solve the following differential equation

**sarah232** · April 10th 2009, 12:38 PM

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.

**Danny** · April 10th 2009, 01:03 PM

Originally Posted by sarah232

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 $y = z^3 u$ .

So $y' = z^3 u' + 3z^2 u$ and $y'' = z^3u'' + 6z^2u' + 6z u$ . Now substitute into your ODE and it becomes

$z(z^3u'' + 6z^2u' + 6z u)+(z-2)(z^3 u' + 3z^2 u) - 3 z^3 u = 0$

which becomes, after some simplification

$z u'' + (z+4)u' = 0$

If you let $v = u'$ then we obatin $z v' + (z+4)v = 0$ (separable). Integrate, put back in $u'$ and integrate again to get u. Then $y = z^3 u$ .

**sarah232** · April 10th 2009, 02:48 PM

Do you mind showing me the Wronskian method? I believe I will need to know this for my exam. Thanks!

**Danny** · April 11th 2009, 04:56 AM

Originally Posted by sarah232

Do you mind showing me the Wronskian method? I believe I will need to know this for my exam. Thanks!

Here a post from this forum on that http://www.mathhelpforum.com/math-he...e-problem.html

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