Differential equation of $y=\sec \left(\tan ^{-1} x\right)$ is

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

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

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

If integrating factor of $x\left( {1 - {x^2}} \right)dy + \left( {2{x^2}y - y - a{x^3}} \right)dx = 0$ is ${e^{\smallint Pdx}}$, then $P$ is equal to

$\frac{{2{x^2} - a{x^3}}}{{x\left( {1 - {x^2}} \right)}}$

$\frac{{2{x^2} - 1}}{{a{x^3}}}$

If integrating factor of $x\left( {1 - {x^2}} \right)dy + \left( {2{x^2}y - y - a{x^3}} \right)dx = 0{\rm{\,\, is \,\,}}{e^{\int P dx}}$ then $P$ is equal to

$\frac{2 x^2-a x^3}{x\left(1-x^2\right)}$

The solution of $\left( {1 + {x^2}} \right)\frac{{dy}}{{dx}} + 2xy - 4{x^2} = 0$ is

$3x\left( {1 + {y^2}} \right) = 4{y^3} + c$

$3x\left( {1 - {y^2}} \right) = 4{y^3} + c$

$3y\left( {1 + {y^2}} \right) = 4{x^3} + c$

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JEE • MATHEMATICS • DIFFERENTIAL EQUATIONS

Differential equations practice · Easy

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These sets use single-correct-answer MCQs. They do not cover every JEE Main or JEE Advanced question format and are not full-length mock papers.

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181 Easy

181 of 494

Application +4 | -1

The integrating factor of

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

View answer and explanation

Correct option: D

x{e^x}

Option D (

x{e^x}

) is correct because it is the integrating factor obtained after rewriting the differential equation in standard form and integrating the coefficient of y.

182 Easy

182 of 494

Application +4 | -1

Differential equation of

y=\sec \left(\tan ^{-1} x\right)

View answer and explanation

Correct option: C

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

Option C

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

is correct because it is the differential equation obtained by differentiating

y=\sec \left(\tan ^{-1} x\right)

and eliminating the trigonometric functions.

183 Easy

183 of 494

Application +4 | -1

The differential equation of the rectangular hyperbola whose axes are the asymptotes of the hyperbola, is

View answer and explanation

Correct option: B

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

Option B (

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

) is correct because it represents the differential equation of a rectangular hyperbola whose axes are the asymptotes.

184 Easy

184 of 494

Application +4 | -1

If integrating factor of

x\left( {1 - {x^2}} \right)dy + \left( {2{x^2}y - y - a{x^3}} \right)dx = 0

{e^{\int Pdx}}

, then

P

is equal to

View answer and explanation

Correct option: D

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

Option D (

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

) is correct because it represents the coefficient of

y

when the given differential equation is written in the standard form

\frac{{dy}}{{dx}} + Py = Q

185 Easy

185 of 494

Application +4 | -1

The differential equation of all ellipses with centres at the origin and the ends of one axis of symmetry at

(\pm 1,0)

, is

View answer and explanation

Correct option: A

\left(x^2-1\right) y^{\prime}-x y=0

Option A

\left(x^2-1\right) y^{\prime}-x y=0

is correct. It is derived by eliminating the arbitrary constant from the equation of the ellipse.

186 Easy

186 of 494

Application +4 | -1

If integrating factor of

x\left( {1 - {x^2}} \right)dy + \left( {2{x^2}y - y - a{x^3}} \right)dx = 0\mathrm{\,\, is \,\,}{e^{\int P dx}}

then

P

is equal to

View answer and explanation

Correct option: D

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

Option D (

\frac{2 x^2-1}{x(1-x^2)}

) is correct because it represents the coefficient of 'y' in the standard form of the given linear differential equation, which is required to find the integrating factor.

187 Easy

187 of 494

Analyzing +4 | -1

The slope of the tangent to a curve

C: y=y(x)

at any point

(x, y)

on it is

\frac{2 e^{2 x}-6 e^{-x}+9}{2+9 e^{-2 x}}

. If

C

passes through the points

\left(0, \frac{1}{2}+\frac{\pi}{2 \sqrt{2}}\right)

and

\left(\alpha, \frac{1}{2} e^{2 \alpha}\right)

, then

e^\alpha

is equal to

View answer and explanation

Correct option: B

\frac{3}{\sqrt{2}}\left(\frac{3+\sqrt{2}}{3-\sqrt{2}}\right)

Option B is correct. The value of

e^\alpha

is found by solving the differential equation and using the given points to find the constant of integration and then solving for

e^\alpha

188 Easy

188 of 494

Understanding +4 | -1

The order and degree of the differential equation

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

are respectively

View answer and explanation

Correct option: B

\left( {3,2} \right)

Option B (

\left( {3,2} \right)

) is correct because the highest order derivative in the equation is

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

(order 3) and its power is 2 (degree 2).

