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Say I have an bvp that involve the equation below

h''' = f(r)*h'' + h'

If I was to have enough boundary conditions to satisfy h(r) ODE, would it be possible to solve the f(r) equation simultaneously in order to get a solution for the h(r) equation?

Ideally f(r) would be a constant coefficient, but as seen with the solutions I am currently getting I'm wondering if the physics involving the currently considered constant coefficient equation h''' = C*h'' + h' describes the physics incorrectly. I would like to see how the now variable coefficient "f(r)" changes in order to get the h(r) to fit my boundary conditions. Any information or examples would be greatly appreciated!

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  • $\begingroup$ As you can always set $f=\frac{h'''-h'}{h''}$, you just have to avoid zeros in the second derivative but are else free to construct $h$ any way you like. $\endgroup$ – LutzL Mar 5 at 10:04

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