#### Provide solution for  RD Sharma maths Class 12 Chapter 21 Differential Equation Exercise Revision Exercise (RE) Question 57 textbook solution.

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

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

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

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

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

Comparing with,

$\frac{d x}{d y}+P x=Q, we get$

$P=\frac{1}{1+y^{2}} \quad, Q=\frac{e^{-\tan ^{-1} y}}{1+y^{2}}$

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

So, the solution is

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