Question

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

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

Answer 1

Answer: (D)

Given differential equation can be written as $\frac{dy}{dx}-\left(\frac{\tan 2x}{{\cos }^{2}x}\right)y={\cos }^{2}x$
Here,
$P=-\frac{\tan 2x}{{\cos }^{2}x}=\frac{-\sin 2x}{\cos 2x\left(\frac{\cos 2x+1}{2}\right)}$
and $Q={\cos }^{2}x$
$\because IF=e^{\int Pdx}=e^{-\int \frac{2\sin 2x}{\cos 2x(\cos 2x+1)}dx}$
Put $\cos 2x=t\Rightarrow -2\sin 2xdx=dt$
$\therefore IF=e^{{\int }_t^1\frac{1}{t}\left(\frac{1}{t+1}\right)dt}$
$=e^{\int \left(\frac{1}{t}-\frac{1}{t+1}\right)dt}$
$=e^{\log t-\log (t+1)]}=e^{\log \frac{t}{t+1}}$
$=e^{{\log }_{\cos 2z-1}}=\frac{\cos 2x}{\cos 2x+1}$
Now, solution is
$y\times \frac{\cos 2x}{\cos 2x+1}=\int \frac{\cos 2x}{\cos 2x+1}\times {\cos }^{2}xdx+C$
$=\int \frac{\cos 2x}{2{\cos }^{2}x}\times {\cos }^{2}xdx+C$
$=\frac{1}{2}\int \cos 2xdx+C$
$=\frac{1}{2}\times \frac{\sin 2x}{2}+C$
$\Rightarrow y\frac{\cos 2x}{\cos 2x+1}=\frac{1}{4}\sin 2x+C$ ... (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
$y\frac{\cos 2x}{\cos 2x+1}=\frac{1}{4}\sin 2x+0$
$\Rightarrow y=\frac{1}{4}\frac{\sin 2x}{\frac{\cos 2x}{\cos 2x+1}}$
$=\frac{1}{4}\frac{\sin 2x}{\frac{{\cos }^{2}x-{\sin }^{2}x}{2{\cos }^{2}x}}$
$=\frac{1}{2}\cdot \frac{\sin 2x}{1-{\tan }^{2}x}$