Question

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

• A. $-\frac{8}{9}{\pi }^2$
• B. $-\frac{4}{9}{\pi }^2$
• C. $\frac{4}{9\sqrt{3}}{\pi }^2$
• D. $\frac{-8}{9\sqrt{3}}{\pi }^2$

Answer 1

The correct option is A $-\frac{8}{9}{\pi }^2$

Given differential equation:

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

$\therefore \frac{dy}{dx}+y\frac{\cos x}{\sin x}=\frac{4x}{\sin x}$

$\frac{dy}{dx}+\cot xy=\frac{4x}{\sin x}$

Comparing with $\frac{dy}{dx}+P\left(x\right)y=Q\left(x\right)$

$\therefore P\left(x\right)=\cot \left(x\right),Q\left(x\right)=\frac{4x}{\sin x}$

I.F $=e^{\int P\left(x\right)dx}$

$=e^{\int \cot xdx}$

$=e^{\int \cot xdx}=e^{\int log\left(|\sin x|\right)}dx=\sin x$

$\therefore$ solution is given by

$yI.F=\int Q\left(x\right)IFdx+c$

$y\sin x=\int \frac{4x}{\sin x}\sin xdx+c$

$=\int 4xdx+c$

$=\frac{4x^2}{2}+c$

$=2x^2+c$

$\therefore y\sin x=2x^2+c$..................(A)

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

i.e when $x=\frac{\pi }{2},y=0$

Equation becomes : $0=2{\left(\frac{\pi }{2}\right)}^2+c$

$0=2\times \frac{{\pi }^2}{4}+c$

$\therefore c=\frac{{\pi }^2}{2}$

put this value in (A) $\therefore y\sin x=2x^2-\frac{{\pi }^2}{2}$

Next to find $y\left(\frac{\pi }{6}\right)$ take $x=\frac{\pi }{6}$

$\therefore y\sin \frac{\pi }{6}=2{\left(\frac{\pi }{6}\right)}^2-\frac{{\pi }^2}{2}$

$y=\frac{1}{2}=2.\frac{{\pi }^2}{36}-\frac{{\pi }^2}{2}$

$\frac{y}{2}=\frac{{\pi }^2}{18}-\frac{{\pi }^2\times 9}{2\times 9}$

$\frac{y}{2}=\frac{{\pi }^2-9{\pi }^2}{18}$

$y=\frac{-8{\pi }^2}{9}$

$\therefore y\left(\frac{\pi }{6}\right)=-\frac{8}{9}{\pi }^2$