The solution of the differential equation $y^3 x^2 \cos \left(x^3\right) d x+\sin \left(x^3\right) y^2 d y=\frac{x}{3} d x$ is $2 \sin \left(x^3\right) y^k=x^2+C$ (where ${C}$ is an arbitrary constant) then the value of ${k}$ is equal to

JEE MATHEMATICS MCQ: DIFFERENTIAL EQUATIONS

1 Easy

1 of 1

Analyzing +4 | -1

The solution of the differential equation

y^3 x^2 \cos \left(x^3\right) d x+\sin \left(x^3\right) y^2 d y=\frac{x}{3} d x

2 \sin \left(x^3\right) y^k=x^2+C

(where

{C}

is an arbitrary constant) then the value of

{k}

is equal to

View answer and explanation

Correct option: A

Option A (3) is correct because by recognizing the exact differential and integrating, the value of k in the solution is found to be 3.

Practice All Chapter MCQs →

y=y(x)

\left(3 y^{2}-5 x^{2}\right) y d x+2 x\left(x^{2}-y^{2}\right) d y=0

y(1)=1

\left|(y(2))^{3}-12 y(2)\right|

x

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

\left( {y' = \frac{{dy}}{{dx}},y'' = \frac{{{d^2}y}}{{d{x^2}}}} \right)

x = A\cos 4t + B\sin 4t,

\frac{{{d^2}x}}{{d{t^2}}}

y=y_{1}(x)

y=y_{2}(x)

\frac{d y}{d x}=y+7

y_{1}(0)=0

y_{2}(0)=1

y=y_{1}(x)

y=y_{2}(x)

y=y(t)

\frac{d y}{d t}+\alpha y=\gamma e^{-\beta t}

\alpha>0, \beta>0

\gamma>0

\lim \limits_{{x \rightarrow \infty}} y(t)