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(C) The given differential equation is $\frac{dy}{dx} + P(x)y = Q(x)$,where $P(x) = \frac{\sqrt{2}}{2\cos^4 x - \cos 2x}$.First,simplify the denominat

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(C) The given differential equation is $\frac{dy}{dx} + P(x)y = Q(x)$,where $P(x) = \frac{\sqrt{2}}{2\cos^4 x - \cos 2x}$.First,simplify the denominator of $P(x)$:$2\cos^4 x - \cos 2x = 2\cos^4 x - (2\cos^2 x - 1) = 2\cos^4 x - 2\cos^2 x + 1 = \cos^4 x + \sin^4 x$.Now,calculate the integrating factor $IF = e^{\int P(x) dx}$:$\int P(x) dx = \int \frac{\sqrt{2} dx}{\cos^4 x + \sin^4 x} = \int \frac{\sqrt{2} \sec^4 x dx}{1 + 	an^4 x} = \int \frac{\sqrt{2}(1 + 	an^2 x) \sec^2 x dx}{1 + 	an^4 x}$.Let $	an x = u$,then $du = \sec^2 x dx$:$\int \frac{\sqrt{2}(1 + u^2) du}{1 + u^4} = \sqrt{2} \int \frac{1 + 1/u^2}{u^2 + 1/u^2} du = \sqrt{2} \int \frac{d(u - 1/u)}{(u - 1/u)^2 + 2} = 	an^{-1}\left(\frac{u - 1/u}{\sqrt{2}}ight) = 	an^{-1}\left(\frac{	an x - \cot x}{\sqrt{2}}ight) = 	an^{-1}\left(\frac{-2\cot 2x}{\sqrt{2}}ight) = -	an^{-1}(\sqrt{2}\cot 2x)$.Thus,$IF = e^{-	an^{-1}(\sqrt{2}\cot 2x)}$.The solution is $y \cdot IF = \int Q(x) \cdot IF dx = \int x dx = \frac{x^2}{2} + C$.Given $y(\pi/4) = \pi^2/32$,we have $\frac{\pi^2}{32} e^{-	an^{-1}(0)} = \frac{(\pi/4)^2}{2} + C \implies \frac{\pi^2}{32} = \frac{\pi^2}{32} + C \implies C = 0$.So,$y = \frac{x^2}{2} e^{	an^{-1}(\sqrt{2}\cot 2x)}$.At $x = \pi/3$,$y(\pi/3) = \frac{(\pi/3)^2}{2} e^{	an^{-1}(\sqrt{2}\cot(2\pi/3))} = \frac{\pi^2}{18} e^{	an^{-1}(\sqrt{2}(-1/\sqrt{3}))} = \frac{\pi^2}{18} e^{-	an^{-1}(\sqrt{2/3})}$.Comparing with $y(\pi/3) = \frac{\pi^2}{18} e^{-	an^{-1}(\alpha)}$,we get $\alpha = \sqrt{2/3}$,so $3\alpha^2 = 3(2/3) = 2$.

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 $y(\pi/4) = \pi^2/32$. If $y(\pi/3) = \frac{\pi^2}{18} e^{-\tan^{-1}(\alpha)}$,then the value of $3\alpha^2$ is equal to

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