# Let y = y(x) be solution of the following

Question:

Let $\mathrm{y}=\mathrm{y}(\mathrm{x})$ be solution of the following differential equation

$e^{y} \frac{d y}{d x}-2 e^{y} \sin x+\sin x \cos ^{2} x=0, y\left(\frac{\pi}{2}\right)=0$

If $y(0)=\log _{e}\left(\alpha+\beta e^{-2}\right)$, then $4(\alpha+\beta)$ is equal to_______.

Solution:

$\operatorname{Let} e^{y}=t$

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

I.F. $=e^{2 \cos x}$

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

$\Rightarrow \mathrm{e}^{\mathrm{y}} \cdot \mathrm{e}^{2 \cos \mathrm{x}}=\int \mathrm{e}^{2 \mathrm{z}} \cdot \mathrm{z}^{2} \mathrm{~d} \mathrm{z}, \mathrm{z}=\mathrm{e}^{2 \cos \mathrm{x}}$

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

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

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

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

Put $x=0$

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