#### Provide solution for  RD Sharma maths Class 12 Chapter 21 Differential Equation Exercise Revision Exercise (RE) Question 59 textbook solution.

Answer : $y=\cos x-2 \cos ^{2} x$

Hint : you must know the rules of solving differential equation and integration

Given :  $2 \cos x \frac{d y}{d x}+4 y \sin x=\sin 2 x$,                           $y=0, x=\frac{\Pi}{3}$

Solution : $2 \cos x \frac{d y}{d x}+4 y \sin x=\sin 2 x$

\begin{aligned} &\frac{d y}{d x}+4 y \frac{\sin x}{2 \cos x} \quad=\frac{2 \sin x \cos x}{2 \cos x} \\ &\frac{d y}{d x}+2 y \tan x \quad=\sin x \end{aligned}

Comparing with, $\frac{d y}{d x}+P y=Q$ , we get

Where, $P=2\tan x, Q = \sin x$

Now,

\begin{aligned} \text { I.F } &=e^{2 \int \tan x d x} \\ &=e^{2 \log |\sec x|} \\ &=\sec ^{2} x \end{aligned}

So, the solution is,

\begin{aligned} &y \times \text { I.F }=\int(Q \times I . F) d x+c \\ &y \sec ^{2} x=\int \sin x \sec ^{2} x d x+c \\ &y \sec ^{2} x=\int \tan x \sec x d x+c \end{aligned}

\begin{aligned} &y \sec ^{2} x=\sec x+c \\ &y=\cos x+C \cos ^{2} x \\ &\text { now } x=\frac{\pi}{3} \quad, y=0 \end{aligned}

Therefore,

\begin{aligned} &0=\cos \frac{\pi}{3}+C \cos ^{2} \frac{\pi}{3} \\ &0=\frac{1}{2}+C\left(\frac{1}{4}\right) \end{aligned}

$C=-2$

Putiing value of C,

$y=\cos x-2 \cos ^{2} x$