ODE from diffusion eqn - Separation of Variables help!

jchung

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Aug 13, 2009
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i'm trying to solve this equation
\(\displaystyle U''+\frac{1}{x}U'+\frac{\lambda}{\alpha}U = 0\)

which I think is the spatial problem from separation of variables of the diffusion equation \(\displaystyle \frac{\partial^{2}T}{\partial x^{2}} + \frac{1}{x}\frac{\partial T}{\partial x} = \frac{1}{\alpha}\frac{\partial T}{\partial t}\) where \(\displaystyle T(x,t) = U(x)V(t)\)

So for \(\displaystyle U''+\frac{1}{x}U'+\frac{\lambda}{\alpha}U = 0\)

what would the characteristic equation look like for this? does the form \(\displaystyle e^{Ax}\) work?

I tried it and i got \(\displaystyle A^2+\frac{1}{x}A+\frac{\lambda}{\alpha} = 0\) but the \(\displaystyle \frac{1}{x}\) has me stuck..
 
jchung said:
i'm trying to solve this equation
\(\displaystyle U''+\frac{1}{x}U'+\frac{\lambda}{\alpha}U = 0\)

which I think is the spatial problem from separation of variables of the diffusion equation \(\displaystyle \frac{\partial^{2}T}{\partial x^{2}} + \frac{1}{x}\frac{\partial T}{\partial x} = \frac{1}{\alpha}\frac{\partial T}{\partial t}\) where \(\displaystyle T(x,t) = U(x)V(t)\)

So for \(\displaystyle U''+\frac{1}{x}U'+\frac{\lambda}{\alpha}U = 0\)

what would the characteristic equation look like for this? does the form \(\displaystyle e^{Ax}\) work?

No - characteristic equation works when you have constant coefficients

I tried it and i got \(\displaystyle A^2+\frac{1}{x}A+\frac{\lambda}{\alpha} = 0\) but the \(\displaystyle \frac{1}{x}\) has me stuck..

Try series solution
 
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