Question

\[\frac{dy}{dx} = y^2 + 2y + 2\]

Answer 1

We have,

\[\frac{dy}{dx} = y^2 + 2y + 2\]

\[ \Rightarrow \frac{dy}{dx} = y^2 + 2y + 1 + 1\]

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

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

Integrating both sides, we get

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

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

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