#### Provide solution for  RD Sharma maths Class 12 Chapter 21 Differential Equation Exercise Revision Exercise (RE) Question 60 textbook solution.

Answer : $x e^{\tan ^{-1} y}=e^{\tan ^{-1} y}\left(\tan ^{-1} y-1\right)+c$

Hint               : You must know the rules of solving differential equation and integration

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

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

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

Comparing with, $\frac{dx}{dy}+Px=Q,$

\begin{aligned} &P=\frac{1}{1+y^{2}}, Q=\frac{\tan ^{-1} y}{1+y^{2}} \\ &\text { Now } \quad \text { I. } F=e^{\int \frac{1}{1+y^{2}} d y}=e^{\tan ^{-1} y} \end{aligned}

So, the solution is ,

\begin{aligned} x \times I . F &=\int(I . F \times Q) d y+c \\ x e^{\tan ^{-1} y} &=\int \frac{\tan ^{-1} y}{1+y^{2}} \times e^{\tan ^{-1} y} d y+c \\ x e^{\tan ^{-1} y} &=I+C \end{aligned}

Now, $I=\int \frac{\tan ^{-1} y}{1+y^{2}} \times e^{\tan ^{-1} y} d y$

\begin{aligned} &\text { Put }_{\mu} \text { tan }^{-1} y=t \text { and differentiate, }\\ &\begin{gathered} \quad \frac{1}{1+y^{2}} d y=d t \\ I=\int e^{t} t d t \end{gathered} \end{aligned}

Integrating by parts,

\begin{aligned} &I=t \times \int e^{t} d t-\int\left\{\frac{d t}{d t} \int e^{t} d t\right\} d t \\ &I=t e^{t}-\int e^{t} d t \\ &I=t e^{t}-e^{t} \end{aligned}

\begin{aligned} &I=\tan ^{-1} y e^{\tan ^{-1} y}-e^{\tan ^{-1} y} \\ &I=e^{\tan ^{-1} y}\left(\tan ^{-1} y-1\right) \end{aligned}

By putting value of I

We get the required solution,

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