Question

Let $y=y\left(x\right)$ be solution of the following differential equation $e^y\frac{dy}{dx}-2e^y\sin x+\sin x{\cos }^{2}x=0,$ $y\left(\frac{\pi }{2}\right)=0$. If $y\left(0\right)={\log }_e(\alpha +\beta e^{–2}),$ then $4(\alpha +\beta )$ is equal to

Answer 1

$e^y\frac{dy}{dx}-2e^y\sin x+\sin x{\cos }^{2}x=0$
Let $e^y=t\Rightarrow \frac{dt}{dx}-2t\sin x=-\sin x\cdot {\cos }^{2}x$
I.F.$=e^{\int -2\sin xdx}=e^{2\cos x}$
$\Rightarrow t\cdot e^{2\cos x}=-\int e^{2\cos x}(\sin x\cdot {\cos }^{2}x)dx$
Let $I=-\int e^{2\cos x}(\sin x\cdot {\cos }^{2}x)dx$
Let $\cos x=u\Rightarrow -\sin xdx=du$
$I=\int e^{2u}{u}^{2}du$
$=u^2\frac{e^{2u}}{2}-\int u{e}^{2u}du=u^2\frac{e^{2u}}{2}-u\frac{e^{2u}}{2}+\frac{e^{2u}}{4}=e^{2\cos x}(\frac{{\cos }^{2}x}{2}-\frac{\cos x}{2}+\frac{1}{4})=\frac{1}{4}{e}^{2\cos x}(2{\cos }^{2}x-2\cos x+1)$
We get,
$e^{2\cos x}\cdot e^y=\frac{1}{4}{e}^{2\cos x}(2{\cos }^{2}x-2\cos x+1)+C$
$\because y\left(\frac{\pi }{2}\right)=0\Rightarrow C=\frac{3}{4}$
Now, at $x=0,$
$e^2\cdot e^{y\left(0\right)}=\frac{e^2}{4}(2-2+1)+\frac{3}{4}$
$\Rightarrow y\left(0\right)=\ln (\frac{1}{4}+\frac{3}{4}{e}^{-2})$
$\Rightarrow \alpha =\frac{1}{4}$ and $\beta =\frac{3}{4}$