# Thread: Non Linear homogenous differential equation

1. ## Non Linear homogenous differential equation

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....

$\displaystyle \frac{x} {{A - \ln |x|}} - x$

2. Originally Posted by dankelly07
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.