Question

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

Answer 1

We have,

\[\frac{dy}{dx} = \left( x + y \right)^2 . . . . . \left( 1 \right)\]

Let `x + y = v`

\[ \Rightarrow 1 + \frac{dy}{dx} = \frac{dv}{dx}\]

\[ \Rightarrow \frac{dy}{dx} = \frac{dv}{dx} - 1\]

Therefore, (1) becomes

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

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

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

Integrating both sides, we get

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

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

\[ \Rightarrow v = \tan\left( x + C \right)\]

\[ \Rightarrow x + y = \tan\left( x + C \right) \]