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(A) The given differential equation is of the form $\frac{dy}{dx} + Py = Q$,where $P = \sec^2 x$ and $Q = 	an x \sec^2 x$.First,we find the Integrati

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$y = 	an x - 1 + c e^{-	an x}$

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(A) The given differential equation is of the form $\frac{dy}{dx} + Py = Q$,where $P = \sec^2 x$ and $Q = 	an x \sec^2 x$.First,we find the Integrating Factor $(I.F.)$:$I.F. = e^{\int P \, dx} = e^{\int \sec^2 x \, dx} = e^{	an x}$.The general solution is given by $y \cdot (I.F.) = \int Q \cdot (I.F.) \, dx + c$.Substituting the values:$y e^{	an x} = \int (	an x \sec^2 x) e^{	an x} \, dx + c$.Let $u = 	an x$,then $du = \sec^2 x \, dx$.The integral becomes $\int u e^u \, du$.Using integration by parts: $\int u e^u \, du = u e^u - \int e^u \, du = u e^u - e^u = e^u(u - 1)$.Substituting back $u = 	an x$:$y e^{	an x} = e^{	an x}(	an x - 1) + c$.Dividing both sides by $e^{	an x}$:$y = 	an x - 1 + c e^{-	an x}$.

The solution of the differential equation $\frac{dy}{dx} + y \sec^2 x = \tan x \sec^2 x$ is

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