Question

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

Answer 1

We have, 

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

\[\text{ Putting }x + y + 1 = v\]

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

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

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

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

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

Integrating both sides, we get

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

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

\[ \Rightarrow \tan^{- 1} \left( x + y + 1 \right) = x + C\]