Let $y = y\left( x \right)$ be the solution of the differential equation $\sin x\frac{{dy}}{{dx}} + y\cos x = 4x,x \in \left( {0,\pi } \right)$ If $y\left( {\frac{\pi }{2}} \right) = 0$, then $y\left( {\frac{\pi }{6}} \right)$ is equal to

$\frac{4}{{9\sqrt 3 }}{\pi ^2}$

$\frac{{ - 8}}{{9\sqrt 3 }}{\pi ^2}$

JEE MATHEMATICS MCQ: DIFFERENTIAL EQUATIONS

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Application +4 | -1

Let

y = y\left( x \right)

be the solution of the differential equation

\sin x\frac{{dy}}{{dx}} + y\cos x = 4x,x \in \left( {0,\pi } \right)

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

, then

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

is equal to

View answer and explanation

Correct option: C

- \frac{8}{9}{\pi ^2}

Option C (

- \frac{8}{9}{\pi ^2}

) is correct because it is the value of

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

obtained by solving the given differential equation and applying the initial condition.

Practice All Chapter MCQs →

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)