189 Easy

189 of 494

Application +4 | -1

If the function

y=\log \frac{1}{1+x}

satisfies the relation

x y^{\prime}

+1=f(y)

then

f(y)

is equal to

View answer and explanation

Correct option: C

e^y

Option D (

e^{-y}

) is correct because it satisfies the given relation after substituting the derivative of the function.

190 Easy

190 of 494

Application +4 | -1

If the solution of

\frac{d y}{d x}=\frac{a x+3}{2 y+f}

represents a circle, then the value of

a

View answer and explanation

Correct option: A

Option A (2) is correct because for the given differential equation to represent a circle, the coefficient of x² and y² must be equal, leading to a = -2.

191 Easy

191 of 494

Application +4 | -1

The order and degree respectively of the differential equation of all tangent lines to parabola

x^2=

2 y

View answer and explanation

Correct option: A

1,2

Option A (1,2) is correct because the differential equation representing the tangent lines to the parabola has an order of 1 and a degree of 2.

192 Easy

192 of 494

Application +4 | -1

If the differential equation representing the family of all circles touching

y

-axis at the origin is

x^2-y^2

=f(x) y \frac{d x}{d y}

then

f(x)

is equal to

View answer and explanation

Correct option: A

2 x

Option A (

2x

) is correct because it is the function

f(x)

that satisfies the differential equation representing the family of circles touching the y-axis at the origin.

193 Easy

193 of 494

Understanding +4 | -1

Solution of the differential equation

xdy - ydx = 0

represents a

View answer and explanation

Correct option: D

Straight line

Option D (Straight line) is correct because the solution to the differential equation

xdy - ydx = 0

y = cx

, which represents a straight line passing through the origin.

194 Easy

194 of 494

Application +4 | -1

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

, then the solution of the equation is

View answer and explanation

Correct option: B

\log \left( {\frac{y}{x}} \right) = cx

Option B (

\log \left( {\frac{y}{x}} \right) = cx

) is correct because it is the solution obtained by solving the given homogeneous differential equation.

195 Easy

195 of 494

Application +4 | -1

Let the solution curve

y=y(x)

of the differential equation

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

pass through the point

\left(0, \frac{\pi}{2}\right)

. Then,

\lim\limits_{x \to \infty } {e^x}y(x)

is equal to

View answer and explanation

Correct option: B

\frac{3 \pi}{4}

Option B (

\frac{{3 \pi}}{{4}}

) is correct because after solving the differential equation and applying the given conditions, the limit evaluates to this value.

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196 Easy

196 of 494

Application +4 | -1

The solution of

y^{\prime}+x y=x y^{-1}, y(0)=2

is given by

View answer and explanation

Correct option: C

y^2=3+e^{-x^2}

Option C (

y^2=3+e^{-x^2}

) is correct because it is the solution to the differential equation

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

with the initial condition

y(0)=2

197 Easy

197 of 494

Application +4 | -1

\cos x\frac{{dy}}{{dx}} - y\sin x = 6x,\left( {0 < x < \frac{\pi }{2}} \right)\,\,{\text{and}}\,\,y\left( {\frac{\pi }{3}} \right) = 0,{\text{then}}\,\,y\left( {\frac{\pi }{6}} \right)

is equal to:

View answer and explanation

Correct option: C

-\frac{\pi^2}{2 \sqrt{3}}

Option C (

-\frac{\pi^2}{2 \sqrt{3}}

) is correct because it is the value of y(

\frac{\pi}{6}

) obtained after solving the given differential equation and applying the initial condition y(

\frac{\pi}{3}

) = 0.

198 Easy

198 of 494

Application +4 | -1

The solution of

\left( {1 + {x^2}} \right)\frac{{dy}}{{dx}} + 2xy - 4{x^2} = 0

View answer and explanation

Correct option: B

3y\left( {1 + {x^2}} \right) = 4{x^3} + c

Option B (

3y\left( {1 + {x^2}} \right) = 4{x^3} + c

) is correct because it is the solution obtained by solving the linear differential equation using an integrating factor.

