#### Please Solve R.D.Sharma class 12 Chapter 21  Differential Equations Exercise 21.8 Question 4 Maths textbook Solution.

Answer:-    $x+y=\tan \left ( x+c \right )$

Given:$\frac{dy}{dx}=\left ( x+y \right )^{2}$

Hint : - Differential equation of the form $\frac{dy}{dx}=\int \left ( ax+by+c \right )$can be reduced to variable separable form by substitution $ax+by+c=v$

Solution : -  We have,

$\frac{dy}{dx}=\left ( x+y \right )^{2}$

Let $x+y=v$

Differentiating with respect to x, we get,

\begin{aligned} \; \; \; \; \; \; &\quad \frac{d }{d x}(\mathrm{x}+\mathrm{y})=\frac{d}{d x}(v) \\ &\Rightarrow \quad 1+\frac{d y}{d x}=\frac{d v}{d x} \\ &\Rightarrow \frac{d y}{d x}=\frac{d v}{d x}-1 \end{aligned}

Putting the value of $\frac{dy}{dx}$ in equation (i), we get

$\frac{d v}{d x}-1=(\mathrm{x}+\mathrm{y})^{2} \\$

$\Rightarrow \frac{d v}{d x}-1=\mathrm{v}^{2}$                                            $(\because \mathrm{x}+\mathrm{y}=\mathrm{v}) \\$

$\Rightarrow \frac{d v}{d x}=\mathrm{v}^{2}+1 \\$

$\Rightarrow \frac{d v}{v^{2}+1}=\mathrm{d} \mathrm{x}$

Integrating on both the sides, we get,

$\int \frac{d v}{v^{2}+1}=\int \mathrm{d} \mathrm{x}$

$\Rightarrow \tan ^{-1}v=x+c$                                                                $\left ( \because \frac{dx^{2}}{x^{2}+1}=\tan ^{-1}x \right )$

Putting $v=x+y,$ we get

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

$\Rightarrow x+y=\tan \left ( x+c \right )$

(This is the required solution).