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(A) Given the differential equation: $\cos^2 x \frac{dy}{dx} + y = 	an x$. Dividing by $\cos^2 x$,we get: $\frac{dy}{dx} + y \sec^2 x = 	an x \sec^2

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$e$

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(A) Given the differential equation: $\cos^2 x \frac{dy}{dx} + y = 	an x$. Dividing by $\cos^2 x$,we get: $\frac{dy}{dx} + y \sec^2 x = 	an x \sec^2 x$. This is a linear differential equation of the form $\frac{dy}{dx} + Py = Q$,where $P = \sec^2 x$ and $Q = 	an x \sec^2 x$. The integrating factor $IF = e^{\int P dx} = e^{\int \sec^2 x dx} = e^{	an x}$. The general solution is $y(IF) = \int Q(IF) dx + C$. $y e^{	an x} = \int 	an x \sec^2 x e^{	an x} dx + C$. Let $u = 	an x$,then $du = \sec^2 x dx$. $y e^{	an x} = \int u e^u du + C = e^u(u - 1) + C$. $y e^{	an x} = e^{	an x}(	an x - 1) + C$. Dividing by $e^{	an x}$,we get $y = 	an x - 1 + Ce^{-	an x}$. Given $y(\frac{\pi}{4}) = 1$,we substitute $x = \frac{\pi}{4}$ and $y = 1$: $1 = 	an(\frac{\pi}{4}) - 1 + Ce^{-	an(\frac{\pi}{4})}$. $1 = 1 - 1 + Ce^{-1}$. $1 = Ce^{-1} \Rightarrow C = e$.

If the general solution of the differential equation $\cos^2 x \frac{dy}{dx} + y = \tan x$ is $y = \tan x - 1 + Ce^{-\tan x}$ and it satisfies $y(\frac{\pi}{4}) = 1$,then $C =$

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