JEE MATHEMATICS MCQ: DIFFERENTIAL EQUATIONS

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Application +4 | -1

x = x\left( y \right)

is the solution of the differential equation

y\frac{{dx}}{{dy}} = 2x + {y^3}\left( {y + 1} \right){e^y},\,x\left( 1 \right) = 0;

then

x\left( e \right)

is equal to:

View answer and explanation

Correct option: A

{e^3}\left( {{e^e} - 1} \right)

Option A (

{e^3}({e^e} - 1)

) is correct because it is the solution to the given differential equation with the specified initial condition, evaluated at y = e.

More Questions from DIFFERENTIAL EQUATIONS

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y=y(x)

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

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

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

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y=y_{2}(x)

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

y_{1}(0)=0

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y=y_{1}(x)

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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)