Undetermined Coefficients

Undetermined coefficients is simply a guess-and-check way to find how a system responds to an outside push. The name says it: you guess the shape of the answer with some numbers left blank (coefficients you haven't determined yet), and the equation fills them in. All we're really saying here is that the system's motion is what it would do on its own plus its reaction to the push. In $y = y_c + y_p$, the pieces are $y_c$ (the system's own natural motion, which you already know how to find) and $y_p$ (one response that matches the push exactly). The new work is only $y_p$, and for the common kinds of pushes you can basically guess it. It looks like a lot of new machinery, but it's two simple parts added together; bear with it.

Now what? Take a look at the equation we've been working with:

It looks just like the equations we solved with the characteristic equation, except the right side isn't zero anymore. It's some function $g(t)$, so the equation is nonhomogeneous. For a spring, $g(t)$ is an outside force pushing on it; we'll see what that does to the motion later in this unit.

The old trick doesn't work by itself here. Plugging in $y = e^{rt}$ gives $(ar^2 + br + c)e^{rt}$, and no choice of $r$ turns that into a $g(t)$ like $t^2$ or $\cos t$.

Solve it in two parts: $y = y_c + y_p$. The usual way is to split the answer in two. $y_c$ is the solution we already know how to get from the characteristic equation. $y_p$, the particular solution, is any one function that solves the full equation, right side included. Put them together and the left side becomes

Think about why that works. $y_c$ was built to make the left side $0$; that was the whole point of the characteristic equation. So when you plug in $y_c + y_p$, the $y_c$ part contributes nothing, and $y_p$ has to supply all of $g(t)$ on its own. $0 + g(t)$ is just $g(t)$. That's also our hint for finding $y_p$: if $ay_p'' + by_p' + cy_p$ has to come out as exactly $g(t)$, it makes sense that $y_p$ should look like $g(t)$ to begin with.

It works in reverse too: if $y$ is any solution, then $y - y_p$ makes the left side $g - g = 0$, so it's one of the homogeneous solutions. Every solution is $y_c + y_p$, nothing more.

For example, take $y'' - y = 3$. A constant does the job: $y_p = -3$ gives $0 - (-3) = 3$. The homogeneous equation $y'' - y = 0$ has $y_c = c_1e^t + c_2e^{-t}$, so every solution is

Picture a spring: $y_c$ is the motion it makes on its own, and $y_p$ is its response to the outside push $g(t)$. There's an animation further down that shows the two adding up to the real motion.

Finding $y_p$: guess something shaped like $g$. When the right side $g(t)$ is built from polynomials, exponentials, sines and cosines, or products of these like $(t+1)e^{2t}$, guess a $y_p$ of the same shape with unknown coefficients. Substitute it and match like terms: the equation decides the coefficients, which is where the name undetermined coefficients comes from.

The duplication rule. If a term of your guess already appears in $y_c$, it solves the homogeneous equation, so substituting it gives $0$, never $g(t)$. Multiply the guess by $t$, or by $t^2$ when the root is repeated, until nothing overlaps $y_c$.

A common mistake. In $y'' - 4y' - 12y = 3e^{5t}$, it's tempting to think $5$ is a root because it's sitting in the exponent on the right. It isn't. The roots only ever come from the characteristic equation, here $r^2 - 4r - 12 = 0$, which gives $6$ and $-2$. The right side tells you the shape of the guess; the characteristic equation tells you whether that guess collides with $y_c$.

The recipe:

1. Find $y_c$ from the characteristic equation.
2. Write the guess for $y_p$ from the table, then fix any overlap with $y_c$.
3. Substitute, match the coefficients of like terms, and solve for the unknowns.
4. Write $y = y_c + y_p$, then apply initial conditions to the whole thing, not to $y_c$ alone.

Why only these kinds of $g$? Substituting a guess takes its first and second derivatives. The method works when those derivatives stay inside a short list of functions, so finitely many unknown coefficients can match $g$. Polynomials, exponentials, sines and cosines are exactly the functions whose derivatives never leave their family: $(e^{3t})' = 3e^{3t}$, $(\sin 2t)' = 2\cos 2t$, $(t^2)' = 2t$, and products of them stay products of the same kinds. Now try $g(t) = \ln t$: its derivatives are $\frac1t, -\frac1{t^2}, \frac2{t^3}, \dots$, a new kind of term every time, so no finite guess can ever close up. For forcing like $\ln t$, $\tan t$ or $\frac1t$ there's variation of parameters, the next lesson.

Forced vibration problems in engineering, a bridge pushed by periodic wind or a circuit driven by an AC voltage, are solved with exactly these guesses. The particular solution is the steady response engineers design for; the complementary part is the transient that dies away.