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

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

Answers (1)

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

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