(x + 2y)\(\frac{dy}{dx}=x\) 

⇒ \(\frac{dy}{dx} = \frac x{x+2y} = \cfrac1{1+\frac{2y}x}\) 

Let y/x = v ⇒ y = vx

⇒ \(\frac{dy}{dx} = v+x\frac{dv}{dx}\)

∴ v + \(x\frac{dv}{dx}\) = \(\frac1{1+2v}\) 

⇒ \(x\frac{dv}{dx}\) = \(\frac1{1+2v}-v\) = \(\frac{1-v-2v^2}{1+2v}\) 

⇒ \(\frac{1+2v}{1-v-2v^2}dv=\frac{dx}x\) 

⇒ \(\int\frac12(\frac{2+4v}{1-v-2v^2})dv=\int\frac{dx}x\)

⇒ \(\int\frac12(\frac{1+4v}{1-v-2v^2}+\frac1{1-v-2v^2})dv=\int\frac{dx}x\)

⇒ -\(\frac12\)log|1 - v - 2v²|  - \(\frac14\int\frac1{(v+\frac14)^2-(\frac34)^2} = log x+c\) 

⇒ -\(\frac12\)log|1 - v - 2v²| - \(\frac14\times\frac1{2\times3/4}\) log\(|\frac{v+1/4-3/4}{v+1/4+3/4}|=\log x + c\)

⇒ -\(\frac12\)log|1 - v - 2v²| - \(\frac16\)log|\(\frac{uv-2}{uv+4}\)| = log x + c 

⇒ -\(\frac12\)log(1 - v - 2v²) - \(\frac16\) log|\(\frac{2v-1}{2v+2}\)| = log x + c

⇒ -\(\frac12\)log(1 - yx - 2y²/x²) - 1/6 log|\(\frac{2y-x}{2y+2x}\)|  = log x + c

\((\because \forall=\frac yx)\)

which is solution of given homogenous differential equation.