If y = f(x) is the solution of the differenti | vedclass

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

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$e - 2$

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(C) The given differential equation is $\frac{dy}{dx} + y \sec^2 x = 	an x \sec^2 x$. 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 is $IF = e^{\int \sec^2 x dx} = e^{	an x}$. The solution is $y \cdot e^{	an x} = \int (	an x \sec^2 x) e^{	an x} dx + C$. Let $t = 	an x$,then $dt = \sec^2 x dx$. The integral becomes $\int t e^t dt = t e^t - e^t + C$. So,$y e^{	an x} = e^{	an x} (	an x - 1) + C$. Given $y(0) = 0$,we have $0 = e^0 (0 - 1) + C$,which implies $C = 1$. Thus,$y = 	an x - 1 + e^{-	an x}$. For $x = -\frac{\pi}{4}$,$	an x = -1$. $y\left( -\frac{\pi}{4} ight) = -1 - 1 + e^{-(-1)} = e - 2$.

If $y = f(x)$ is the solution of the differential equation $\frac{dy}{dx} = (\tan x - y) \sec^2 x$,$x \in \left( -\frac{\pi}{2}, \frac{\pi}{2} \right)$,such that $y(0) = 0$,then $y\left( -\frac{\pi}{4} \right)$ is equal to

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