Question

Solution of the equation
$\cos ^{2} x \frac{d y}{d x}-(\tan 2 x) y=\cos ^{4} x,|x|<\frac{\pi}{4},$ where $y\left(\frac{\pi}{6}\right)=\frac{3 \sqrt{3}}{8}$, is given by

• A. $y \frac{\tan 2 x}{1-\tan ^{2} x}=0$
• B. $y\left(1-\tan ^{2} x\right)=C$
• C. $y=\sin 2 x +C$
• D. $y=\frac{1}{2} \cdot \frac{\sin 2 x}{1-\tan ^{2} x}$

Answer 1

Answer: (D)

Given differential equation can be written as $\frac{d y}{d x}-\left(\frac{\tan 2 x}{\cos ^{2} x}\right) y=\cos ^{2} x$
Here,
$P=-\frac{\tan 2 x}{\cos ^{2} x}=\frac{-\sin 2 x}{\cos 2 x\left(\frac{\cos 2 x+1}{2}\right)}$
and $Q=\cos ^{2} x$
$\because I F=e^{\int P d x}=e^{-\int \frac{2 \sin 2 x}{\cos 2 x(\cos 2 x+1)} d x}$
Put $\cos 2 x=t \Rightarrow-2 \sin 2 x d x=d t$
$\therefore I F=e^{\int_{t}^{1} \frac{1}{t}\left(\frac{1}{t+1}\right) d t}$
$=e^{\int\left(\frac{1}{t}-\frac{1}{t+1}\right) d t}$
$=e^{\log t-\log (t+1)]}=e^{\log \frac{t}{t+1}}$
$=e^{\log _{\cos 2 z-1}}=\frac{\cos 2 x}{\cos 2 x+1}$
Now, solution is
$y \times \frac{\cos 2 x}{\cos 2 x+1}=\int \frac{\cos 2 x}{\cos 2 x+1} \times \cos ^{2} x d x +C$
$=\int \frac{\cos 2 x}{2 \cos ^{2} x} \times \cos ^{2} x d x+ C$
$=\frac{1}{2} \int \cos 2 x d x+ C$
$=\frac{1}{2} \times \frac{\sin 2 x}{2}+C$
$\Rightarrow y \frac{\cos 2 x}{\cos 2 x+1}=\frac{1}{4} \sin 2 x +C$ ... (i)
But $y\left(\frac{\pi}{6}\right)=\frac{3 \sqrt{3}}{8}$
$\therefore \frac{3 \sqrt{3}}{8} \times \frac{\cos \left(2 \times \frac{\pi}{6}\right)}{\cos \left(2 \times \frac{\pi}{6}\right)+1}$
$=\frac{1}{4} \sin 2\left(\frac{\pi}{6}\right)+C$
$\Rightarrow \frac{3 \sqrt{3}}{8 \frac{1}{2}+1}=\frac{1}{4} \times \frac{\sqrt{3}}{2}+C$
$\Rightarrow \frac{3 \sqrt{3}}{2 \times 8 \times \frac{3}{2}}=\frac{\sqrt{3}}{8}+C$
$\Rightarrow \frac{\sqrt{3}}{8}=\frac{\sqrt{3}}{8}+C$
$\Rightarrow C=0$
From Eq. (i), we get
$y \frac{\cos 2 x}{\cos 2 x+1}=\frac{1}{4} \sin 2 x+0$
$\Rightarrow y=\frac{1}{4} \frac{\sin 2 x}{\frac{\cos 2 x}{\cos 2 x+1}}$
$=\frac{1}{4} \frac{\sin 2 x}{\frac{\cos ^{2} x-\sin ^{2} x}{2 \cos ^{2} x}}$
$=\frac{1}{2} \cdot \frac{\sin 2 x}{1-\tan ^{2} x}$