#### Explain Solution R.D.Sharma Class 12 Chapter 21 Differential Equations Exercise Revision Exercise Question 35 Maths Textbook Solution.

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

Hint: you must know the rules of solving differential equation and integrations.

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

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

Let $\left ( x+y \right )=u$ and differentiate both sides ,we get,

\begin{aligned} &1+\frac{d y}{d x}=\frac{d u}{d x} \\ &\frac{d y}{d x}=\frac{d u}{d x}-1 \quad(x+y)=u \quad\left[\because(x+y)^{2}=u^{2]}\right. \\ &U^{2}=\frac{d u}{d x}-1 \quad\left[\because \frac{d u}{d x}=u^{2}\right] \\ &\frac{d u}{d x}=u^{2}+1 \\ &\frac{d u}{u 2+1}=d x \end{aligned}

Now, integrating both sides,

\begin{aligned} &\int \frac{\mathrm{du}}{\mathrm{u} 2+1}=\int \mathrm{dx} \\ &\tan ^{-1}=\mathrm{x}+\mathrm{C} \quad\left[\int \frac{1}{x^{2}+1} d x=\tan ^{-1} \mathrm{x}+\mathrm{C}\right] \\ &\therefore \tan ^{-1}(\mathrm{x}+\mathrm{y})=\mathrm{x}+\mathrm{c} \text { where, } \mathrm{u}=\mathrm{x}+\mathrm{y} \end{aligned}

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