Question

Let $y=y(x)$ be the solution of the differential equation
$\frac{dy}{dx}+\frac{\sqrt{2}y}{2{\cos }^{4}x-\cos 2x}=x{e}^{{\tan }^{-1}(\sqrt{2}\cot 2x)},0<x<\pi /2$ 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

Answer: 2

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