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

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

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

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

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JEE • MATHEMATICS • DIFFERENTIAL EQUATIONS

Differential equations practice · Easy

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Showing 481-494 of 494 questions

481 Easy

481 of 494

Application +4 | -1

The integrating factor of differential equation

\frac{d y}{d x}(1+x)-x y=1-x

View answer and explanation

Correct option: B

(1+x) e^{-x}

Option B (

(1+x)e^{-x}

) is correct because it is the integrating factor of the given differential equation after rewriting it in standard form.

482 Easy

482 of 494

Application +4 | -1

Solution of

x \frac{d y}{d x}+y-x+x y \cot x=0 \,\,\,(x \neq 0)

View answer and explanation

Correct option: A

x y=\frac{c}{\sin x}+1-x \cot x

Option A (

x y=\frac{{c}}{{\sin x}}+1-x \cot x

) is correct because it represents the general solution of the given differential equation after finding the integrating factor and integrating.

483 Easy

483 of 494

Application +4 | -1

Solution of

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

View answer and explanation

Correct option: A

\frac{1}{y}=x^2-x+C

Option A (

\frac{1}{y}=x^2-x+C

) is correct because the given differential equation is separable, and integrating both sides leads to this solution.

484 Easy

484 of 494

Understanding +4 | -1

The solution of

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

View answer and explanation

Correct option: C

y=c e^{-\int p d x}

Option C (

y=c e^{-\int p d x}

) is the correct solution to the given first-order linear homogeneous differential equation, derived by integrating factor method.

485 Easy

485 of 494

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.

486 Easy

486 of 494

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.

487 Easy

487 of 494

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.

488 Easy

488 of 494

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.

489 Easy

489 of 494

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.

490 Easy

490 of 494

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.

491 Easy

491 of 494

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.

492 Easy

492 of 494

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.

493 Easy

493 of 494

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.

494 Easy

494 of 494

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.

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

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