Differential Equations jee mains questions | JEE Main Maths Differential Equations Previous Year Question

Q 11 :

The general solution of $\frac{d y}{d x} + y \tan x = \sec x$ is:

$y \sec x = \tan x + C$
$y \tan x = \sec x + C$
$\tan x = y \tan x + C$
$x \sec x = \tan y + C$

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Ans. $y \sec x = \tan x + C$

Explanation:

Given differential equation is

$\frac{d y}{d x} + y \tan x = \sec x$

which is a linear differential equation.

Here, $P = \tan x,$ $Q = \sec x$

$∴$ $IF = e^{\int \tan x d x} = e^{\log | \sec x |} = \sec x$

The general solution is

$y \cdot \sec x = \int \sec x \cdot \sec x d x + C$

$\Rightarrow$ $y \cdot \sec x = \int \sec^{2} x d x + C$

$\Rightarrow$ $y \sec x = \tan x + C$

Q 12 :

The solution of differential equation $\frac{d y}{d x} + \frac{2 x y}{1 + x^{2}} = \frac{1}{{(1 + x^{2})}^{2}}$ is:

$y (1 + x^{2}) = C + \tan^{- 1} x$
$\frac{y}{1 + x^{2}} = C + \tan^{- 1} x$
$y \log (1 + x^{2}) = C + \tan^{- 1} x$
$y (1 + x^{2}) = C + \sin^{- 1} x$

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Ans. $y (1 + x^{2}) = C + \tan^{- 1} x$

Explanation:

Given that,

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

Here, $P = \frac{2 x}{1 + x^{2}}$ and $Q = \frac{1}{{(1 + x^{2})}^{2}}$

which is a linear differential equation.

$∴$ $IF = e^{\int \frac{2 x}{1 + x^{2}} d x}$

Put $1 + x^{2} = t \Rightarrow 2 x d x = d t$

$∴$ $IF = e^{\int \frac{d t}{t}} = e^{\log t} = e^{\log (1 + x^{2})} = 1 + x^{2}$

The general solution is

$y (1 + x^{2}) = \int (1 + x^{2}) \frac{1}{{(1 + x^{2})}^{2}} d x + C$

$\Rightarrow$ $y (1 + x^{2}) = \int \frac{1}{1 + x^{2}} d x + C$

$\Rightarrow$ $y (1 + x^{2}) = \tan^{- 1} x + C$

Q 13 :

Which of the following equation is linear?

$\frac{d y}{d x} + x y^{2} = 1$
$x^{2} \frac{d y}{d x} + y = e^{x}$
$\frac{d y}{d x} + 3 y = x y^{2}$
$x \frac{d y}{d x} + y^{2} = \sin x$

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Ans. $x^{2} \frac{d y}{d x} + y = e^{x}$

Explanation:

$x^{2} \frac{d y}{d x} + y = e^{x}$ can be written as

$\frac{d y}{d x} + \frac{y}{x^{2}} = \frac{e^{x}}{x^{2}},$ which is a linear equation.

Q 14 :

Integrating factor of the equation $(x^{2} + 1) \frac{d y}{d x} + 2 x y = x^{2} - 1$ is:

$x^{2} + 1$
$\frac{2 x}{x^{2} + 1}$
$\frac{x^{2} - 1}{x^{2} + 1}$
$\frac{x^{2} + 1}{x^{2} - 1}$

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Ans. $x^{2} + 1$

Explanation:

$\Rightarrow$ $\frac{d y}{d x} + \frac{2 x}{1 + x^{2}} y = \frac{x^{2} - 1}{x^{2} + 1}$

$I . F . = e^{\int \frac{2 x}{1 + x^{2}} d x}$

$= e^{\log (1 + x^{2})}$

$= 1 + x^{2}$

Q 15 :

The solution of $\frac{d y}{d x} + P (x) y = 0$ is:

$y = C e^{\int P d x}$
$x = C e^{- \int P d y}$
$y = C e^{- \int P d x}$
$x = C e^{\int P d y}$

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Ans. $y = C e^{- \int P d x}$

Explanation:

$\frac{d y}{d x} + P (x) y = 0$

Here, $Q = 0$

$∴$ $I . F . = e^{\int P d x}$

$∴$ Solution of the given equation is

$y (I . F .) = \int Q (I . F .) d x + C$

$\Rightarrow$ $y e^{\int P d x} = 0 + C$

$\Rightarrow$ $y = C e^{- \int P d x}$

Q 16 :