199 Easy

199 of 494

Application +4 | -1

Solution of

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

, is

View answer and explanation

Correct option: B

xy = {e^x}\left( {x - 1} \right) + C

Option B (

xy = {e^x}\left( {x - 1} \right) + C

) is correct because it is the solution obtained by integrating the given differential equation.

200 Easy

200 of 494

Application +4 | -1

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

and

y=1

when

x=1

, then

y=2

when

5 x=\ldots (x \neq 0)

View answer and explanation

Correct option: A

Option A (8) is correct because solving the differential equation and using the given condition, we find that 5x = 8 when y = 2.

201 Easy

201 of 494

Understanding +4 | -1

The order and degree of the differential equation

\frac{d^2 y}{d x^2}+\left(\frac{d y}{d x}\right)^{1 / 3}+x^{1 / 4}=0

View answer and explanation

Correct option: B

order -2 , degree -3

Option B (order -2 , degree -3) is correct because the order is the highest derivative and the degree is the power of the highest derivative after removing radicals.

202 Easy

202 of 494

Understanding +4 | -1

The order of the differential equation

\frac{{{d^2}y}}{{d{x^2}}} = \sqrt {1 + {{\left( {\frac{{dy}}{{dx}}} \right)}^3}}

, is

View answer and explanation

Correct option: A

Option A (2) is correct because the highest order derivative in the equation is the second derivative, making the order 2.

203 Easy

203 of 494

Application +4 | -1

The differential equation of the family of curves represented by the equation

x^2 y=a

, is

View answer and explanation

Correct option: A

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

Option A (

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

) is correct because differentiating

x^2y=a

and eliminating 'a' leads to this differential equation.

204 Easy

204 of 494

Application +4 | -1

e^y+x y=e

, then the ordered pair

\left(\frac{d y}{d x}, \frac{d^2 y}{d x^2}\right)

x=0

is equal to

View answer and explanation

Correct option: B

\left(-\frac{1}{\mathrm{e}}, \frac{1}{\mathrm{e}^2}\right)

Option B (

\left(-\frac{{1}}{{\mathrm{e}}}, \frac{{1}}{{\mathrm{e}^2}}\right)

) is correct because it is the ordered pair of the first and second derivatives evaluated at x=0.

205 Easy

205 of 494

Application +4 | -1

\frac{d y}{d x}=y+3 > 0

and

y(0)=2

, then

y(\ln 2)

is equal to:

View answer and explanation

Correct option: D

Option D (7) is correct because solving the differential equation and applying the initial condition yields y(\ln 2) = 7.

206 Easy

206 of 494

Application +4 | -1

Solution of the differential equation

x \mathrm{~d} y-y \mathrm{~d} x-\sqrt{x^2+y^2} \mathrm{~d} x=0

View answer and explanation

Correct option: B

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

Option B (

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

) is correct because it is the solution obtained by solving the homogeneous differential equation using the substitution y=vx.

207 Easy

207 of 494

Application +4 | -1

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

satisfies the relation

\left(1-x^2\right) y^{\prime}-x y

=k

then the value of

k

is1

View answer and explanation

Correct option: A

Option A (1) is correct because it is the value of k that satisfies the given relation after substituting the derivative of the function.

208 Easy

208 of 494

Application +4 | -1

The differential equation

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

, has the solution

View answer and explanation

Correct option: D

\left( {y + 1} \right)\left( {{e^x} + 1} \right) = c{e^y}

Option D (

\left( {y + 1} \right)\left( {{e^x} + 1} \right) = c{e^y}

) is correct because it is the solution obtained by separating variables and integrating the given differential equation.

209 Easy

209 of 494

Application +4 | -1

The solution of

y-x y^{\prime}=2\left(x+y y^{\prime}\right), y(1)=1

View answer and explanation

Correct option: B

\tan ^{-1}(y / x)+\log \left(x^2+y^2\right)=\pi / 4+\log 2

Option B (

\tan ^{-1}(y / x)+\log \left(x^2+y^2\right)=\pi / 4+\log 2

) is correct as it is the solution to the given differential equation with the specified initial condition.

210 Easy

210 of 494

Application +4 | -1

The equation of curves which intersect the hyperbola

x y=4

at an angle

\pi / 2

View answer and explanation

Correct option: C

y^2-x^2=C

Option C (

y^2-x^2=C

) is correct because it represents the orthogonal trajectories to the given hyperbola.

Revision notes and key takeaways

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