Just need a bit of help finishing this one, not sure if I'm going about this correctly...

\(\displaystyle

\begin{gathered}

\frac{{dy}}

{{dx}} = \frac{{y^2 + 3xy + x^2 }}

{{x^2 }} \hfill \\

\frac{{dy}}

{{dx}} = \frac{{v^2 + 3v}}

{x} - \frac{v}

{x} \hfill \\

\frac{{dy}}

{{dx}} = v^2 + 2v/x \hfill \\

\hfill \\

\int {\frac{{dv}}

{{v(v + 2v)}} = \int {\frac{{dx}}

{x} = > \int {(\frac{1}

{v} - \frac{1}

{{1 + v}})dv = \int {\frac{{dx}}

{x}} } } } \hfill \\

\end{gathered}

\)

\(\displaystyle

\ln |v/2 + v| = \ln |x| + A

\)

......

I need to know how to get to this....

\(\displaystyle

\frac{x}

{{A - \ln |x|}} - x

\)

\(\displaystyle \frac{dy}{dx} = \frac{y^2 + 3xy + x^2 }{x^2}\)

With \(\displaystyle y = vx\) substitution:

\(\displaystyle v + x\frac{dv}{dx} = v^2 + 3v + 1\)

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

\(\displaystyle \int \frac{dv}{(v+1)^2} = \int \frac{dx}{x} \)

\(\displaystyle -\frac1{v+1} = \ln C|x|\)

\(\displaystyle v = -\frac1{\ln C|x|} - 1 \Rightarrow y = -\frac{x}{\ln C|x|} - x\)

So the solution is \(\displaystyle y = \frac{x}{A - \ln|x|} - x\) where \(\displaystyle A = - \ln C\), where A is as arbitrary as C is.*

*- These are Mr.Fantastic's words... I like it.