The solution of $(x + \tan y) d y = \sin 2 y d x$ is:

$x = \tan y + c \sqrt{\tan y}$
$x = \tan^{- 1} x + c$
$x = \cot y + c \sqrt{\tan y}$
$x = \tan y + c$

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Ans. $x = \tan y + c \sqrt{\tan y}$

Explanation:

$(x + \tan y) d y = \sin 2 y d x$

$\frac{d x}{d y} = \frac{\sin 2 y}{x + \tan y}$

$\frac{d y}{d x} = \frac{x + \tan y}{\sin 2 y}$

$\frac{d x}{d y} - (\frac{1}{\sin 2 y}) x = \frac{\tan y}{\sin 2 y}$

$I . F . = e^{- \int \frac{1}{\sin 2 y} d y}$

$= e^{- \int c o s e c 2 y d y} = e^{- \frac{1}{2} \log (\tan y)}$

$= \frac{1}{\sqrt{\tan y}}$

The general solution is:

$x \cdot \frac{1}{\sqrt{\tan y}} = \int \frac{1}{\sqrt{\tan y}} \cdot \frac{\tan y}{\sin 2 y} d x$

$\frac{x}{\sqrt{\tan y}} = \frac{1}{2} \int \frac{\sec^{2} y}{\sqrt{\tan y}} d y$

$\frac{x}{\sqrt{\tan y}} = \frac{1}{2} \sqrt{\tan y} + c$

$x = \tan y + c \sqrt{\tan y}$

Q 17 :

The solution of the differential equation $x d y - y d x = 0$ represent family of:

Circles passing through origin
Straight line passing through (−1, 6)
Straight line passing through the origin
Circle whose centre is at the origin

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Ans. Straight line passing through the origin

Explanation:

$x d y - y d x = 0$

$\frac{1}{y} d y = \frac{1}{x} d x$

On integrating both side,

$\int \frac{1}{y} d y = \int \frac{1}{x} d x$

$\log y = \log x + \log c$

$\log y = \log (c x)$

$y = c x$

It is equation of straight line passing through origin.

Q 18 :

The order of the differential equation whose general solution is $y = e^{x} (a \cos x + b \sin x),$ where $a$ and $b$ are arbitrary constant is:

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6

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Ans. 3

Explanation:

$y = e^{x} [a \cos x + b \sin x]$

The equation has 3 arbitrary constant so order of its differential equation is 3.

Q 19 :

Match List-I with List-II. Match the integrating factors:

List-I List-II

(Differential Equation) (Integrating factor)

(A) $\frac{d y}{d x} + 3 y = e^{- 2 x}$ (I) $\frac{1}{x}$

(B) $x \frac{d y}{d x} + y = 3 x^{2}$ (II) $e^{- x}$

(C) $x \frac{d y}{d x} - y = 3 x^{2}$ (III) $x$

(D) $\frac{d y}{d x} - y = x$ (IV) $e^{3 x}$

Choose the correct answer from the options given below:

(A)-(IV), (B)-(III), (C)-(I), (D)-(II)
(A)-(III), (B)-(IV), (C)-(I), (D)-(II)
(A)-(II), (B)-(III), (C)-(IV), (D)-(I)
(A)-(I), (B)-(II), (C)-(III), (D)-(IV)

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Ans. (A)-(IV), (B)-(III), (C)-(I), (D)-(II)

Explanation:

$(A)$ $\frac{d y}{d x} + 3 y = e^{- 2 x}$

$\Rightarrow$ $I . F . = e^{\int 3 d x} = e^{3 x}$

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

$\Rightarrow$ $\frac{d y}{d x} + \frac{1}{x} y = 3 x$

$I . F . = e^{\int \frac{1}{x} d x}$

$= e^{\log x} = x$

$(C)$ $x \frac{d y}{d x} - y = 3 x^{2}$

$\Rightarrow$ $\frac{d y}{d x} - \frac{1}{x} y = 3 x$

$I . F . = e^{\int \frac{- 1}{x} d x} = e^{- \log x} = \frac{1}{x}$

$(D)$ $\frac{d y}{d x} - y = x$

$\Rightarrow$ $I . F . = e^{\int - 1 d x} = e^{- x}$

Q 20 :

