Let y = y(x) be the solution of the differential equation ${\rm{x}}\,\,{\rm{tan}}\left( {\frac{{\rm{y}}}{{\rm{x}}}} \right){\rm{dy = }}\left( {{\rm{y tan}}\left( {\frac{{\rm{y}}}{{\rm{x}}}} \right){\rm{ - x}}} \right){\rm{dx}}$, ${\rm{ - 1}} \le {\rm{x}} \le 1,{\rm{ y}}\left( {\frac{1}{2}} \right) = \frac{\pi }{6}$ Then, the area of the region bounded by the curves x = 0, ${\rm{x = }}\frac{1}{{\sqrt 2 }}$ and y = y(x) in the upper half plane is

$\frac{1}{{12}}\left( {\pi - 3} \right)$

$\frac{1}{4}\left( {\pi - 2} \right)$

$\frac{1}{6}\left( {\pi - 1} \right)$

Back to DIFFERENTIAL EQUATIONS

Your Progress

0 / 975 Answered

Correct

Incorrect

Accuracy

JEE • MATHEMATICS • DIFFERENTIAL EQUATIONS

Differential equations practice · Mixed difficulty

Choose an answer to check it, or open “View answer” to study the explanation. Sign in to save progress to your account.

Practice format and chapter quiz

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.

Take a chapter quiz →

Showing 961-975 of 975 questions

Preparing practice and saved progress…

961 Easy

961 of 975

Application +4 | -1

x^3 d y+x y d x=x^2 d y+2 y d x ; y(2)=e

and

x > 1

, then

y(4)

is equal to :

View answer and explanation

Correct option: B

\frac{3}{2} \sqrt{e}

Option B (

\frac{3}{2} \sqrt{e}

) is correct because it is the value of y(4) obtained after solving the given differential equation and applying the initial condition y(2) = e.

962 Easy

962 of 975

Application +4 | -1

If a curve

y = f(x)

passes through the point (1, – 1) and satisfies the differential equation,

y\left( {1 + xy} \right)dx = xdy,

then

f\left( { - \frac{1}{2}} \right)\,

is equal to

View answer and explanation

Correct option: D

\frac{4}{5}

Option D (

\frac{4}{5}

) is correct because solving the differential equation and using the given point (1, -1) allows us to find the function and evaluate it at -1/2.

963 Easy

963 of 975

Application +4 | -1

y=\left(x+\sqrt{1+x^{2}}\right)^{n}

, then

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

View answer and explanation

Correct option: A

n^{2} y

Option A (

n^2 y

) is correct because the second derivative of the given function satisfies the provided differential equation with

n^2 y

as the result.

964 Easy

964 of 975

Application +4 | -1

The integrating factor of

\frac{x d y}{d x}-y=x^4-3 x

View answer and explanation

Correct option: C

\frac{1}{x}

Option C (1/x) is correct because it is the integrating factor obtained by solving the differential equation after rewriting it in standard form.

965 Easy

965 of 975

Application +4 | -1

The degree of

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

is ......

View answer and explanation

Correct option: B

Option B is correct because after raising both sides to the power of 6 to eliminate the radicals, the highest order derivative (d²y/dx²) has a power of 2.

966 Medium

966 of 975

Application +4 | -1

The tangent at point

(x, y)

on the curve

y=f(x)

makes an angle of measure

\tan ^{-1}(2 x+3 y)

with positive direction of

\mathrm{X}

-axis. If the curve is passing through the point

(1,2)

, then its equation is .......

View answer and explanation

Correct option: A

6 x+9 y+2=26 e^{3(x-1)}

Option B (

6x + 9y - 2 = 26e^{3(x-1)}

) is correct because it is the solution to the differential equation given the initial condition (1, 2).

967 Easy

967 of 975

Application +4 | -1

The degree of the differential equation

y_2^{3 / 2}

-y_1^{1 / 2}-4=0

View answer and explanation

Correct option: A

Option A (6) is correct because after rationalizing the powers, the highest order derivative (

y_2

) has a power of 6.

