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Provide Solution for RD Sharma Class 12 Chapter 21 Differential Equation Exercise 21.10 Question 41

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

Give:  \begin{aligned} & & \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} \end{aligned}

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

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

 \begin{aligned} & &=\frac{d x}{d y}+(\cot y) x=2 y+y^{2} \cot y \end{aligned}

This is a linear differential equation of the form

 \frac{d x}{d y}+P x=Q

P=\cot x \text { and } Q=2 y+y^{2} \cot y

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|} \qquad\qquad\qquad\left[\int \cot x d x=\log |\sin x|+C\right] \\ &=\sin y \quad\qquad\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 \end{aligned}

 \begin{aligned} &=x \sin y=\int 2 y \sin y+y^{2} \frac{\cos y}{\sin y} \sin y d y+C \\\\ &=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) \qquad\left[\frac{1}{\sin y}=\operatorname{cosec} y\right] \end{aligned}

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

\begin{aligned} &=0=\left(\frac{\pi}{2}\right)^{2}+C \operatorname{cosec} \frac{\pi}{2} \end{aligned}

\begin{aligned} &=0=\frac{\pi^{2}}{4}+C(1) \qquad\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

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