Mathematics - Differential Equation Question with Solution | TestHub

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

I.F.$=e^{\int \cot x}=e^{\ln \sin x}=\sin x$

$\therefore y \sin x=\int \frac{4x}{\sin x}.\sin x dx$

$y\sin x=2x^2+c$

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

$\therefore 0\times 1=2\times {\left(\frac{\pi }{2}\right)}^2+c$

$\Rightarrow c=-\frac{{\pi }^2}{2}$

$y\left(\frac{\pi }{6}\right)=\frac{2\times {\left(\frac{\pi }{6}\right)}^2-\left(\frac{{\pi }^2}{2}\right)}{\frac{1}{2}}=\left[\left(\frac{{\pi }^2}{18}\right)-\left(\frac{{\pi }^2}{2}\right)\right]\times 2$

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

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