# JEE Main

Explore popular questions from Differential Equations for JEE Main. This collection covers Differential Equations previous year JEE Main questions hand picked by experienced teachers.

## Mathematics

Differential Equations

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Q 1. The differential equation which represents the family of curves {tex} y = c _ { 1 } e ^ { c _ { 2 } x } {/tex} where {tex} c _ { 1 } {/tex} and {tex} c _ { 2 } {/tex} are arbitrary constants is

A

{tex} y ^ { \prime } = y ^ { 2 } {/tex}

B

{tex} y ^ { \prime \prime } = y ^ { \prime } y {/tex}

C

{tex} y y ^ { \prime \prime } = y ^ { \prime } {/tex}

{tex} y y ^ { \prime \prime } = \left( y ^ { \prime } \right) ^ { 2 } {/tex}

##### Explanation

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Q 2. Solution of the differential equation {tex} \cos x d y = y ( \sin x - y ) d x , 0 < x < \frac { \pi } { 2 } {/tex} is

A

{tex} y \sec x = \tan x + c {/tex}

B

{tex} y \tan x = \sec x + c {/tex}

C

{tex} \tan x = ( \sec x + c ) y {/tex}

{tex} \sec x = ( \tan x + c ) y {/tex}

##### Explanation

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Q 3. Let {tex} f :( - 1,1 ) \rightarrow R {/tex} be a differentiable function with {tex} f ( 0 ) = - 1 {/tex} and {tex} f ^ { \prime } ( 0 ) = 1 . {/tex} Let {tex} g ( x ) = [ f ( 2 f ( x ) + 2 ) ] ^ { 2 } . {/tex} Then {tex} g ^ { \prime } ( 0 ) = {/tex}

-4

B

0

C

-2

D

4

##### Explanation

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Q 4. If {tex} \frac { d y } { d x } = y + 3 > 0 {/tex} and {tex} y ( 0 ) = 2 , {/tex} then {tex} y ( \ln 2 ) {/tex} is equal to

A

5

B

13

C

-2

7

##### Explanation

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Q 5. Let {tex} I {/tex} be the purchase value of an equipment and {tex} V ( t ) {/tex} be the value after it has been used for {tex} t {/tex} years. The value {tex} V ( t ) {/tex} depreciates at a rate given by differential equation {tex} \frac { d V ( t ) } { d t } = - k ( T - t ) , {/tex} where {tex} k > {/tex} 0 is a constant and {tex} T {/tex} is the total life in years of the equipment. Then the scrap value {tex} V ( T ) {/tex} of the equipment is

{tex} I - \frac { k T ^ { 2 } } { 2 } {/tex}

B

{tex} I - \frac { k ( T - t ) ^ { 2 } } { 2 } {/tex}

C

{tex} e ^ { - k T } {/tex}

D

{tex} T ^ { 2 } - \frac { 1 } { k } {/tex}

##### Explanation

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Q 6. At present, a firm is manufacturing {tex}2000{/tex} items. It is estimated that the rate of change of production {tex} P {/tex} w.r.t. additional number of workers {tex} x {/tex} is given by {tex} \frac { d P } { d x } = 100 - 12 \sqrt { x } {/tex} . If the firm employs {tex}25{/tex} more workers, then the new level of production of items is

A

3000

3500

C

4500

D

2500

##### Explanation

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Q 7. Let the population of rabbits surviving at a time {tex} t {/tex} be governed by the differential equation {tex} \frac { d p ( t ) } { d t } = \frac { 1 } { 2 } p ( t ) - 200 . {/tex} If {tex} p ( 0 ) = 100 {/tex} , then {tex} p ( t ) {/tex} equals

A

{tex} 600 - 500 e ^ { t / 2 } {/tex}

B

{tex} 400 - 300 e ^ { - t / 2 } {/tex}

{tex} 400 - 300 e ^ { t / 2 } {/tex}

D

{tex} 300 - 200 e ^ { - t / 2 } {/tex}

##### Explanation

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Q 8. If the differential equation representing the family of all circles touching {tex} x {/tex} -axis at the origin is {tex} \left( x ^ { 2 } - y ^ { 2 } \right) \frac { d y } { d x } = g ( x ) y , {/tex} then {tex} g ( x ) {/tex} equals

