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(A) The given differential equation is $\frac{dy}{dx} + (2 \sec^2 x)y = 2 \sec^2 x + 3 	an x \sec^2 x$.This is a linear differential equation of the

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$21$

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(A) The given differential equation is $\frac{dy}{dx} + (2 \sec^2 x)y = 2 \sec^2 x + 3 	an x \sec^2 x$.This is a linear differential equation of the form $\frac{dy}{dx} + P(x)y = Q(x)$,where $P(x) = 2 \sec^2 x$ and $Q(x) = 2 \sec^2 x + 3 	an x \sec^2 x$.Integrating factor ($I$.$F$.) $= e^{\int 2 \sec^2 x dx} = e^{2 	an x}$.The solution is $y \cdot e^{2 	an x} = \int (2 \sec^2 x + 3 	an x \sec^2 x) e^{2 	an x} dx$.Let $u = 	an x$,then $du = \sec^2 x dx$. The integral becomes $\int (2 + 3u) e^{2u} du$.Using integration by parts: $\int (2 + 3u) e^{2u} du = (2 + 3u) \frac{e^{2u}}{2} - \int 3 \frac{e^{2u}}{2} du = (1 + \frac{3}{2}u) e^{2u} - \frac{3}{4} e^{2u} + C = (\frac{3}{2}u + \frac{1}{4}) e^{2u} + C$.So,$y \cdot e^{2 	an x} = (\frac{3}{2} 	an x + \frac{1}{4}) e^{2 	an x} + C$.Dividing by $e^{2 	an x}$,we get $y = \frac{3}{2} 	an x + \frac{1}{4} + C e^{-2 	an x}$.Given $y(0) = \frac{5}{4}$,so $\frac{5}{4} = 0 + \frac{1}{4} + C \implies C = 1$.Thus,$y(x) = \frac{3}{2} 	an x + \frac{1}{4} + e^{-2 	an x}$.At $x = \frac{\pi}{4}$,$y(\frac{\pi}{4}) = \frac{3}{2}(1) + \frac{1}{4} + e^{-2} = \frac{7}{4} + e^{-2}$.Therefore,$12(y(\frac{\pi}{4}) - e^{-2}) = 12(\frac{7}{4}) = 21$.

Let $y=y(x)$ be the solution of the differential equation $\frac{dy}{dx}+2y \sec^2 x = 2 \sec^2 x + 3 \tan x \cdot \sec^2 x$ such that $y(0)=\frac{5}{4}$. Then $12\left(y\left(\frac{\pi}{4}\right)-e^{-2}\right)$ is equal to . . . . . . .

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