Non Linear homogenous differential equation

Sep 2007
81
0
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....

\(\displaystyle
\frac{x}
{{A - \ln |x|}} - x
\)
 
Dec 2007
1,465
674
IISc, Bangalore
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.