Question

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

• A. 2.00
• B. 2
• C. 2.000
• D. 2.0

Answer 1

Answer: (A), (B), (C), (D)

$\frac{dy}{dx}+\frac{2\sqrt{2}y}{1+{\cos }^{2}2x}=x{e}^{{\tan }^{-1}\left(\sqrt{2}\cot 2x\right)}$

$I.F.=\exp \left(\int \frac{2\sqrt{2}dx}{1+{\cos }^{2}2x}\right)=\exp \left(\sqrt{2}\int \frac{2{\sec }^{2}2x}{2+{\tan }^{2}2x}dx\right)$

$=\exp \left({\tan }^{-1}\left(\frac{\tan 2x}{\sqrt{2}}\right)\right)$
Solution is
$y\cdot e^{{\tan }^{-1}}\left(\frac{\tan 2x}{\sqrt{2}}\right)=\int x{e}^{{\tan }^{-1}}\left(\sqrt{2}\cot 2x\right)\cdot e^{{\tan }^{-1}}\left(\frac{\tan 2x}{\sqrt{2}}\right)dx+c$

$\Rightarrow y.e^{{\tan }^{-1}}\left(\frac{\tan 2x}{\sqrt{2}}\right)=e^{\pi /2}\cdot \frac{x^2}{2}+c$

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 $\frac{{\pi }^2}{18}{e}^{-{\tan }^{-1}\alpha }\cdot e^{-{\tan }^{-1}(-\sqrt{\frac{3}{2}})}=e^{\pi /2}\cdot \frac{{\pi }^2}{18}$
$\Rightarrow -{\tan }^{-1}\alpha +{\tan }^{-1}\left(\sqrt{\frac{3}{2}}\right)=\frac{\pi }{2}$

$\Rightarrow {\tan }^{-1}(-\alpha )={\tan }^{-1}\left(\sqrt{\frac{2}{3}}\right)$

$\Rightarrow \alpha =-\sqrt{\frac{2}{3}}\Rightarrow 3{\alpha }^2=2$