Question

$Le{t}_y-y\left(x\right)$ be the solution of the differential
equation $sinx\frac{dy}{dx}-ycosx-4x{,}_xϵ(0,\pi ).ify\left(\frac{\pi }{2}\right)$ = 0,
then y $\left(\frac{\pi }{6}\right)$ is equal to

• A. $-\frac{4}{9}{\pi }^2$
• B. $-\frac{4}{9\sqrt{3}}{\pi }^2$
• C. $-\frac{-8}{9\sqrt{3}}{\pi }^2$
• D. None of these

Answer 1

The correct option is D None of these

$\Rightarrow \frac{dy}{dx}-\cot x\cdot y=\frac{4x}{\sin x}$

This is a linear D.E

Integrating factor(I.F)$=e^{\int -\cot xdx}=e^{-ln|\sin x|}$

$\Rightarrow I.F=\frac{1}{\sin x}$

$\Rightarrow \frac{y}{\sin x}=\int \frac{4x}{{\sin }^{2}x}dx=\int \underset{I}{4x}\underset{II}{cose{c}^2}xdx$

Applying integration by parts,

$\frac{y}{\sin x}=4\left[x\right(-\cot x)-\int (-\cot x)dx]$

$=4[-x\cot x+ln\sin x]+c$

$\Rightarrow y=-4x\sin x+4\sin x\cdot ln\sin x+c\sin x$

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

$\Rightarrow 0$

$=-2\pi +0+c$

$\Rightarrow c=2\pi |ln|=0$

$\Rightarrow y=-4x\sin x+4\sin x\cdot ln\sin x+2\pi \sin x$

$y\left(\frac{\pi }{6}\right)\Rightarrow y=\frac{-2\pi }{3}\left(\frac{1}{2}\right)+4\left(\frac{1}{2}\right)(-ln2)+\pi$

$y=\frac{2\pi }{3}-2ln2$.