Solution of differential equation $x d y - y d x = 0$ represents:

family of straight lines passing through origin
family of parabolas whose vertex is at origin
family of circles whose centre is at origin
family of straight lines passing through (1, 1)

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Ans. family of straight lines passing through origin

Explanation:

Given, $x d y - y d x = 0$

$x d y = y d x$

$\Rightarrow$ $\frac{1}{y} d y = \frac{1}{x} d x$

On integrating both sides, we get

$\log y = \log x + \log c$

$\log y = \log c x$

$\Rightarrow$ $y = c x$

A straight line passes through origin.

Topic Question Set

Q 11 :

The general solution of $\frac{d y}{d x} + y \tan x = \sec x$ is:

Q 12 :

The solution of differential equation $\frac{d y}{d x} + \frac{2 x y}{1 + x^{2}} = \frac{1}{{(1 + x^{2})}^{2}}$ is:

Q 13 :

Which of the following equation is linear?

Q 14 :

Integrating factor of the equation $(x^{2} + 1) \frac{d y}{d x} + 2 x y = x^{2} - 1$ is:

Q 15 :

The solution of $\frac{d y}{d x} + P (x) y = 0$ is:

Q 16 :

The solution of $(x + \tan y) d y = \sin 2 y d x$ is:

Q 17 :

The solution of the differential equation $x d y - y d x = 0$ represent family of:

Q 18 :

The order of the differential equation whose general solution is $y = e^{x} (a \cos x + b \sin x),$ where $a$ and $b$ are arbitrary constant is:

Q 20 :

Solution of differential equation $x d y - y d x = 0$ represents:

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	List-I		List-II
	(Differential Equation)		(Integrating factor)
(A)	$\frac{d y}{d x} + 3 y = e^{- 2 x}$	(I)	$\frac{1}{x}$
(B)	$x \frac{d y}{d x} + y = 3 x^{2}$	(II)	$e^{- x}$
(C)	$x \frac{d y}{d x} - y = 3 x^{2}$	(III)	$x$
(D)	$\frac{d y}{d x} - y = x$	(IV)	$e^{3 x}$

Topic Question Set

Q 11 : The general solution of dydx+ytanx=secx is:

Q 12 : The solution of differential equation dydx+2xy1+x2=1(1+x2)2 is:

Q 13 : Which of the following equation is linear?

Q 14 : Integrating factor of the equation (x2+1)dydx+2xy=x2-1 is:

Q 15 : The solution of dydx+P(x)y=0 is:

Q 16 : The solution of (x+tany) dy=sin2y dx is:

Q 17 : The solution of the differential equation xdy-ydx=0 represent family of:

Q 18 : The order of the differential equation whose general solution is y=ex(acosx+bsinx), where a and b are arbitrary constant is:

Q 19 : Match List-I with List-II. Match the integrating factors: List-I List-II (Differential Equation) (Integrating factor) (A) dydx+3y=e-2x (I) 1x (B) xdydx+y=3x2 (II) e-x (C) xdydx-y=3x2 (III) x (D) dydx-y=x (IV) e3x Choose the correct answer from the options given below:

Q 20 : Solution of differential equation xdy−ydx=0 represents:

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Q 11 :

The general solution of $\frac{d y}{d x} + y \tan x = \sec x$ is:

Q 12 :

The solution of differential equation $\frac{d y}{d x} + \frac{2 x y}{1 + x^{2}} = \frac{1}{{(1 + x^{2})}^{2}}$ is:

Q 13 :

Which of the following equation is linear?

Q 14 :

Integrating factor of the equation $(x^{2} + 1) \frac{d y}{d x} + 2 x y = x^{2} - 1$ is:

Q 15 :

The solution of $\frac{d y}{d x} + P (x) y = 0$ is:

Q 16 :

The solution of $(x + \tan y) d y = \sin 2 y d x$ is:

Q 17 :

The solution of the differential equation $x d y - y d x = 0$ represent family of:

Q 18 :

The order of the differential equation whose general solution is $y = e^{x} (a \cos x + b \sin x),$ where $a$ and $b$ are arbitrary constant is:

Q 20 :

Solution of differential equation $x d y - y d x = 0$ represents: