JEE MATHEMATICS MCQ: DIFFERENTIAL EQUATIONS

1 Easy

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Application +4 | -1

The solution of differential equation

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

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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.

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y=y(x)

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x

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

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y=y(t)

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

\alpha>0, \beta>0

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