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

Let e^y = t<br><br>$\Rightarrow {{dt} \over {dx}} - (2\sin x)t = - \sin x{\cos ^2}x$<br><br>$I.F. = {e^{2\cos x}}$<br><br>$\Rightarrow t.{e^{2\cos x}} = \int {{e^{2\cos x}}.( - \sin x{{\cos }^2}x)dx}$<br><br>$\Rightarrow {e^y}.{e^{2\cos x}} = \int {{e^{2z}}.{z^2}dz,z = {e^{2\cos x}}}$<br><br>$$ \Rightarrow {e^y}.{e^{2\cos x}} = {1 \over 2}.{\cos ^2}x.{e^{2\cos x}} - {1 \over 2}\cos x.{e^{2\cos x}} + {{{e^{2\cos x}}} \over 4} + C$$<br><br>at $x = {\pi \over 2},y = 0 \Rightarrow C = {3 \over 4}$<br><br>$$ \Rightarrow {e^y} = {1 \over 2}{\cos ^2}x - {1 \over 2}\cos x + {1 \over 4} + {3 \over 4}.{e^{ - 2\cos x}}$$<br><br>$$ \Rightarrow y = \log \left[ {{{{{\cos }^2}x} \over 2} - {{\cos x} \over 2} + {1 \over 4} + {3 \over 4}{e^{ - 2\cos x}}} \right]$$<br><br>Put x = 0<br><br>$$ \Rightarrow y = \log \left[ {{1 \over 4} + {3 \over 4}{e^{ - 2}}} \right] \Rightarrow \alpha = {1 \over 4},\beta = {3 \over 4}$$