(adsbygoogle = window.adsbygoogle || []).push({}); 1. The problem statement, all variables and given/known data

Ugh I feel really stupid for posting this but for some reason I can't remember how to solve it.

I am trying to solve the diff eqn: D*[itex]\frac{d^{2}\phi}{dx^{2}}[/itex] - [itex]\Sigma_{a}[/itex]*[itex]\phi(x)[/itex] = -q[itex]_{t}[/itex]

(thermal diffusion equation for neutrons slowing down to thermal energy)

3. The attempt at a solution

Anyways I've taken the roots of the LHS as if it was homogeneous:

([itex]\phi(x)[/itex][itex]^{(2)}[/itex]-[itex]\frac{\Sigma_{a}}{D}[/itex]) = 0

which gives the 2 roots ±[itex]\sqrt{\frac{\Sigma_{a}}{D}}[/itex]

and therefore [itex]\phi(x)[/itex] = C1*[itex]e^{\sqrt{\frac{\Sigma_{a}}{D}}x}[/itex] + C2*[itex]e^{-\sqrt{\frac{\Sigma_{a}}{D}}x}[/itex]

so the only thing that's bothering me is the -q[itex]_{t}[/itex] on the RHS which I don't know what to do with. I've used Maple to get the solution and it gives me

[itex]\phi(x)[/itex] = C1*[itex]e^{\sqrt{\frac{\Sigma_{a}}{D}}x}[/itex] + C2*[itex]e^{-\sqrt{\frac{\Sigma_{a}}{D}}x}[/itex] + [itex]\frac{q_{t}}{\Sigma_{a}}[/itex]

I would just like to know where that last term comes from and how to get it.

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# 2nd order Differential Equation

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