Higher Order Nonhomogeneous Equations

A nonhomogeneous equation is simply one with something outside pushing on the system: a right side $g(t)$ that isn't zero. ("Homogeneous" means every term has $y$ in it; the extra $g(t)$ has no $y$, which is what makes it nonhomogeneous.) All we're really saying here is that for higher order equations you find the answer the same way you did in second order. The answer $y = y_c + y_p$ is just two pieces: $y_c$ (what the system does on its own) and $y_p$ (one response to the push, guessed in the push's shape). The only new thing is that your first guess can collide with $y_c$ more than twice, so you may need to multiply it by $t$ a few extra times. It sounds like a lot, but it's one old rule with a bigger number in it; bear with it.

Everything from undetermined coefficients in second order still applies: $y = y_c + y_p$, $y_c$ comes from the characteristic polynomial, and $y_p$ is a guess shaped like the forcing. The guess table is the same too.

The duplication rule now depends on multiplicity. If the forcing's form matches a root of multiplicity $k$, multiply the guess by $t^k$. In second order you only ever needed $t$ or $t^2$, because a root could repeat at most twice. Think about why the power has to be $k$ now. A root repeated $k$ times puts $e^{rt}, te^{rt}, \dots, t^{k-1}e^{rt}$ all into $y_c$, and each of those gives $0$ when substituted. You have to climb past all of them, to $t^ke^{rt}$, before the guess can produce anything nonzero.

A common mistake. In $y''' - y' = 2t$, it's tempting to think the guess $At + B$ is safe because $y_c = c_1 + c_2e^t + c_3e^{-t}$ has no $t$ in it. It isn't: $r = 0$ is a root, so constants already solve the homogeneous equation, and the $B$ in the guess would just vanish. Multiply by $t$: guess $At^2 + Bt$. A root at $0$ is easy to miss because $e^{0t} = 1$ doesn't look like an exponential.

A beam under a distributed load satisfies a fourth-order forced equation; its deflection is a particular solution plus homogeneous terms fixed by how the ends are supported.