Question

(x + y) (dx − dy) = dx + dy

Answer 1

We have,
(x + y) (dx − dy) = dx + dy

$$\begin{gathered} \Rightarrow xdx+ydx-xdy-ydy=dx+dy \\ \Rightarrow \left(x+y-1\right)dx=\left(x+y+1\right)dy \\ \Rightarrow \frac{dy}{dx}=\frac{x+y-1}{x+y+1} \\ \mathrm{Let}x+y=v \\ \therefore 1+\frac{dy}{dx}=\frac{dv}{dx} \\ \Rightarrow \frac{dy}{dx}=\frac{dv}{dx}-1 \\ \therefore \frac{dv}{dx}-1=\frac{v-1}{v+1} \\ \Rightarrow \frac{dv}{dx}=\frac{v-1}{v+1}+1 \\ \Rightarrow \frac{dv}{dx}=\frac{v-1+v+1}{v+1} \\ \Rightarrow \frac{dv}{dx}=\frac{2v}{v+1} \\ \Rightarrow \frac{v+1}{2v}dv=dx \\ \mathrm{Integrating}\mathrm{both}\mathrm{sides},\mathrm{we}\mathrm{get} \\ \int \frac{v+1}{2v}dv=\int dx \\ \Rightarrow \frac{1}{2}\int dv+\frac{1}{2}\int \frac{1}{v}dv=\int dx \\ \Rightarrow \frac{1}{2}v+\frac{1}{2}\log \left|v\right|=x+C \\ \Rightarrow \frac{1}{2}\left(x+y\right)+\frac{1}{2}\log \left|x+y\right|=x+C \\ \Rightarrow \frac{1}{2}\left(y-x\right)+\frac{1}{2}\log \left|x+y\right|=C \end{gathered}$$