# Differential equation substitution

• May 23rd 2008, 11:38 AM
dankelly07
Differential equation substitution
If anyone could explicitly show me the steps of how this substitution works I'd be v. grateful

$
\frac{{dy}}
{{dx}} = \frac{{x^2 + y^2 + 9xy}}
{{9x^2 }} \Rightarrow \frac{{dy}}
{{dx}} = \frac{1}
{9}(1 + (\frac{y}
{x})^2 ) + \frac{y}
{x}
$
• May 23rd 2008, 12:20 PM
Krizalid
Now substitute $y=xz\implies y'=z+xz'$ and replace this into the equation.
• May 24th 2008, 03:44 AM
dankelly07
is this substituion always valid for homogenous equations?
• May 24th 2008, 05:59 AM
Krizalid
As far as I know, yes.

Unless someone wants to give another opinion.
• May 25th 2008, 09:22 AM
dankelly07
just quickly one more thing...

Where does this v^2 come from in the last equation????

$
\frac{{dy}}
{{dx}} = \frac{{3y - x}}
{{x + y}}
$

so use
$
y = vx
$

$
v + xv' = \frac{{3v - 1}}
{{1 + v}}
$

$
xv' = \frac{{3v - 1 - v - v^2 }}
{{1 + v}}
$
• May 25th 2008, 10:50 AM
Krizalid
$\frac{3v-1}{1+v}-v=\frac{3v-1-v-v^{2}}{1+v}.$

It's just algebra.