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If $-\frac{\pi}{4} < x < \frac{\pi}{4}$, then the general solution of the differential equation $\cos ^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
$y=\frac{1}{2}\left[\frac{\tan 2 x+c}{1-\tan ^2 x}\right]$ $y=\frac{1}{2}\left[\frac{\cos 2 x+c}{1-\tan ^2 x}\right]$ $y=\frac{1}{2}\left[\frac{\sin 2 x+c}{1-\tan ^2 x}\right]$ $y=\frac{1}{2}\left[\frac{\sin x+c}{1-\tan ^2 x}\right]$ Solution
$\begin{aligned} & \cos ^2 x \frac{d y}{d x}-\tan 2 x \cdot y=\cos ^4 x \\ & \Rightarrow \frac{d y}{d x}-\frac{\tan 2 x}{\cos ^2 x} y=\cos ^2 x\end{aligned}$
the above differential equation is in the form of.
$
\frac{d y}{d x}+p y=q \text {, }
$
where $p=\frac{-\tan 2 x}{\cos ^2 x}$
and $q=\cos ^2 x$
Now, IF $=e^{\int p d x}=e^{\int-\frac{\tan 2 x}{\cos ^2 x} d x}$
Let $\begin{aligned} I & =\int-\frac{\tan 2 x}{\cos ^2 x} d x=\int \frac{-\sin 2 x}{\cos 2 x \cdot \cos ^2 x} d x \\ & =\int \frac{-2 \sin 2 x}{\cos 2 x(1+\cos 2 x)} d x\end{aligned}$
put $\cos 2 x=t$
$-2 \sin 2 x d x=d t$
$
\begin{aligned}
I & =\int \frac{d t}{t(t+1)}=\int \frac{d t}{t^2+t+\left(\frac{1}{2}\right)^2-\left(\frac{1}{2}\right)^2} \\
& =\int \frac{d t}{\left(t+\frac{1}{2}\right)^2-\left(\frac{1}{2}\right)^2}=\log \left|\frac{t}{t+1}\right|
\end{aligned}
$
Asked in: AP EAMCET 2018 (23 Apr Shift 1)
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