A

{tex} \frac { 1 } { 2 } x {/tex}

B

2{tex} x ^ { 2 } {/tex}

2{tex} x {/tex}

D

{tex} \frac { 1 } { 2 } x ^ { 2 } {/tex}

##### Explanation

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Q 9. If the general solution of the differential equation {tex}y^{\prime} = \frac{y}{x} + \phi \left(\frac{x}{y}\right){/tex} for some function {tex}\phi{/tex}, is given by y In |cx| = x, where c is an arbitary constant, then {tex}\phi(2){/tex} is equal to

A

{tex}4{/tex}

B

{tex} \frac { 1 } { 4 } {/tex}

C

{tex} - 4 {/tex}

{tex} - \frac { 1 } { 4 } {/tex}

##### Explanation

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Q 10. The general solution of the differential equation {tex} \sin 2 x \left( \frac { d y } { d x } - \sqrt { \tan x } \right) - y = 0 {/tex} is

A

{tex} y \sqrt { \tan x } = x + c {/tex}

B

{tex} y \sqrt { \cot x } = \tan x + c {/tex}

C

{tex} y \sqrt { \tan x } = \cot x + c {/tex}

{tex} y \sqrt { \cot x } = x + c {/tex}

##### Explanation

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Q 11. If {tex} \frac { d y } { d x } + y \tan x = \sin 2 x {/tex} and {tex} y ( 0 ) = 1 , {/tex} then {tex} y ( \pi ) {/tex} is equal to

A

1

B

- 1

-5

D

5

##### Explanation

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Q 12. If {tex} y ( x ) {/tex} is the solution of the differential equation {tex} ( x + 2 ) \frac { d y } { d x } = x ^ { 2 } + 4 x - 9 , x \neq - 2 {/tex} and {tex} y ( 0 ) = 0 , {/tex} then {tex} y ( - 4 ) {/tex} is equal to

0

B

1

C

-1

D

2

##### Explanation

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Q 13. The solution of the differential equation {tex} y d x - \left( x + 2 y ^ { 2 } \right) d y = 0 {/tex} is {tex} x = f ( y ) . {/tex} If {tex} f ( - 1 ) = 1 , {/tex} then {tex} f ( 1 ) {/tex} is equal to

A

4

3

C

2

D

1

##### Explanation

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Q 14. If a curve {tex} y = f ( x ) {/tex} passes through the point {tex} ( 1 , - 1 ) {/tex} and satisfies the differential equation, {tex} y ( 1 + x y ) d x = x d y , {/tex} then {tex} f \left( - \frac { 1 } { 2 } \right) {/tex} is equal to

{tex} \frac { 4 } { 5 } {/tex}

B

{tex}- \frac { 2 } { 5 } {/tex}

C

{tex} - \frac { 4 } { 5 } {/tex}

D

{tex} \frac { 2 } { 5 } {/tex}

##### Explanation

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Q 15. If the tangent at a point {tex} P , {/tex} with parameter {tex} t , {/tex} on the curve {tex} x = 4 t ^ { 2 } + 3 , y = 8 t ^ { 3 } - 1 , t \in \mathbb { R } {/tex} meets the curve again at a point {tex} Q , {/tex} then the coordinates of {tex} Q {/tex} are

A

{tex} \left( 16 t ^ { 2 } + 3 , - 64 t ^ { 3 } - 1 \right) {/tex}

B

{tex} \left( 4 t ^ { 2 } + 3 , - 8 t ^ { 3 } - 1 \right) {/tex}

C

{tex} \left( t ^ { 2 } + 3 , t ^ { 3 } - 1 \right) {/tex}

{tex} \left( t ^ { 2 } + 3 , - t ^ { 3 } - 1 \right) {/tex}

##### Explanation

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Q 16. For {tex}x \in R, x \neq 0,{/tex} if y(x) is a differentiable function such that {tex}x\int\limits^{x}_{1} y(t)dt = (x+1)\int\limits^{x}_{1}ty(t)dt{/tex} then y(x) equals (where C is a constant)

