Tricky....

Reduce u'' + (ax+b)u = 0 to u'' + tu = 0 by a suitable change of variables.

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- Feb 26th 2010, 07:27 AMjoeyjoejoeChange of Variables
Tricky....

Reduce u'' + (ax+b)u = 0 to u'' + tu = 0 by a suitable change of variables. - Feb 26th 2010, 08:36 AMsashikanth
Let ax+b = t, then by differenciating u(x) with respect to t twice using the chain rule you will get (a^2)u'' + tu = 0. Is this what you wanted?

- Feb 26th 2010, 08:39 AMMauritzvdworm
this can surely not be the whole story, since by just chosing $\displaystyle t=ax+b$ your problem will already be solved...

can you give a bit more information?

I suppose it should read like this:

$\displaystyle \frac{d^{2}u(x)}{dx^{2}}+(ax+b)u(x)=0$

and you want to make a substitution such that you end up with

$\displaystyle \frac{d^{2}u(z)}{dz^{2}}+tu(z)=0$

??