#### Please solve RD Sharma Class 12 Chapter 21 Differential Equation Exercise 21.10 Question 37 Subquestion (v) maths textbook solution.

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

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

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

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

\begin{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} \end{aligned}

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

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

Now $y(0)=0$

\begin{aligned} &=0 e^{\tan ^{-1}(0)}=\tan ^{-1}(0)+C \\ &=C=0 \quad\left[\tan ^{-1}(0)=0\right] \end{aligned}

Substituting in (i)

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