Question

(x + y) (dx − dy) = dx + dy

Answer 1

We have,
(x + y) (dx − dy) = dx + dy
\[\Rightarrow x dx + y dx - x dy - y dy = dx + dy\]
\[ \Rightarrow \left( x + y - 1 \right)dx = \left( x + y + 1 \right)dy\]
\[ \Rightarrow \frac{dy}{dx} = \frac{x + y - 1}{x + y + 1}\]
Let x + y = v
\[ \therefore 1 + \frac{dy}{dx} = \frac{dv}{dx}\]
\[ \Rightarrow \frac{dy}{dx} = \frac{dv}{dx} - 1\]
\[ \therefore \frac{dv}{dx} - 1 = \frac{v - 1}{v + 1}\]
\[ \Rightarrow \frac{dv}{dx} = \frac{v - 1}{v + 1} + 1\]
\[ \Rightarrow \frac{dv}{dx} = \frac{v - 1 + v + 1}{v + 1}\]
\[ \Rightarrow \frac{dv}{dx} = \frac{2v}{v + 1}\]
\[ \Rightarrow \frac{v + 1}{2v}dv = dx\]
Integrating both sides, we get
\[\int\frac{v + 1}{2v}dv = \int dx\]
\[ \Rightarrow \frac{1}{2}\int dv + \frac{1}{2}\int\frac{1}{v}dv = \int dx\]
\[ \Rightarrow \frac{1}{2}v + \frac{1}{2}\log\left| v \right| = x + C\]
\[ \Rightarrow \frac{1}{2}\left( x + y \right) + \frac{1}{2}\log\left| x + y \right| = x + C\]
\[ \Rightarrow \frac{1}{2}\left( y - x \right) + \frac{1}{2}\log\left| x + y \right| = C\]