Question

Let $y=y(x)$ be the solution of the differential equation
$\frac{d y}{d x}+\frac{\sqrt{2} y}{2 \cos ^{4} x-\cos 2 x}= xe ^{\tan ^{-1}(\sqrt{2} \cot 2 x )}, 0 < x < \pi / 2$ with $y\left(\frac{\pi}{4}\right)=\frac{\pi^{2}}{32}$
If $y\left(\frac{\pi}{3}\right)=\frac{\pi^{2}}{18} e^{-\tan ^{-1}(\alpha)}$, then the value of $3 \alpha^{2}$ is equal to ______.

Answer 1

$\frac{d y}{d x}+\frac{\sqrt{2}}{2 \cos ^{4} x-\cos 2 x} y=x e^{\tan ^{-1}(\sqrt{2} \cot 2 x)}$
$\int \frac{d x}{2 \cos ^{4} x-\cos 2 x}$
$=\int \frac{d x}{\cos ^{4} x+\sin ^{4} x}=\int \frac{\operatorname{cosec}^{4} x d x}{1+\cot ^{4} x}$
$=-\int \frac{t^{2}+1}{t^{4}+1} d t=-\int \frac{1}{\left(t-\frac{1}{t}\right)^{2}+2} d t$
$=\frac{-1}{\sqrt{2}} \tan ^{-1}\left(\frac{t-\frac{1}{t}}{\sqrt{2}}\right)$
$\operatorname{Cotx}=t$
$=-\frac{1}{\sqrt{2}} \tan ^{-1}(\sqrt{2} \cot 2 x)$
$\therefore IF =e^{-\tan ^{-1}(\sqrt{2} \cot 2 x )}$
$y e^{-\tan ^{-1}(\sqrt{2} \cot 2 x)}=\int x d x$
$y e^{-\tan ^{-1}(\sqrt{2} \cot 2 x)}=\frac{x^{2}}{2}+c$
$y\left(\frac{\pi}{4}\right)=\frac{\pi^{2}}{32}+c \Rightarrow c=0$
$y=\frac{x^{2}}{2} e^{\tan ^{-1}(\sqrt{2} \cot 2 x)}$
$y\left(\frac{\pi}{3}\right)=\frac{\pi^{2}}{18} e^{\tan ^{-1}\left(\sqrt{2} \cot \frac{2 \pi}{3}\right)}$
$=\frac{\pi^{2}}{18} e^{-\tan ^{-1}\left(\sqrt{\frac{2}{3}}\right)}$
$ \alpha=\frac{2}{3} \Rightarrow 3 \alpha^{2}=2$