Why does forcing in the homogeneous solution break undetermined coefficients?

[EvelynR]

I keep hitting a wall trying to understand why the method of undetermined coefficients for solving non-homogeneous linear ODEs fails when the forcing function is part of the complementary solution. My textbook just states the rule about the particular solution needing to be linearly independent, but I can't visualize the "why" behind that algebraic roadblock.

[Stella_J]

I hit this when I tried y'' - y' = e^{t}. The complementary solution has e^{t} in it, so a guess y_p = A e^{t} just sits in the same space and you end up with 0 on substitution. So I tried y_p = A t e^{t}. After plugging in: y_p' = A e^{t}(t+1), y_p'' = A e^{t}(t+2). Then L[y_p] = y_p'' - y_p' = A e^{t}, so you can match the forcing by taking A = 1. Then y_p = t e^{t}. It feels clunky until you realize you’re lifting out of the homogeneous space by multiplying by t.

[ScottVW]

Intuition: the operator L kills anything in its kernel. If your trial solution sits in that kernel, L[y_trial] is zero, but the right side g is not zero. So you can't solve for coefficients because you'd be trying to equal zero to a nonzero thing. Multiply by t or switch to a different form to push your trial out of the kernel, and suddenly you can match g.

[Zoey_R]

One thing I'm not sure about: sometimes you think the forcing is outside the homogeneous space but it actually isn't in the correct combination, and you waste time chasing the wrong form. Do you double check whether g is genuinely independent of the complementary solution, or if it might be a piece that the homogeneous part can already produce with some choice of constants?

[James.J]

I remember trying annihilator ideas for a while, and it felt like a detour; eventually I circled back to the simple reason: you need a trial that's outside the kernel. It helps when the roots are repeated or complex; multiply by t or by x etc. Then the failure case disappears.