A

{tex} Cx ^ { 3 } \frac { 1 } { e ^ { x } } {/tex}

B

{tex} \frac { C } { x ^ { 2 } } e ^ { - \frac { 1 } { x } } {/tex}

C

{tex} \frac { c } { x } e ^ { - \frac { 1 } { x } } {/tex}

{tex} \frac { c } { x ^ { 3 } } e ^ { - \frac { 1 } { x } } {/tex}

##### Explanation

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Q 17. The differential equation of all non-vertical lines in a plane is

{tex} \frac { d ^ { 2 } y } { d x ^ { 2 } } = 0 {/tex}

B

{tex} \frac { d ^ { 2 } x } { d y ^ { 2 } } = 0 {/tex}

C

{tex} \frac { d y } { d x } = 0 {/tex}

D

{tex} \frac { d x } { d y } = 0 {/tex}

##### Explanation

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Q 18. The differential equation of all non-horizontal lines in a plane is

A

{tex} \frac { d ^ { 2 } y } { d x ^ { 2 } } = 0 {/tex}

{tex} \frac { d ^ { 2 } x } { d y ^ { 2 } } = 0 {/tex}

C

{tex} \frac { d y } { d x } = 0 {/tex}

D

{tex} \frac { d x } { d y } = 0 {/tex}

##### Explanation

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Q 19. A solution of the differential equation {tex} \left( \frac { d y } { d x } \right) ^ { 2 } - x \frac { d y } { d x } + y = 0 {/tex} is

A

{tex} y = 2 {/tex}

B

{tex} y = 2 x {/tex}

{tex} y = 2 x - 4 {/tex}

D

{tex} y = 2 x ^ { 2 } - 4 {/tex}

##### Explanation

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Q 20. Let F denotes the family of ellipses whose centre is at the origin and major axis is the y-axis. Then, equation of the family F is

A

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

B

$\text{xy}\ \frac{d^{2}y}{dx^{2}} - \frac{\text{dy}}{\text{dx}}\left( x\ \frac{\text{dy}}{\text{dx}} - y \right) = 0$

$\text{xy}\ \frac{d^{2}y}{dx^{2}} + \frac{\text{dy}}{\text{dx}}\left( x\ \frac{\text{dy}}{\text{dx}} - y \right) = 0$

D

$\frac{d^{2}y}{dx^{2}} - \frac{\text{dy}}{\text{dx}}\left( x\ \frac{\text{dy}}{\text{dx}} - y \right) = 0$

##### Explanation

Equation of family of ellipse is $\frac{x^{2}}{a^{2}} + \frac{y^{2}}{b^{2}} = 1$

⇒ $\frac{2x}{a^{2}} + \frac{2y}{b^{2}}.\frac{\text{dy}}{\text{dx}} = 0$

⇒ $\frac{x}{a^{2}} + \frac{y}{b^{2}}.\frac{\text{dy}}{\text{dx}} = 0$ …(i)

⇒ $\frac{1}{a^{2}} + \frac{y}{b^{2}}.\frac{d^{2}y}{dx^{2}} + \left( \frac{\text{dy}}{\text{dx}} \right)^{2}\ \frac{1}{b^{2}} = 0$

⇒ $\frac{b^{2}}{a^{2}} + y\left( \frac{d^{2}y}{dx^{2}} \right) + \left( \frac{\text{dy}}{\text{dx}} \right)^{2} = 0$

⇒ $- \frac{y}{x}\text{.\ }\frac{\text{dy}}{\text{dx}} + y\ \frac{d^{2}y}{dx^{2}} + \left( \frac{\text{dy}}{\text{dx}} \right)^{2} = 0$

[from Eq. (i), $\frac{b^{2}}{a^{2}} = - \frac{y}{x}.\frac{\text{dy}}{\text{dx}}$]

⇒ $\text{xy}\ \frac{d^{2}y}{dx^{2}} + \frac{\text{dy}}{\text{dx}}\left( x\ \frac{\text{dy}}{\text{dx}} - y \right) = 0$

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Q 21. The order and degree of the differential equation $\rho = \frac{\left\lbrack 1 + \left( \frac{\text{dy}}{\text{dx}} \right)^{2} \right\rbrack^{3/2}}{\frac{d^{2}y}{dx^{2}}}$ are respectively

2, 2

B

2, 3

C

2, 1

D

None of these

##### Explanation

The given equation can be rewritten as

$\rho \bullet \frac{d^{2}y}{dx^{2}} = \left\lbrack 1 + \left( \frac{\text{dy}}{\text{dx}} \right)^{2} \right\rbrack^{3/2}$

On squaring both sides, we get

$\left( \rho \bullet \frac{d^{2}y}{dx^{2}} \right) = \left\lbrack 1 + \left( \frac{\text{dy}}{\text{dx}} \right)^{2} \right\rbrack^{3}$

⇒ order = 2, degree = 2.

