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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}
$
Now substitute $\displaystyle y=xz\implies y'=z+xz'$ and replace this into the equation.
is this substituion always valid for homogenous equations?
As far as I know, yes.
Unless someone wants to give another opinion.
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}}
$
$\displaystyle \frac{3v-1}{1+v}-v=\frac{3v-1-v-v^{2}}{1+v}.$
It's just algebra.
