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

Correct Answer: 2

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

\(⇒α=-\sqrt\frac{2}{3}⇒3a^2=2\)