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Suppose we have a linear differential of the form $\frac{dy}{dt} = a(t)y + b(t)$ with $b(t) \neq 0$. I understand that I need to solve the corresponding homogeneous equation $\frac{dy}{dt} = a(t)y$ and find a particular solution to the nonhomogenous equation and combine them to form the general solution.

I am actually reviewing this for a tutoring session. I vaguely recall from my own ODE class, that there are heuristics for finding a particular solution to the original nonhomogenous equation based on the form of $b(t)$, but I forget exactly what they are.

Based on the notes from my student's class, I have deduced the following:

  1. If $b(t)$ is of the form $b(t) = Ae^{kt}$ (with constants $A$ and $k$) and the solution to the corresponding homogenous equation is not a multiple of $e^{kt}$, we choose $y(t) = Be^{kt}$ (solving for $B$ if we have an initial condition).

  2. If $b(t)$ is of the form $b(t) = Ae^{kt}$ (with constants $A$ and $k$) and the solution to the corresponding homogenous equation is a multiple of $e^{kt}$, we choose $y_p(t) = Bte^{kt}$ (solving for $B$ if we have an initial condition).

  3. If $b(t)$ is of the form $b(t) = A\sin{t} + B\cos{t}$ (with constants $A$ and $B$, one of which may be zero) and the solution to the corresponding homogenous equation is not a linear combination of $sin{t}$ and $cos{t}$, we choose $y_p(t) = C\sin{t} + D\cos{t}$ (again, solving for $C$ and $D$ if we have an initial condition.

  4. If $b(t)$ is of the form $b(t) = A\sin{t} + B\cos{t}$ (with constants $A$ and $B$, one of which may be zero) and the solution to the corresponding homogenous equation is a linear combination of $sin{t}$ and $cos{t}$, we choose $y_p(t) = Ct\sin{t} + Dt\cos{t}$ (again, solving for $C$ and $D$ if we have an initial condition.

Am I missing any other forms of $b(t)$ that may arise in an undergraduate ODE course?

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The general solution takes the form

$$y \left( t \right) = \left( \int _{0}^{t}\!b \left( \tau \right) { {\rm e}^{-\int _{0}^{\tau}\!a \left( \sigma \right) {d\sigma}}}{d\tau} +y(0) \right) {{\rm e}^{\int _{0}^{t}\!a \left( \sigma \right) {d\sigma}}} $$

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  • $\begingroup$ We haven't covered integrating factors yet, so I don't think this solution will be appropriate for my students. $\endgroup$ – Code-Guru Sep 15 '14 at 23:43
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A little bit more research tracked down this paper. It looks like the only "basic" form I missed are polynomials. More complicated problems could combine any of these forms by adding or multiplying functions of the possible forms. However, I highly doubt my students will see something like that on an exam.

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