Question

Let $\frac{dy}{dx}-2y\cot x=\cos x$ such that $y\left(\frac{\pi }{2}\right)=0$. If the maximum value of $y$ is $k,$ then the value of $k$ is

Answer 1

$\frac{dy}{dx}-2y\cot x=\cos x$
This is a linear differential equation in $x.$
$\text{I.F.}=e^{-2\int \cot xdx}=\frac{1}{{\sin }^{2}x}$
General solution is
$\frac{y}{{\sin }^{2}x}=\int \frac{\cos x}{{\sin }^{2}x}dx+C$
$\Rightarrow y=-\sin x+C{\sin }^{2}x$
Since $y\left(\frac{\pi }{2}\right)=0,$ therefore $C=1$
$\therefore y={\sin }^{2}x-\sin x$
$\Rightarrow y={(\sin x-\frac{1}{2})}^2-\frac{1}{4}$
$\therefore$ Maximum value of $y$ is ${(-1-\frac{1}{2})}^2-\frac{1}{4}=2$