JEE Main 2021 — Differential Equations Question with Solution

Let $e^y=t\Rightarrow e^y\frac{dy}{dx}=\frac{dt}{dx}$

So,

$e^y\frac{dy}{dx}-2e^y\sin x+\sin x{\cos }^{2}x=0$

$\Rightarrow \frac{dt}{dx}-\left(2\sin x\right)t=-\sin x{\cos }^{2}x$

So, the integrating factor $I.F.=e^{\int -2\sin xdx}={\mathrm{e}}^{2\cos x}$

So, the solution of differential equation is 

$t·e^{2\cos x}=\int e^{2\cos x}·\left(-\sin x{\cos }^{2}x\right)dx$

$\Rightarrow e^y·e^{2\cos x}=\int e^{2z}·z^2 dz$  (Assume, $z=\cos x\Rightarrow dz=-\sin x$)

Using integration by parts,

$\Rightarrow {\mathrm{e}}^y·{\mathrm{e}}^{2\cos x}=\frac{1}{2}·z^2·e^{2z}-\frac{1}{2}z·{\mathrm{e}}^{2z}+\frac{{\mathrm{e}}^{2z}}{4}+\mathrm{C}$

$\Rightarrow {\mathrm{e}}^y·{\mathrm{e}}^{2\cos x}=\frac{1}{2}·{\cos }^{2}x·e^{2\cos x}-\frac{1}{2}\cos x·{\mathrm{e}}^{2\cos x}+\frac{{\mathrm{e}}^{2\cos x}}{4}+\mathrm{C}$

at $x=\frac{\pi }{2},y=0\Rightarrow C=\frac{3}{4}$

$\Rightarrow {\mathrm{e}}^y=\frac{1}{2}{\cos }^{2}x-\frac{1}{2}\cos x+\frac{1}{4}+\frac{3}{4}·{\mathrm{e}}^{-2\cos x}$

Put $x=0$

$\Rightarrow y=\log \left[\frac{1}{4}+\frac{3}{4}{\mathrm{e}}^{-2}\right]\Rightarrow \alpha =\frac{1}{4},\beta =\frac{3}{4}$

So, $\alpha +\beta =\frac{1}{4}+\frac{3}{4}=1$