#### Please solve RD Sharma Class 12 Chapter 21 Differential Equation Exercise 21.10 Question 30 maths textbook solution.

Answer : $y \sin x=\frac{2}{3} \sin ^{3} x+C$

Hint : To solve this equation we use $\frac{dy}{dx}+Py=Q$ where P,Q are constants.

Give : $(\sin x) \frac{d y}{d x}+y \cos x=2 \sin ^{2} x \cos x$

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

\begin{aligned} &=\frac{d y}{d x}+\frac{y \cos x}{\sin x}=\frac{2 \sin ^{2} x \cos x}{\sin x} \\ &=\frac{d y}{d x}+y \cot x=2 \sin x \cos x \ldots(i) \\ &=\frac{d y}{d x}+P y=Q \\ &P=\cot x, Q=2 \sin x \cos x \end{aligned}

$I\; f$ of differential equation is

\begin{aligned} &I f=e^{\int P d x} \\ &=e^{\int \cot x d x} \\ &=e^{\log |\sin x|} \\ &=\sin x \end{aligned}

\begin{aligned} &y I f=\int Q I f d x+C \\ &=y \sin x=\int 2 \sin ^{2} x \cos x d x \\ &=2 \int \sin ^{2} x \cos x d x \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \quad[\sin x=t, \cos x d x=d t] \end{aligned}

\begin{aligned} &=2 \int t^{2} d t \\ &=2\left[\frac{t^{3}}{3}\right]+C \\ &=\frac{2}{3} t^{3}+C \end{aligned}

\begin{aligned} &=\frac{2}{3} \sin ^{3} x+C \\ &y \sin x=\frac{2}{3} \sin ^{3} x+C \end{aligned}