Let $y=y(x)$ be the solution of the differential equation $\sin x \frac{\mathrm{~d} y}{\mathrm{~d} x}+y \cos…
Let $y=y(x)$ be the solution of the differential equation $\sin x \frac{\mathrm{~d} y}{\mathrm{~d} x}+y \cos x=4 x, x \in(0, \pi)$.
If $y\left(\frac{\pi}{2}\right)=0$, then $y\left(\frac{\pi}{6}\right)$ is equal to
$-\frac{4}{9} \pi^2$
$\frac{4}{9 \sqrt{3}} \pi^2$
$\frac{-8}{9 \sqrt{3}} \pi^2$
$-\frac{8}{9} \pi^2$
Solution
$\begin{aligned}
& \sin x \frac{\mathrm{~d} y}{\mathrm{~d} x}+y \cos x=4 x \\
& \therefore \quad \frac{\mathrm{~d} y}{\mathrm{~d} x}+y \cot x=\frac{4 x}{\sin x} \\
& \text { Here, } \mathrm{P}(x)=\cot x, \mathrm{Q}(x)=\frac{4 x}{\sin x} \\
& \text { Integrating factor (I.F.) }=\mathrm{e}^{\int \mathrm{P}(x) d x} \\
& =\mathrm{e}^{\int \operatorname{oot} x d x},=\mathrm{e}^{\log |\sin x|}=\sin x
\end{aligned}$
$\begin{array}{ll}
\therefore & y(\text { I.F. })=\int \mathrm{Q}(\text { I.F. }) \mathrm{d} x+\mathrm{c} \\
\therefore & y(\sin x)=\int \frac{4 x}{\sin x} \times \sin x \mathrm{~d} x=\mathrm{C} \\
\therefore & y \sin x=4 \int x \mathrm{~d} x+\mathrm{C} \\
\therefore & y=\frac{2 x^2+\mathrm{C}}{\sin x} \\
& y\left(\frac{\pi}{2}\right)=0
\end{array}$...(i)
$\begin{aligned} & \Rightarrow \frac{2\left(\frac{\pi}{2}\right)^2+C}{\sin \left(\frac{\pi}{2}\right)}=0 \\ & \Rightarrow \frac{\pi^2}{2}+C=0 \\ & \Rightarrow C=\frac{-\pi^2}{2}\end{aligned}$
$\Rightarrow y=\frac{2 x^2-\frac{\pi^2}{2}}{\sin x}$
is the solution of the given differential equation.
$\begin{aligned}
\therefore y\left(\frac{\pi}{6}\right) & =\frac{2\left(\frac{\pi}{6}\right)^2-\frac{\pi^2}{2}}{\sin \left(\frac{\pi}{6}\right)} \\
& =\frac{\frac{2 \pi^2}{36}-\frac{\pi^2}{2}}{\frac{1}{2}}=\frac{-8}{9} \pi^2
\end{aligned}$