JEE MATHEMATICS MCQ: DIFFERENTIAL EQUATIONS

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Application +4 | -1

y=0

when

x=1

, then a particular solution of the equation

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

View answer and explanation

Correct option: C

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

Option D (

y(1+x^2)=\operatorname{{Tan}}^{-1} x+\frac{{\pi}}{{4}}

) is incorrect. The correct answer is C (

y(1+x^2)=\frac{{\pi}}{{4}}-\operatorname{{Cot}}^{-1} x

). The error likely stems from incorrect handling of the initial condition or trigonometric identities.

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)