Differential equation substitution

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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.

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