#### Need solution for RD Sharma maths Class 12 Chapter 21 Differential Equation Exercise 21.10 Question 41 textbook solution.

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

Give : $\frac{d x}{d y}+x \cot y=2 y+y^{2} \cot y, y \neq 0, x=0 \text { when } y=\frac{\pi}{2}$

Hint : Using integration by parts and $\int x^{n} d x$

Explanation: $\frac{d x}{d y}+x \cot y=2 y+y^{2} \cot y$

$=\frac{d x}{d y}+ (\cot y)\; x=2 y+y^{2} \cot y$

This is a linear differential equation of the form

\begin{aligned} &\frac{d x}{d y}+P x=Q \\ & P=\cot x \text { and } Q=2 y+y^{2} \cot y \end{aligned}

The integrating factor $If$  of this differential equation is

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

Hence, the solution is

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

\begin{aligned} &=x \sin y=\int 2 y \sin y d y+\int y^{2} \cos y d y+C \\ &=x \sin y=2 \int y \sin y d y+\int y^{2} \cos y d y+C \ldots(i) \end{aligned}

Using integration by parts

\begin{aligned} &=2 \int y \sin y d y \\ &=2\left[\sin y \frac{y^{2}}{2}-\int \cos y \frac{y^{2}}{2} d y\right] \\ &=\sin y y^{2}-\int \cos y y^{2} d y \end{aligned}

Substituting on (i)

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

Divide by $\sin y$

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

Now  $x=0 \text { when } y=\frac{\pi}{2}$

$=0=\left(\frac{\pi}{2}\right)^{2}+C \operatorname{cosec} \frac{\pi}{2}$

\begin{aligned} &=0=\frac{\pi^{2}}{4}+C(1)\; \; \; \; \; \; \; \; \quad\left[\operatorname{cosec} \frac{\pi}{2}=1\right] \\ &=C=-\frac{\pi^{2}}{4} \end{aligned}

Substituting in (ii)

$=x=y^{2}-\frac{\pi^{2}}{4} \operatorname{cosec} y$