The Resonance Modification Rule

When the educated guess is already a solution

The previous guide taught you to attack a y'' + b y' + c y = f(x) by writing down a trial solution shaped like the forcing f and solving for its coefficients. That works beautifully — until it spectacularly does not. Try y'' - 3 y' + 2 y = e^(2x). The forcing is e^(2x), so the natural guess is y_p = A e^(2x). Substitute it and every term collapses: 4A - 6A + 2A = 0, which says 0 = 1. There is no A on Earth that satisfies that. The method has not made an error; it has hit a wall.

Look at why. The characteristic equation r^2 - 3r + 2 = 0 has roots r = 1 and r = 2, so the complementary solution is y_c = C1 e^(x) + C2 e^(2x). Your guess A e^(2x) is literally one of those homogeneous pieces. Feeding a homogeneous solution into the left-hand side can only ever produce zero — that is what 'homogeneous solution' means. So it can never equal a nonzero forcing e^(2x). The guess was doomed the moment it overlapped y_c.

The fix: multiply by t

Here is the whole rule. When the trial guess duplicates a term of y_c, multiply that guess by x (or by t, whichever your variable is) and try again. For the example, change y_p = A e^(2x) into y_p = A x e^(2x). This modification rule — also called the duplication rule — is the entire content of this guide. Everything else is understanding why one extra factor of x is exactly the right repair, no more and no less.

Run it through. With y_p = A x e^(2x), the product rule gives y_p' = A e^(2x) (1 + 2x) and y_p'' = A e^(2x) (4 + 4x). Substitute into y'' - 3 y' + 2 y: the e^(2x) factors out, and the bracket becomes (4 + 4x) - 3(1 + 2x) + 2x = 4 + 4x - 3 - 6x + 2x = 1. Notice the x-terms cancel completely — they had to, because A x e^(2x) sits half inside the homogeneous world — and what survives is a clean constant. So A e^(2x) = e^(2x) forces A = 1, and y_p = x e^(2x). The wall is gone.

forcing f(x)        bare guess           if it appears in y_c, modify to
-----------------   ------------------   -------------------------------
  e^(2x)              A e^(2x)             A x e^(2x)
  sin(w x)            A cos + B sin        x (A cos + B sin)
  3x + 1              A x + B              x (A x + B)
  e^x  (double root)  A e^x                A x^2 e^x      <- multiply by x^2

Why one t, and when it must be t-squared

The factor of x is not magic; it is the same x that showed up in the repeated-root case of the homogeneous theory, where a double root r gave the second solution x e^(rx). The mechanism is identical. A simple root contributes e^(rx) to y_c; if the forcing pushes at that exact same rate r, the response is one notch 'deeper' — it grows with an extra factor of x. One overlap, one factor of x. The depth of the overlap sets the power.

That is why the general statement is: multiply by x^s, where s is the multiplicity of the matching root. If the forcing e^(rx) hits a root that the characteristic equation has as a DOUBLE root, then both e^(rx) and x e^(rx) are already in y_c, so a single x is not enough — x e^(rx) is still homogeneous — and you must go to x^2 e^(rx). For a plain second-order equation s is only ever 0, 1, or 2; s = 0 means no overlap and the unmodified guess from the last guide is correct.

What the algebra is really telling you: resonance

The algebraic collision has a vivid physical name. When the forcing matches a natural mode of the system — when you push a swing at exactly its own rhythm — you have resonance, and a resonant forcing term is precisely a forcing that overlaps y_c. The extra factor of x is the mathematics announcing that the response does not settle to a steady oscillation of fixed size; instead its amplitude grows. The bounded e^(2x)-style answer has been replaced by an x e^(2x) answer that climbs without bound.

The cleanest picture is an undamped oscillator y'' + w^2 y = cos(w x): a spring pushed at exactly its own natural frequency w. The complementary solution is cos(w x) and sin(w x), so the forcing cos(w x) is resonant, and the modification rule turns the guess into x times sines and cosines. The particular solution is proportional to x sin(w x) — an oscillation whose envelope x grows linearly forever. This is pure resonance: the textbook glass shattered by a sustained note, the amplitude marching upward without ceiling.

One honest caveat that this picture invites you to forget: resonance does NOT require zero damping. Pure, unbounded, linear-in-x growth is the idealized, frictionless extreme. Add even a little damping (a y'' + b y' + c y with b > 0) and the roots gain a negative real part, the forcing frequency no longer exactly matches a root, the bare guess succeeds without any x, and the response stays bounded — but it can still peak sharply when the driving frequency is near the natural one. Real resonance is usually a large finite response, not an infinite one; the runaway x is the special undamped case.

The recipe and its honest limits

1. Solve the homogeneous problem first and write down y_c — you cannot apply the rule without knowing it.
2. Form the natural trial guess for the forcing f(x), exactly as in the previous guide (no x-factor yet).
3. Compare: does any term of the guess already appear in y_c? If not, s = 0 and you are done modifying.
4. If there is an overlap, multiply the WHOLE guess by x^s, where s is the multiplicity of the matching root (1 for a simple root, 2 for a double root).
5. Substitute the modified guess, solve for the coefficients, and add y_c + y_p for the general solution.

Carry the boundaries with you. This rule, like the particular solution method it repairs, lives entirely inside the constant-coefficient, special-forcing world: f must be a polynomial, an exponential, a sine or cosine, or a product of those. Hand it a forcing like tan(x), ln(x), or 1/x and undetermined coefficients has no finite guess to modify at all — that is the territory of the next guide, variation of parameters, which handles any forcing but costs more work.

Two reminders to keep you honest. First, x^s is the minimum bump that works; using x^2 when x already suffices does not break the method but loads you with extra dead coefficients that all come out zero — wasted algebra, not an error. Second, the modification touches only the resonant block: if a forcing is a sum like e^(2x) + sin(3x) and only e^(2x) is resonant, multiply only that block by x and leave the sine alone. Treat each piece of the forcing on its own merits, which is exactly the superposition idea the final guide of this rung makes precise.