I am supposed to solve the following
zy'' + (z-2)y' -3y = 0 and y=z^3 is a solution.
I have a feeling I should use the Wronskian but I don't know how because my text book is horrible for explaining anything.
You can but an easy way is let $\displaystyle y = z^3 u$.
So $\displaystyle y' = z^3 u' + 3z^2 u$ and $\displaystyle y'' = z^3u'' + 6z^2u' + 6z u$. Now substitute into your ODE and it becomes
$\displaystyle z(z^3u'' + 6z^2u' + 6z u)+(z-2)(z^3 u' + 3z^2 u) - 3 z^3 u = 0$
which becomes, after some simplification
$\displaystyle z u'' + (z+4)u' = 0$
If you let $\displaystyle v = u'$ then we obatin $\displaystyle z v' + (z+4)v = 0$ (separable). Integrate, put back in $\displaystyle u'$ and integrate again to get u. Then $\displaystyle y = z^3 u$.
Here a post from this forum on that http://www.mathhelpforum.com/math-he...e-problem.html