JEE MATHEMATICS MCQ: DIFFERENTIAL EQUATIONS

1 Easy

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Application +4 | -1

Let

y = y\left( x \right)

be the solution of the differential equation,

\frac{{dy}}{{dx}} + y = x{\log _e}x,\,\left( {x > 1} \right).

2y\left( 2 \right) = {\log _e}\,4 - 1,

then

y(e)

is equal to

View answer and explanation

Correct option: C

\frac{e}{4}

Option C (

\frac{e}{4}

) is correct because it's the value of y(e) obtained by solving the given differential equation and using the 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

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x

{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)

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)