Answer
$(2y-x)(y+x)=2$
Work Step by Step
Given:
$$\frac{dy}{dx}=\frac{2(2y-x)}{x+y}$$
Let $y=vx \rightarrow \frac{dy}{dx}=v+x\frac{dv}{dx}$
Then it becomes:
$$v+x\frac{dv}{dx}=\frac{2-v}{1+4v}$$
$$x\frac{dv}{dx}=-\frac{2(v^2+v-1)}{4v+1}$$
$$\frac{4v+1}{-4v^2-2v+2}dv=\frac{1}{x}dx$$
Integrating both sides:
$\int \frac{4v+1}{-4v^2-2v+2}dv=\int \frac{1}{x}dx$
$$\int (-\frac{1}{2v-1}-\frac{1}{2v+2})dv=\int \frac{1}{x}dx$$
$$\frac{-1}{2}\ln(2v-1)-\frac{1}{2}\ln(v+1)=\ln x +\ln C$$
where $C$ is a constant of integration.
$$(2v-1)-1)(v+1)x^2=C$$
Subtitute when $y=vx $
$$\rightarrow (2y-x)(y+x)=C$$
Since $y(1)=1$
$$\rightarrow (2\times1-1)(1+1)=C$$
$$C=2$$
The solution to the differential equation is:
$$(2y-x)(y+x)=2$$
