Question

The solution of differential equation ${\cos }^{2}x\frac{dy}{dx}-\left(\tan 2x\right)y={\cos }^{4}x,|x|<\frac{\pi }{4},$ where $y\left(\frac{\pi }{6}\right)=\frac{3\sqrt{3}}{8}$

• A. $y=\tan 2x{\cos }^{2}x$
• B. $y=\cot 2x{\cos }^{2}x$
• C. $2y=\tan 2x{\cos }^{2}x$
• D. $2y=\cot 2x{\cos }^{2}x$

Answer 1

The correct option is A $y=\tan 2x{\cos }^{2}x$

Given that:-

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

To find:- Solution of the given equation.

Solution:- $\because {\cos }^{2}x\left(\frac{dy}{dx}\right)-\left(\tan 2x\right)y={\cos }^{4}x$

$\Rightarrow {\cos }^{2}x\left(\frac{dy}{dx}\right)-\tan 2xy={\cos }^{4}x$

$\Rightarrow \frac{dy}{dx}-\frac{\tan 2x}{{\cos }^{2}x}.y={\cos }^{4}x$

Now, we will compat the eqn i to

$\frac{dy}{dx}+Py=B$

$\therefore P=f\left(x\right)$

$B=f\left(x\right)$

So, Integration factor (I.F) $=e^{\int Pdx}$

Here we will find the value of $-\frac{\tan 2x}{{\cos }^{2}x}$

$-\int \frac{\tan 2x}{{\cos }^{2}x}dx$

$=-\int \frac{2\sin 2x}{\cos 2x\left(2{\cos }^{2}x\right)}dx$

$=-\int \frac{2\sin 2xdx}{\cos 2x(1+\cos 2x)}$

Let, $t=\cos 2x$

$\Rightarrow dt=-2\sin xdx$

$\Rightarrow =\int \frac{dt}{t(1+t)}$

$=\int \frac{1}{t}-\frac{1}{1+t}dt$

$=ln\left|\frac{t}{1+t}\right|$

$=ln\left|\frac{\cos 2x}{1+\cos 2x}\right|$

Now, putting $ln\left|\frac{\cos 2x}{1+\cos 2x}\right|$ in I.F

$\therefore e^{Pdx}$

$=e^{ln}\left|\frac{\cos 2x}{1+\cos 2x}\right|$

$=\frac{\cos 2x}{1+\cos 2x}$

$=\frac{\cos 2x}{2{\cos }^{2}x}$

Since, multiplying eqn i by I.F both side, we get.

$\frac{\cos 2x}{2{\cos }^{2}x}.\frac{dy}{dx}-\frac{\sin 2x}{\cos 4x}y=\frac{\cos 2x}{2}$

$\frac{\cos 2x}{2{\cos }^{2}x}.dy-\left(\frac{\sin 2x}{\cos 4x}\right)dx.y=\frac{\cos 2x}{2}dx$

$\Rightarrow \int d\left(y\frac{\cos 2x}{2{\cos }^{2}x}\right)=\int \frac{\cos 2x}{2}dx$

$\Rightarrow y.\frac{\cos 2x}{2{\cos }^{2}x}=\frac{\sin 2x}{4}+C$...........(II)

$\Rightarrow \frac{3\sqrt{3}}{8}\left(\frac{1/2}{2.\frac{3}{4}}\right)=\frac{\sqrt{3}}{8}+c$

$\Rightarrow x=\frac{\pi }{6}y=\frac{3\sqrt{3}}{8}$

$\Rightarrow \frac{\sqrt{3}}{8}=\frac{\sqrt{3}}{8}+c$

$\therefore c=0$

putting the value of c in eq ii.

$\therefore y.\frac{\cos 2x}{2{\cos }^{2}x}=\frac{\sin 2x}{4}+0$

$\Rightarrow y=\frac{\sin 2x.2{\cos }^{2}x}{\cos 2x.4}$

$\therefore y=\frac{\tan 2x.{\cos }^{2}x}{2}$

$\therefore 2y=\tan 2x.{\cos }^{2}x$

hence, the required solution is

$2y=\tan 2x.{\cos }^{2}x$