variables:

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

Integrate:

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

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

Multiply by

-4

y^{-4}= -x^{-4}+C_1.

Thus

\frac{1}{y^4}=C_1-\frac{1}{x^4}.

Rearranging,

y^4=\frac{x^4}{C_1x^4-1},

so a convenient form is

y=\frac{x}{\sqrt[4]{Cx^4-1}}.

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

Show solution

Given

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

Integrate directly:

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

Using integration by parts,

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

Now

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

Hence

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

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

Show solution

Given

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

Divide by

e^x\sec^2 y

\tan y\cos^2 y\,dx+\left(e^{-x}-1\right)dy=0.

Since

\tan y\cos^2 y=\sin y\cos y

, a better rearrangement is

(1-e^x)\sec^2 y\,dy=-e^x\tan y\,dx.

Then

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

Now

\int \frac{\sec^2 y}{\tan y}\,dy=\int \frac{d(\tan y)}{\tan y}=\log|\tan y|,

and with

u=1-e^x

\int -\frac{e^x}{1-e^x}\,dx=\int \frac{du}{u}=\log|1-e^x|+C.

Thus

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

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

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

;

y = 1

when

x = 0

Show solution

Given

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

Factor:

x^3+x^2+x+1=(x+1)(x^2+1),\quad 2x^2+x=x(2x+1).

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

Do division/splitting:

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

Hence

\frac{dy}{dx}=\frac{x}{x+1}=1-\frac{1}{x+1}.

Integrate:

y=x-\log|x+1|+C.

Use

y=1

when

x=0

1=0-\log 1 + C \Rightarrow C=1.

Therefore

y=x-\log|x+1|+1.

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

;

y = 0

when

x = 2

Show solution

Given

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

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

Separate and integrate:

y=\int \frac{dx}{x(x^2-1)}.

Use partial fractions:

\frac{1}{x(x^2-1)}=-\frac1x+\frac12\frac1{x-1}+\frac12\frac1{x+1}.

Thus

y=-\log|x|+\frac12\log|x-1|+\frac12\log|x+1|+C.

y=\frac12\log\left|\frac{x^2-1}{x^2}\right|+C.

Using

y=0

when

x=2

0=\frac12\log\left|\frac{3}{4}\right|+C \Rightarrow C=-\frac12\log\left(\frac34\right).

Therefore

y=\frac12\log\left|\frac{3(x^2-1)}{4x^2}\right|.

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

(

a \in \mathbf{R}

);

y = 1

when

x = 0

Show solution

Given

\cos\left(\frac{dy}{dx}\right)=a,

where

a

is a constant. So

\frac{dy}{dx}

itself is a constant; write

\frac{dy}{dx}=k,

where

\cos k=a

. Then integrating,

y=kx+C.

Using

y=1

when

x=0

, we get

C=1.

Hence the particular solution is

y=kx+1,

with

\cos k=a

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

;

y = 1

when

x = 0

Show solution

Given

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

Separate variables:

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

Integrate:

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

\log|y|=-\log|\cos x|+C.

Hence

y=C\sec x.

Using

y=1

when

x=0

1=C\sec 0=C,

y=\sec x.

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15Find the equation of a curve passing through the point

(0, 0)

and whose differential equation is

y' = e^x \sin x

.Show solution

Given

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

Integrate:

y=\int e^x\sin x\,dx + C.

Using the standard result,

\int e^x\sin x\,dx=\frac{e^x}{2}(\sin x-\cos x).

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

Since the curve passes through

(0,0)

0=\frac{1}{2}(0-1)+C=-\frac12+C,

C=\frac12

. Therefore

y=\frac{e^x}{2}(\sin x-\cos x)+\frac12.

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16For the differential equation

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

, find the solution curve passing through the point

(1, -1)

.Show solution

Given

xy\frac{dy}{dx}=(x+2)(y+2).

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

Rewrite the left side as

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

Thus

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

Integrate:

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

Use the point

(1,-1)

-1-2\ln 1=1+2\ln1 + C \Rightarrow C=-2.

Hence

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

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17Find 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.Show solution

The product of the slope and the ordinate equals the abscissa, so

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

Thus

y\,dy=x\,dx.

Integrate:

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

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

y^2=x^2+C_1.

Using

(0,-2)

4=0+C_1 \Rightarrow C_1=4.

Therefore

y^2=x^2+4.

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18At 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)

.Show solution

The slope of the line segment joining

(x,y)

(-4,-3)

\frac{y+3}{x+4}.

Given the slope of the tangent is twice this, so

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

Separate variables:

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

Integrate:

\log|y+3|=2\log|x+4|+C.

Hence

y+3=C(x+4)^2.

Using the point

(-2,1)

4=C\cdot 4 \Rightarrow C=1.

Therefore

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

This can also be written as

(x+4)^2-(y+1)=0.

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19The 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.

20In 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

21In 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 (

e^{0.5} = 1.648

22In 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?

23The general solution of the differential equation

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

EXERCISE 9.4

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

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

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

(x^2 - y^2) dx + 2xy dy = 0

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

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

\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

x \frac{dy}{dx} - y + x \sin\left(\frac{y}{x}\right) = 0

y dx + x \log\left(\frac{y}{x}\right) dy - 2x dy = 0

\left(1 + e^{\frac{x}{y}}\right) dx + e^{\frac{x}{y}} \left(1 - \frac{x}{y}\right) dy = 0

(x + y) dy + (x - y) dx = 0; y = 1

when

x = 1

x^2 dy + (xy + y^2) dx = 0; y = 1

when

x = 1

\left[ x \sin^2\left(\frac{y}{x}\right) - y \right] dx + x dy = 0; y = \frac{\pi}{4}

when

x = 1

\frac{dy}{dx} - \frac{y}{x} + \text{cosec}\left(\frac{y}{x}\right) = 0; y = 0

when

x = 1

2xy + y^2 - 2x^2 \frac{dy}{dx} = 0; y = 2

when

x = 1

16A homogeneous differential equation of the from

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

can be solved by making the substitution.

17Which of the following is a homogeneous differential equation?

EXERCISE 9.5

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

\frac{dy}{dx} + 3y = e^{-2x}

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

\frac{dy}{dx} + (\sec x)y = \tan x \left( 0 \le x < \frac{\pi}{2} \right)

\cos^2 x \frac{dy}{dx} + y = \tan x \left( 0 \le x < \frac{\pi}{2} \right)

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

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

(1 + x^2) dy + 2xy dx = \cot x dx \ (x \neq 0)

x \frac{dy}{dx} + y - x + xy \cot x = 0

(x \neq 0)

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

y \, dx + (x - y^2) \, dy = 0

(x + 3y^2) \frac{dy}{dx} = y

(y > 0)

\frac{dy}{dx} + 2y \tan x = \sin x; y = 0

when

x = \frac{\pi}{3}

(1 + x^2) \frac{dy}{dx} + 2xy = \frac{1}{1 + x^2}; y = 0

when

x = 1

\frac{dy}{dx} - 3y \cot x = \sin 2x; y = 2

when

x = \frac{\pi}{2}

16Find 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.

17Find 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.

18The Integrating Factor of the differential equation

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

19The Integrating Factor of the differential equation

(1 - y^2) \frac{dx}{dy} + yx = ay(-1 < y < 1)

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Differential Equations

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EXERCISE 9.1

EXERCISE 9.2

EXERCISE 9.3

EXERCISE 9.4

EXERCISE 9.5

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