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  • Let be the solution of the differential equation <img alt="\frac{\mathrm{d} y}{\mathrm{~d} x}+\frac{\sqrt{2} y}{

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

Option: 1

2


Option: 2

-


Option: 3

-


Option: 4

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Answers (1)

best_answer

\mathrm{\frac{d y}{d x}+\frac{\sqrt{2} y}{2 \cos ^{4} x-\cos 2 x}=x e^{\tan ^{-1}(\sqrt{2} \cot 2 x)}}

Now to get I.F, first find

\mathrm{ \int \frac{\sqrt{2}}{2 \cos ^{4} x-\cos 2 x} d x} \\

\mathrm{ = \int \frac{\sqrt{2}}{2 \cos ^{4} x-2 \cos ^{2} x+1} d x} \\

\mathrm{ = \int \frac{\sqrt{2}}{2 \cos ^{2} x\left(\cos ^{2} x-1\right)+1} d x} \\

\mathrm{= \int \frac{\sqrt{2}}{-2 \sin ^{2} x \cos ^{2} x+1} d x } \\

\mathrm{ = \int \frac{2 \sqrt{2}}{2-\sin ^{2}(2 x)} d x}

Dividing by \mathrm{ \cos^{2}2x}

\mathrm{=\int \frac{2 \sqrt{2} \sec ^{2} 2 x}{2 \sec ^{2} 2 x-\tan ^{2}(2 x)} d x}

\mathrm{=\int \frac{2 \sqrt{2} \sec ^{2} 2 x}{2+\tan ^{2}(2 x)} d x} \\

\mathrm{\text { Let } \tan 2 x=t \Rightarrow 2 \sec ^{2} 2 x d x=d t }\\

\mathrm{=\int \frac{\sqrt{2} d t}{t^{2}+2}=\tan ^{-1}\left(\frac{t}{\sqrt{2}}\right)} \\

\mathrm{=\tan ^{-1}\left(\frac{1}{\sqrt{2}} \tan 2 x\right)} \\

\mathrm{\therefore \text { I.F. }=e^{\tan ^{-1}\left(\frac{1}{\sqrt{2}} \tan 2 x\right)}}

 

Solution of equation is

\mathrm{y \cdot e^{\tan ^{-1}\left(\frac{1}{\sqrt{2}} \tan 2 x\right)}=\int x e^{\tan ^{-1}(\sqrt{2} \cot 2 x)} \cdot e^{\tan ^{-1}\left(\frac{1}{\sqrt{2}} \tan 2 x\right)} d x} \\

\mathrm{y \cdot e^{\tan ^{-1}\left(\frac{1}{\sqrt{2}} \tan 2 x\right)}=\int x e^{\pi / 2} d x}

\mathrm{\text { (As } \sqrt{2} \cot x \cdot \frac{1}{\sqrt{2}} \tan x=1 \text { ) }}\\

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

\mathrm{\Rightarrow y=\frac{x^{2}}{2} e^{\frac{\pi}{2}-\cot ^{-1}(\sqrt{2} \cot 2 x)}}+c e^{-cot^{-1}(\sqrt{2} \cot 2 x)}\\

\mathrm{\Rightarrow y=\frac{x^{2}}{2} e^{\tan ^{-1}(\sqrt{2} \cot 2 x)}+c e^{-cot^{-1}(\sqrt{2} \cot 2 x)}}\\

\mathrm{\text { As } y\left(\frac{\pi}{4}\right)=\frac{\pi^{2}}{32} \Rightarrow \frac{\pi^{2}}{32}=\frac{\pi^{2}}{32}+c \cdot e^{-\pi / 2} \Rightarrow c=0 }

\mathrm{Put \: x=\frac{\pi}{3} }\\

\mathrm{y\left(\frac{\pi}{3}\right)= \frac{\pi^{2}}{18} e^{\tan ^{-1}\left(-\frac{\sqrt{2}}{\sqrt{3}}\right)} \Rightarrow \alpha =\sqrt{\frac{2}{3}} \Rightarrow 3 \alpha^{2}=2}

Hence answer is 2

Posted by

Nehul

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