Answer
See below
Work Step by Step
Let the equation be: $\frac{dy}{dx}=\frac{2(2\frac{y}{x}-1)}{1+\frac{y}{x}}$
Let $V=\frac{y}{x}$, we have: $\frac{dy}{dx}=V+x\frac{dV}{dx}$
The equation becomes:
$V+x\frac{dV}{dx}=\frac{2(2V-1)}{1+V}\\
x\frac{dV}{dx}=\frac{4V-2}{1+V}-\frac{V+V^2}{1+V}\\
x\frac{dV}{dx}=\frac{-V^2+3V-2}{1+V}\\
x\frac{dV}{dx}=\frac{-(V-1)(V-2)}{1+V}$
Obtain: $\frac{dx}{x}=\frac{(1+V)dV}{-(V-1)(V-2)}$
Integrate both sides: $\int \frac{-1}{x}dx=\int \frac{(1+V)dV}{(V-1)(V-2)}$
Solve the integral on the left side: $\frac{(1+V)dV}{-(V-1)(V-2)}=\frac{A}{V-1}+\frac{B}{V-2}\\\rightarrow 1+V=A.V-2A+B.V-B\\\rightarrow A+B=1 \land -2A-B=1\\ \rightarrow B=1-A \land -2A-(1-A)=1\\\rightarrow B=3 \land A=-2$
Hence, $\int \frac{(1+V)dV}{(V-1)(V-2)}dV=\int \frac{-2}{V-1}dV+\int \frac{3}{V-2}dV\\
\rightarrow -2\ln|V-1|+3\ln|V-2|+c=-\ln |x|+\ln k\\
\rightarrow \ln\frac{(V-2)^3}{(V-1)^2}=\ln\frac{k}{x}\\
\rightarrow \frac{(V-2)^3}{(V-1)^2}=\frac{k}{x}$
Substitute back: $\frac{(\frac{y}{x}-2)^3}{(\frac{y}{x}-1)^2}=\frac{k}{x}\\
\rightarrow \frac{(y-2x)^3}{(y-x)^2}=k$
Since $y(0)=2$, we have: $k=\frac{(2-2.0)^3}{(2-0)^2}=2$
Hence, $ \frac{(y-2x)^3}{(y-x)^2}=2$
