Question

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

Answer 1

We have, 

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

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

Let x + y = v

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

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

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

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

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

Integrating both sides, we get

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

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

\[ \Rightarrow \int\left( 1 - \frac{1}{v^2 + 1} \right)dv = \int dx\]

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

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

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