If -frac{pi}{4}

Question

(C) Given,$\cos^{2} x \cdot \frac{dy}{dx} - (	an 2x) y = \cos^{4} x$ Dividing by $\cos^{2} x$,we get: $\frac{dy}{dx} - \left(\frac{	an 2x}{\cos^{2}

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$y = \frac{1}{2} \left[ \frac{\sin 2x + c}{1 - 	an^{2} x} ight]$

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(C) Given,$\cos^{2} x \cdot \frac{dy}{dx} - (	an 2x) y = \cos^{4} x$ Dividing by $\cos^{2} x$,we get: $\frac{dy}{dx} - \left(\frac{	an 2x}{\cos^{2} x}ight) y = \cos^{2} x$ This is a linear differential equation of the form $\frac{dy}{dx} + Py = Q$,where $P = -\frac{	an 2x}{\cos^{2} x}$ and $Q = \cos^{2} x$. The Integrating Factor $(I.F.)$ is $e^{\int P dx}$. $\int P dx = -\int \frac{	an 2x}{\cos^{2} x} dx = -\int \frac{2 	an x}{(1 - 	an^{2} x) \cos^{2} x} dx$. Let $	an x = t$,then $\sec^{2} x dx = dt$. $\int P dx = -\int \frac{2t}{1 - t^{2}} dt = \ln |1 - t^{2}| = \ln |1 - 	an^{2} x|$. Thus,$I.F. = e^{\ln |1 - 	an^{2} x|} = 1 - 	an^{2} x$. The general solution is $y \cdot I.F. = \int (Q \cdot I.F.) dx + C$. $y(1 - 	an^{2} x) = \int (\cos^{2} x (1 - 	an^{2} x)) dx + C$. $y(1 - 	an^{2} x) = \int (\cos^{2} x - \sin^{2} x) dx + C$. $y(1 - 	an^{2} x) = \int \cos 2x dx + C$. $y(1 - 	an^{2} x) = \frac{\sin 2x}{2} + C$. $y = \frac{1}{2} \left[ \frac{\sin 2x + 2C}{1 - 	an^{2} x} ight]$. Replacing $2C$ with $c$,we get $y = \frac{1}{2} \left[ \frac{\sin 2x + c}{1 - 	an^{2} x} ight]$.

If $-\frac{\pi}{4} < x < \frac{\pi}{4},$ then the general solution of the differential equation $\cos^{2} x \cdot \frac{dy}{dx} - (\tan 2x) y = \cos^{4} x$ is

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