If -frac{pi}{4}

Question

(C) Given differential equation: $\cos ^2 x \frac{d y}{d x}-(	an 2 x) y=\cos ^4 x$. Dividing by $\cos ^2 x$,we get: $\frac{d y}{d x} - \frac{	an 2 x

Vedclass · Accepted Answer

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

Vedclass · Answer

(C) Given differential equation: $\cos ^2 x \frac{d y}{d x}-(	an 2 x) y=\cos ^4 x$. Dividing by $\cos ^2 x$,we get: $\frac{d y}{d x} - \frac{	an 2 x}{\cos ^2 x} y = \cos ^2 x$. This is a linear differential equation of the form $\frac{d y}{d x} + P y = Q$,where $P = -\frac{	an 2 x}{\cos ^2 x}$ and $Q = \cos ^2 x$. Integrating factor $IF = e^{\int P dx} = e^{\int -\frac{	an 2 x}{\cos ^2 x} dx}$. Let $I = \int -\frac{	an 2 x}{\cos ^2 x} dx = \int -\frac{\sin 2 x}{\cos 2 x \cdot \cos ^2 x} dx$. Using $\cos^2 x = \frac{1+\cos 2 x}{2}$,we get $I = \int -\frac{2 \sin 2 x}{\cos 2 x (1+\cos 2 x)} dx$. Let $\cos 2 x = t$,then $-2 \sin 2 x dx = dt$. $I = \int \frac{dt}{t(t+1)} = \int (\frac{1}{t} - \frac{1}{t+1}) dt = \ln|t| - \ln|t+1| = \ln|\frac{t}{t+1}|$. Thus,$IF = \frac{\cos 2 x}{\cos 2 x + 1} = \frac{\cos 2 x}{2 \cos^2 x}$. The solution is $y \cdot IF = \int Q \cdot IF dx + c$. $y \cdot \frac{\cos 2 x}{2 \cos^2 x} = \int \cos^2 x \cdot \frac{\cos 2 x}{2 \cos^2 x} dx + c = \frac{1}{2} \int \cos 2 x dx + c = \frac{1}{4} \sin 2 x + c$. $y = \frac{2 \cos^2 x}{\cos 2 x} (\frac{\sin 2 x + 4c}{4}) = \frac{1}{2} \frac{\cos^2 x}{\cos 2 x} (\sin 2 x + C)$. Using $\cos 2 x = \cos^2 x - \sin^2 x = \cos^2 x (1 - 	an^2 x)$,we get $y = \frac{1}{2} \frac{\sin 2 x + C}{1 - 	an^2 x}$.

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

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