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

Answer : $y \sin x=2 x^{2}-\frac{\pi^{2}}{2}$

Give : $\frac{d y}{d x}+y \cot x=4 x \operatorname{cosec} x, y\left(\frac{\pi}{2}\right)=0$

Hint : Using $\int \cot x d x \text { and } \int x d x$

Explanation : $\frac{d y}{d x}+y \cot x=4 x \operatorname{cosec} x$

$\frac{d y}{d x}+(\cot x) y=4 x \operatorname{cosec} x$

This is a linear differential equation of the form

\begin{aligned} &\frac{d x}{d y}+P y=Q \\ &P=\cot x \text { and } Q=4 x \operatorname{cosec} x \end{aligned}

The integrating factor $If$  of this differential equation is

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

Hence, the solution is

\begin{aligned} &y I f=\int \text { QIf } d x+C \\ &=y(\sin x)=\int(4 x \operatorname{cosec} x) \sin x d x+C \\ &=y \sin x=4 \int x \frac{1}{\sin x} \sin x d x+C \quad\left[\operatorname{cosec} x=\frac{1}{\sin x}\right] \\ &=y \sin x=4 \int x d x+C \end{aligned}

\begin{aligned} &=y \sin x=4\left(\frac{x^{2}}{2}\right)+C \\ &=y \sin x=2 x^{2}+C \ldots(i) \end{aligned}

Now, $y\left(\frac{\pi}{2}\right)=0 \text { when } x=\frac{\pi}{2}, y=0$

\begin{aligned} &=0 \sin x=2\left(\frac{\pi}{2}\right)^{2}+C \\ &=0=2 \frac{\pi^{2}}{4}+C \\ &=C=-\frac{\pi^{2}}{2} \end{aligned}

Substituting in (i)

$=y \sin x=2 x^{2}-\frac{\pi^{2}}{2}$