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Q 22. The differential equation of the family of curves y = e2x(acos x + bsin x), where a and b are arbitrary constants, is given by

y2 − 4y1 + 5y = 0

B

2y2 − y1 + 5y = 0

C

y2 + 4y1 − 5y = 0

D

y2 − 2y1 + 5y = 0

##### Explanation

Since,

y = e2x(acos x + bsin x) …(i)

⇒ y1 = e2x(−asinx+bcosx) + (acosx+bsinx)2e2x

⇒ y1 = e2x(−asinx+bcosx) + 2y …(ii)

⇒ y2 = e2x(−acosx−bsinx) + (−asinx+bcosx)e2x2 + 2y1

= − y + 2e2x(−asinx+bcosx) + 2y1 (using eq.(ii))

⇒ y2 = − y + 2(y1−2y) + 2y1 (using eq.(ii))

⇒ y2 = − y + 4y1 − 4y

⇒ y2 − 4y1 + 5y = 0

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Q 23. The solution of $\frac{\text{dy}}{\text{dx}} = \frac{\text{ax} + g}{\text{by} + f}$ represents a circle when

A

a = b

a = − b

C

a = − 2b

D

a = 2b

##### Explanation

We have, $\frac{\text{dy}}{\text{dx}} = \frac{\text{ax} + g}{\text{by} + f}$

⇒ (by+f)dy = (ax+g)dx

On integrating, we get

$\frac{by^{2}}{2} + \text{fy} = \frac{ax^{2}}{2} + \text{gx} + c$

⇒ ax2 − by2 + 2gx − 2fy + c = 0

This represents a circle, if a = − b

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Q 24. Solution of the differential equation $\frac{\text{dy}}{\text{dx}} + \frac{y}{x} = x^{2}$ is

A

$y = \frac{x^{2}}{4} + cx^{- 2}$

B

y = x − 1 + cx − 3

$y = \frac{x^{3}}{4} + cx^{- 1}$

D

xy = x2 + c

##### Explanation

Given, $\frac{\text{dy}}{\text{dx}} + \frac{y}{x} = x^{2}$

∴ $\text{IF} = e^{\int_{}^{}{\frac{1}{x}\ \text{dx}}} = e^{\log x} = x$

∴ Complete solution is

y.x = ∫x.x2dx + c

⇒ $y.x = \frac{1}{4}\ x^{4} + c$

⇒ $y = \frac{1}{4}\ x^{3} + cx^{- 1}$

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Q 25. The differential equation satisfied by the family of curves $y = \text{ax}\cos\left( \frac{1}{x} + b \right)$, where a and b are parameters, is

A

x2y2 + y = 0

x4y2 + y = 0

C

xy2 − y = 0

D

x4y2 − y = 0

##### Explanation

Given, $y = \text{ax}\cos\left( \frac{1}{x} + b \right)$

⇒ $y_{1} = - \text{ax}\sin{\left( \frac{1}{x} + b\ \right) \times}\left( - \frac{1}{x^{2}} \right) + a\cos{(\frac{1}{x}} + b)$

⇒ $y_{1} = \frac{a}{x}\sin\left( \frac{1}{x} + b \right) + \text{acos}\left( \frac{1}{x} + b \right)$

⇒ $xy_{1} = \text{asin}{\left( \frac{1}{x} + b \right) + y}$

⇒ $y_{1} + xy_{2} = \text{acos}{\left( \frac{1}{x} + b \right)\left( - \frac{1}{x^{2}} \right) + y_{1}}$

⇒ $x^{3}y_{2} = - \text{acos}\left( \frac{1}{x} + b \right)$

⇒ x4y2 + y = 0