JEE MATHEMATICS MCQ: DIFFERENTIAL EQUATIONS

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Application +4 | -1

y=y(x), x \in\left(0, \frac{\pi}{2}\right)

be the solution curve of the differential equation

\left(\sin ^2 2 x\right) \frac{d y}{d x}+\left(8 \sin ^2 2 x+\right.

2 \sin 4 x) y=2 e^{-4 x}(2 \sin 2 x+\cos 2 x)

, with

y\left(\frac{\pi}{4}\right)=e^{-\pi}

then

y\left(\frac{\pi}{6}\right)

is equal to

View answer and explanation

Correct option: A

\frac{2}{\sqrt{3}} e^{-2 \pi / 3}

Option A (

\frac{2}{\sqrt{3}} e^{-2 \pi / 3}

) is correct because it is the value of y at x=

\frac{\pi}{6}

after solving the given differential equation with the specified 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

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)