Let y = y ( x ) be solution of the following differential equation...

Let ${e}^{y} = t ⇒ {e}^{y}\dfrac{dy}{dx} = \dfrac{dt}{dx}$

So,

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

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

So, the integrating factor $I . F . = {e}^{∫ - 2sinxdx} = {e}^{2\cos x}$

So, the solution of differential equation is

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

$⇒ {e}^{y} \cdot {e}^{2\cos x} = ∫ {e}^{2z} \cdot {z}^{2}dz$ (Assume, $z = \cos x ⇒ dz = - \sin x$)

Using integration by parts,

$⇒ {e}^{y} \cdot {e}^{2\cos x} = \dfrac{1}{2} \cdot {z}^{2} \cdot {e}^{2z} - \dfrac{1}{2}z \cdot {e}^{2z} + \dfrac{{e}^{2z}}{4} + C$

$⇒ {e}^{y} \cdot {e}^{2\cos x} = \dfrac{1}{2} \cdot \cos^{2}x \cdot {e}^{2\cos x} - \dfrac{1}{2}\cos x \cdot {e}^{2\cos x} + \dfrac{{e}^{2\cos x}}{4} + C$

at $x = \dfrac{\pi}{2}, y = 0 ⇒ C = \dfrac{3}{4}$

$⇒ {e}^{y} = \dfrac{1}{2}\cos^{2}x - \dfrac{1}{2}\cos x + \dfrac{1}{4} + \dfrac{3}{4} \cdot {e}^{- 2\cos x}$

Put $x = 0$

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

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