JEE MATHEMATICS MCQ: DIFFERENTIAL EQUATIONS

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Application +4 | -1

The solution of

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

\left.y\right|_{x=0}=1

is given by

View answer and explanation

Correct option: D

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

The correct answer is D (

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

) because it is the solution of the linear differential equation with the given initial condition.

More Questions from DIFFERENTIAL EQUATIONS

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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)