Question

check my work

Answer 1

Solve:
\[(x+y)^2(x\frac{dy}{dx}+y)=xy(1+\frac{dy}{dx})\]\[\frac{x\frac{dy}{dx}+y}{xy}=\frac{1+\frac{dy}{dx}}{(x+y)^2}\]\[xy=t \space \space \space \implies \space x\frac{dy}{dx}+y=\frac{dt}{dx}\]\[x+y=p \space \space \space \implies \space 1+\frac{dy}{dx}=\frac{dp}{dx}\]\[\therefore \frac{1}{t}\frac{dt}{dx}=\frac{1}{p^2}\frac{dp}{dx}\]\[\frac{dt}{t}=\frac{dp}{p^2}\]\[\int\limits \frac{dt}{t}=\int\limits \frac{dp}{p^2}\]\[\log|t|=-\frac{1}{p}+C\]\[\log|xy|=-\frac{1}{x+y}+C\]