#### Need solution for RD Sharma Maths Class 12 Chapter 21 Differential Equation Excercise 21.10 Question 37 subquestion (viii)

Answer:  \begin{aligned} & y \sin x=2 x^{2}-\frac{\pi^{2}}{2}\\ & \end{aligned}

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} \\ & \end{aligned}

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

$x=e^{\log |\sin x|}$                       $\quad\left[\int \cot x d x=\log |\sin x|+C\right] \\$

$=\sin x$                     $\quad\left[e^{\log e^{x}}=x\right]$

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

Now    \begin{aligned} &y\left(\frac{\pi}{2}\right)=0 \text { when } x=\frac{\pi}{2}, y=0 \\ & \end{aligned}

$\quad=0 \sin x=2\left(\frac{\pi}{2}\right)^{2}+C \\$

$=0=2 \frac{\pi^{2}}{4}+C \\$

$=C=-\frac{\pi^{2}}{2}$

Substituting in (i)

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