# 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

$\displaystyle \frac{{dy}} {{dx}} = \frac{{x^2 + y^2 + 9xy}} {{9x^2 }} \Rightarrow \frac{{dy}} {{dx}} = \frac{1} {9}(1 + (\frac{y} {x})^2 ) + \frac{y} {x}$
• May 23rd 2008, 12:20 PM
Krizalid
Now substitute $\displaystyle y=xz\implies y'=z+xz'$ 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????

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

so use
$\displaystyle y = vx$

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

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

It's just algebra.