968 Easy

968 of 975

Application +4 | -1

The orthogonal trajectories of the family of curves

a^{n-1} y=x^n

are given by (

a

is the arbitrary constant)

View answer and explanation

Correct option: B

ny^2+x^2=

constant

Option B (

ny^2+x^2=

constant) is the correct orthogonal trajectory after finding the differential equation and solving the orthogonal trajectory equation.

969 Medium

969 of 975

Application +4 | -1

The differential equation whose solution is

y=c_1 \cos a x+c_2 \sin a x

is

(Where

c_1, c_2

are arbitrary constants)

View answer and explanation

Correct option: B

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

Option B (

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

) is correct because the second derivative of the given solution satisfies this differential equation.

970 Medium

970 of 975

Application +4 | -1

A curve is passing through the point

(0,1)

. If the slope of the curve at any point

(x, y)

is equal to

x+x y

, then the equation of the curve is

View answer and explanation

Correct option: B

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

Option B (

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

) is correct because it is the solution obtained after solving the given differential equation with the initial condition y(0)=1.

971 Easy

971 of 975

Analyzing +4 | -1

The degree of the differential equation of all curves having normal of constant length

c

, is

View answer and explanation

Correct option: D

None of these

The correct answer is None of these, because the degree of the differential equation is 2.

972 Medium

972 of 975

Application +4 | -1

Let a solution

y=y(x)

of the differential equation

e^y d y-(2+\cos x) d x=0

satisfy

y(0)=0

then the value of

f\left(\frac{\pi}{2}\right)

is equal to

View answer and explanation

Correct option: B

\ln (2+\pi)

Option B (

\ln (2+\pi)

) is correct because it is the solution to the separable differential equation with the given initial condition.

973 Medium

973 of 975

Application +4 | -1

y = y(x)

is the solution of the differential equation,

\frac{{dy}}{{dx}} + 2y\tan x = \sin x,\,y\left( {\frac{\pi }{3}} \right) = 0

, then the maximum value of the function

y(x)

over

R

is equal to:

View answer and explanation

Correct option: D

\frac{1}{8}

Option D (

\frac{1}{8}

) is correct because it represents the maximum value of the function y(x) obtained after solving the differential equation and finding the constant of integration.

974 Easy

974 of 975

Application +4 | -1

The solution of differential equation

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

View answer and explanation

Correct option: A

2x{e^{{{\tan }^{ - 1}}y}} = {e^{2{{\tan }^{ - 1}}y}} + k

Option A (

2x{e^{{{\tan }^{ - 1}}y}} = {e^{2{{\tan }^{ - 1}}y}} + k

) is the correct solution to the given differential equation, obtained by finding the integrating factor and solving the resulting integral.

975 Easy

975 of 975

Application +4 | -1

Let y = y(x) be the solution of the differential equation

\mathrm{x}\,\,\mathrm{tan}\left( {\frac{\mathrm{y}}{\mathrm{x}}} \right)\mathrm{dy = }\left( {\mathrm{y tan}\left( {\frac{\mathrm{y}}{\mathrm{x}}} \right)\mathrm{ - x}} \right)\mathrm{dx}

\mathrm{ - 1} \le \mathrm{x} \le 1,\mathrm{ y}\left( {\frac{1}{2}} \right) = \frac{\pi }{6}

Then, the area of the region bounded by the curves x = 0,

\mathrm{x = }\frac{1}{{\sqrt 2 }}

and y = y(x) in the upper half plane is

View answer and explanation

Correct option: A

\frac{1}{8}\left( {\pi - 1} \right)

Option A (

\frac{1}{8}\left( {\pi - 1} \right)

) is correct because solving the differential equation and integrating the resulting function over the given interval yields this area.

Explore Other Chapters in MATHEMATICS

View All →

SEQUENCE OF SERIES 1764 MCQs DEFINITE INTEGRATION 1443 MCQs MATRICES 1363 MCQs INDEFINITE INTEGRATION 1179 MCQs PROBABILITY 1146 MCQs COMPLEX NUMBERS 980 MCQs

Revision notes and key takeaways

Review DIFFERENTIAL EQUATIONS in your textbook or class notes, then use this set to check your understanding.

Showing 961-975 of 975 questions

Your Progress

Differential equations practice · Mixed difficulty

Explore Other Chapters in MATHEMATICS

Continue Your Practice

Your practice