#### Explain solution RD Sharma class 12 Chapter 21 Differential Equation exercise 21.10 question 37 subquestion (v)

Answer:  $x e^{\tan ^{-1} y}=\tan ^{-1} y$

Give:  \begin{aligned} &\left(1+y^{2}\right) d x+\left(x-e^{\tan ^{-1} y}\right) d y=0 \\ & \end{aligned}

Hint: Using  $\int \frac{1}{1+x^{2}} d x$

Explanation:  \begin{aligned} &\left(1+y^{2}\right) d x+\left(x-e^{\tan ^{-1} y}\right) d y=0 \\ & \end{aligned}

$=\left(x-e^{\tan ^{-1} y}\right) d y=-\left(1+y^{2}\right) d x \\$

$=x-e^{\tan ^{-1} y}=-\left(1+y^{2}\right) \frac{d x}{d y} \\$

$=\frac{x-e^{\tan ^{-1} y}}{1+y^{2}}=-\frac{d x}{d y}$

\begin{aligned} &=\frac{x}{1+y^{2}}-\frac{e^{\tan ^{-1} y}}{1+y^{2}}+\frac{d x}{d y}=0 \\ &\end{aligned}

$=\frac{d x}{d y}+\left(\frac{1}{1+y^{2}}\right) x=\frac{e^{\tan ^{-1} y}}{1+y^{2}}$

This is a linear differential equation of the form

\begin{aligned} &\frac{d x}{d y}+P x=Q \\ &P=\frac{1}{1+y^{2}} \text { and } Q=\frac{e^{\tan ^{1} y}}{1+y^{2}} \end{aligned}

The integrating factor $If$  of this differential equation is

\begin{aligned} &\text { If }=e^{\int \operatorname{Pdx}} \\ &=e^{\int \frac{1}{1+y^{2}} d x} \\ &=e^{\tan ^{-1} y} \quad\left[\int \frac{1}{1+x^{2}} d x=\tan ^{-1} x+C\right] \end{aligned}

Hence, the solution is

\begin{aligned} &x I f=\int Q I f d y+C \\ &=x\left(e^{\tan ^{-1} y}\right)=\int \frac{e^{-\tan ^{-1} y}}{1+y^{2}} e^{\tan ^{-1} y} d y+C \\ &=x e^{\tan ^{-1} y}=\int \frac{1}{1+y^{2}} d y+C \end{aligned}          $\quad\left[e^{-\tan ^{-1} y+\tan ^{-1} y}=e^{0}=1\right]$

$\\ =x e^{\tan ^{-1} y}=\tan ^{-1} y+C \ldots \text { (i) }$

$\quad\left[\int \frac{1}{1+y^{2}} d y=\tan ^{-1} y\right]$

Now  \begin{aligned} &y(0)=0 \\ & \end{aligned}

$\quad=0 e^{\tan ^{-1}(0)}=\tan ^{-1}(0)+C \\$

$\quad=C=0$

Substituting in (i)

\begin{aligned} &=x e^{\tan ^{-1} y}=\tan ^{-1} y+0 \\ &=x e^{\tan ^{-1} y}=\tan ^{-1} y \end{aligned}