The solution of differential equation $\left( {1 + x} \right)y{\rm{ }}dx + \left( {1 - y} \right)x{\rm{ }}dy = 0$ is

${\log _e}\left( {\frac{x}{y}} \right) + x + y = c$

${\log _e}\left( {\frac{x}{y}} \right) - x + y = c$

JEE MATHEMATICS MCQ: DIFFERENTIAL EQUATIONS

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Application +4 | -1

The solution of differential equation

\left( {1 + x} \right)y\mathrm{ }dx + \left( {1 - y} \right)x\mathrm{ }dy = 0

View answer and explanation

Correct option: A

{\log _e}\left( {xy} \right) + x - y = c

Option A (

{\log _e}\left( {xy} \right) + x - y = c

) is correct because it is the solution obtained by separating variables and integrating the given differential equation.

Practice All Chapter MCQs →

y=y(x)

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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}}}

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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)