#### Provide solution for RD Sharma maths Class 12 Chapter 21 Differential Equation Exercise 21.10 Question 37 Subquestion (x) textbook solution.

Answer : $y \operatorname{cosec}^{2} x=4 \sin x-2$

Give : $\frac{d y}{d x}-3 y \cot x=\sin 2 x, y=2 \text { when } x=\frac{\pi}{2}$

Hint : Using $\int \operatorname{cosec} x \cot x d x$

Explanation : $\frac{d y}{d x}-3 y \cot x=\sin 2 x$

$\frac{d y}{d x}+(-3 \cot x) y=\sin 2 x$

This is a linear differential equation of the form

\begin{aligned} &\frac{d x}{d y}+P y=Q \\ &P=-3 \cot x \operatorname{and} Q=\sin 2 x \end{aligned}

The integrating factor $If$  of this differential equation is

\begin{aligned} &I f=e^{\int P d x} \\ &=e^{\int-3 \cot x d x} \\ &=e^{-3 \int \cot x d x} \\ &=e^{-3 \log |\sin x|}\; \; \; \; \; \; \; \; \; \; \quad\left[\int \cot x d x=\log |\sin x|+C\right] \end{aligned}

\begin{aligned} &=e^{\log \sin ^{-3} x} \\ &=\sin ^{-3} x\; \; \; \; \; \; \; \; \quad\left[e^{\log e^{x}}=x\right] \\ &=\frac{1}{\sin ^{3} x} \\ &=\operatorname{cosec}^{3} x \end{aligned}

Hence, the solution is

\begin{aligned} &y I f=\int Q I f d x+C \\ &=y \operatorname{cosec}^{3} x=\int\left(\sin 2 x \operatorname{cosec}^{3} x\right) d x+C \\ &=y \operatorname{cosec}^{3} x=\int\left(2 \sin x \cos x \frac{1}{\sin ^{3} x}\right) d x+C \\ &=y \operatorname{cosec}^{3} x=2 \int\left(\cos x \frac{1}{\sin ^{2} x}\right) d x+C \end{aligned}

\begin{aligned} &=y \operatorname{cosec}^{3} x=2 \int\left(\frac{\cos x}{\sin x} \frac{1}{\sin x}\right) d x+C \\ &=y \operatorname{cosec}^{3} x=2 \int \cot x \operatorname{cosec} x d x+C \quad\left[\frac{\cos x}{\sin x}=\cot x, \frac{1}{\sin x}=\operatorname{cosec} x\right] \\ &=y \operatorname{cosec}^{3} x=2(-\operatorname{cosec} x)+C \; \; \; \; \; \; \; \; \; \; \; \quad\left[\int \cot x \operatorname{cosec} x d x=-\operatorname{cosec} x\right] \end{aligned}

Divide by $cosec\; x$

\begin{aligned} &=y \operatorname{cosec}^{2} x=-2+\frac{C}{\operatorname{cosec} x} \\ &=y \operatorname{cosec}^{2} x=-2+C \sin x \ldots(i) \quad\left[\frac{1}{\operatorname{cosec} x}=\sin x\right] \end{aligned}

Now $y=2 \text { when } x=\frac{\pi}{2}$

\begin{aligned} &=2 \operatorname{cosec}^{2}\left(\frac{\pi}{2}\right)=-2+C \sin \frac{\pi}{2} \\ &=2(1)^{2}=-2+C(1) \; \; \; \; \; \; \; \; \; \; \; \; \; \; \quad\left[\operatorname{cosec} \frac{\pi}{2}=\frac{1}{\sin \frac{\pi}{2}}=\frac{1}{1}=1\right] \\ &=2=-2+C \\ &=C=4 \end{aligned}

Substituting in (i)

$=y \operatorname{cosec}^{2} x=+4 \sin x-2$