#### Need solution for RD Sharma maths Class 12 Chapter 21 Differential Equation Exercise 21.10 Question 42 textbook solution.

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

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

Hint : Using integration by parts and $\int \frac{1}{1+x^{2}} d x$

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

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

The integrating factor $If$  of this differential equation is

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

Hence, the solution is

\begin{aligned} &x I f=\int Q I f d y+C \\ &=x\left(e^{\cot ^{-1} y}\right)=\int \frac{\cot ^{-1} y}{1+y^{2}} e^{\cot ^{-1} y} d y \end{aligned}

Put $\cot ^{-1}y=t$

\begin{aligned} &=-\frac{d y}{1+y^{2}}=d t \\ &=\frac{d y}{1+y^{2}}=-d t \end{aligned}

So,

$=-\int t e^{t} d t$

Using integration by parts

\begin{aligned} &=-\left[t e^{t}-\int e^{t} d t\right] \\ &=-\left[t e^{t}-e^{t}-C\right] \\ &=-t e^{t}+e^{t}+C \\ &=x e^{\cot ^{-1} y}=-\cot ^{-1} y e^{\cot ^{-1} y}+e^{\cot ^{-1} y}+C \end{aligned}

Divide by $e^{\cot ^{-1} y}$

\begin{aligned} &=x=-\cot ^{-1} y+1+\frac{c}{e^{\cot ^{-1} y}} \\ &=x+\cot ^{-1} y=1+C e^{\tan ^{-1} y} \; \; \; \; \; \; \; \; \; \quad\left[\cot ^{-1} y=\tan ^{-1}\left(\frac{1}{y}\right)\right] \end{aligned}