#### Explain solution RD Sharma class 12 Chapter 21 Differential Equation exercise 21.10 question 30

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

Hint: To solve this equation we use $\frac{d y}{d x}+P y=Q$  where $P,Q$  are constants.

Give:  \begin{aligned} &(\sin x) \frac{d y}{d x}+y \cos x=2 \sin ^{2} x \cos x \\ & \end{aligned}

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} \\ & \end{aligned}

$=\frac{d y}{d x}+y \cot x=2 \sin x \cos x \ldots(i) \\$

$=\frac{d y}{d x}+P y=Q \\$

$P=\cot x, Q=2 \sin x \cos x$

$If$  of differential equation is

\begin{aligned} &\text { If }=e^{\int P d x} \\ & \end{aligned}

$=e^{\int \cot x d x} \\$

$=e^{\log |\sin x|} \\$

$=\sin x \\$

$y \text { If }=\int \text { QIf } d x+C$

\begin{aligned} &=y \sin x=\int 2 \sin ^{2} x \cos x d x \\ & \end{aligned}

$=2 \int \sin ^{2} x \cos x d x$                           $\quad[\sin x=t, \cos x d x=d t] \\$

$=2 \int t^{2} d t \\$

$=2\left[\frac{t^{3}}{3}\right]+C$

\begin{aligned} &=\frac{2}{3} t^{3}+C \\ & \end{aligned}

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

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