Question

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

Answer 1

We have, 

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

Let x + y = v

$$\Rightarrow 1+\frac{dy}{dx}=\frac{dv}{dx}$$

$$\Rightarrow \frac{dy}{dx}=\frac{dv}{dx}-1$$

$$\therefore \frac{dv}{dx}-1=v^2$$

$$\Rightarrow \frac{dv}{dx}=v^2+1$$

$$\Rightarrow \frac{1}{v^2+1}dv=dx$$

Integrating both sides, we get

$$\int \frac{1}{v^2+1}dv=\int dx$$

$$\Rightarrow {\tan }^{-1}v=x+C$$

$$\Rightarrow v=\tan (x+C)$$

$$\Rightarrow x+y=\tan (x+C)$$