Question

Let y = y(x) be the solution of the differential equation $\frac{dy}{dx}+\frac{√2y}{2cos^4x−cos2x}=xe^{tan−1}(√2cot2x),0<x<\frac{π}{2}$ with $y(\frac{π}{4})=\frac{π^2}{32}$ If $y(\frac{π}{3})=\frac{π^2}{18}e^{−tan−1}(α)$ then the value of $3α^2$ is equal to ______

Answer 1

The correct answer is: 2

$\frac{dy}{dx}+\frac{√2y}{2cos^4x−cos2x}=xe^{tan−1}(√2cot2x)$

$I,F.=e^{∫}\frac{2\sqrt2dx}{1+cos^22x}=e^{\sqrt2\,∫}\frac{2\,sec^22x}{2+tan^22x}$
= $e^{\tan^{-1}\left(\frac{\tan(2x)}{\sqrt{2}}\right)}$
$⇒$ $y \cdot e^{\tan^{-1}\left(\frac{\tan(2x)}{\sqrt{2}}\right)}$ = $\int x e^{\tan^{-1}(\sqrt{2}\cot(2x))} \cdot e^{\tan^{-1}\left(\frac{\tan(2x)}{\sqrt{2}}\right)} \, dx + c$
$⇒$ $y \cdot e^{\tan^{-1}\left(\frac{\tan(2x)}{\sqrt{2}}\right)}$
$= e^{\frac{\pi}{2}}.\frac{x^2}{2}+c$

When

$x=\frac{\pi}{4},y=\frac{\pi^2}{32}$ gives $c=0.$
When
$x=\frac{\pi}{3},y=\frac{\pi^2}{18}e^{tan^{-1}\alpha}$

So, $\frac{\pi^2}{18}e^{tan^{-1}\alpha}.e^{-tan{-1}(-\sqrt{\frac{3}{2}})}=e^{\frac{\pi}{2}}\frac{\pi^2}{18}$
$⇒ tan^{-1}(-\alpha)=tan^{-1}\left(\sqrt{\frac{3}{2}}\right)$

$⇒α=-\sqrt\frac{2}{3}⇒3a^2=2$