Choose a particular solution that does not duplicate the homogeneous basis

Try

For y″−y=e^(2t), try a particular solution yₚ=Ae^(2t). Substitution gives (4A−A)e^(2t)=e^(2t), so A=1/3. The general solution also includes the homogeneous modes Ce^t+De^(−t). A particular solution handles the forcing; it does not usually satisfy the requested initial data by itself.

Now replace the forcing with e^t. The same trial Ae^t is annihilated by y″−y, regardless of A. No choice of the coefficient can repair a trial whose entire left side vanishes.

Predict

For y″+y=cos t, predict whether A cos t+B sin t can supply a nonzero forcing. Both functions already solve the homogeneous equation. Try multiplying the whole candidate by t, and predict that the response may grow even though the unforced system has bounded oscillations.

Experiment

Take yₚ=(t/2)sin t and verify it without relying on a remembered resonance formula.

Quantity	Expression
yₚ	(t/2)sin t
yₚ′	(1/2)sin t+(t/2)cos t
yₚ″	cos t−(t/2)sin t
yₚ″+yₚ	cos t

The terms proportional to t cancel in the equation, leaving the forcing. Yet the solution’s oscillation amplitude grows in proportion to t. Sketch the envelopes y=±t/2 and place several oscillations between them. This ideal undamped resonance is a consequence of the model, not a claim that a real device can grow without limit.

Understand

Undetermined coefficients works efficiently when the forcing belongs to a finite-dimensional family closed under differentiation: polynomials, exponentials, sine and cosine, and suitable products or sums. Choose a trial family broad enough that differentiation stays within it, substitute, and solve for coefficients.

If the trial overlaps the homogeneous family, multiply it by enough powers of t to remove the overlap. For a forcing e^(λt) times a polynomial, the required power corresponds to the multiplicity of λ as a characteristic root. For oscillatory forcing, use both sine and cosine terms because differentiation mixes them, even if the forcing initially contains only one.

For (D−1)²y=e^t, where D means differentiation, the root one has multiplicity two. Set y=e^t v. Direct differentiation shows (D−1)(e^t v)=e^t v′, so applying it twice gives e^t v″. Thus v″=1 and one particular choice is v=t²/2. The particular solution is (t²/2)e^t; multiplying by t only once would still be insufficient.

For y″+y=3, a constant particular solution yₚ=3 works. Its simplicity does not remove the homogeneous sine and cosine terms. Conversely, a forcing such as ln t does not fit the same finite derivative family, so a different method is usually more appropriate.

Apply

Classify several forcings before calculating: constant, e^(2t), cos t, and ln t. For each equation, compare the forcing family with the characteristic roots. State the trial form and explain any time multiplier. This decision is more transferable than memorizing a list of finished solutions.

After finding a particular solution, add the homogeneous family and impose initial conditions. Verify the forcing and the initial values separately. Explain why adding two solutions for the same nonzero forcing doubles the forcing, while adding a homogeneous solution leaves it unchanged.

Continue

Variation of parameters handles broader continuous forcings. It uses the homogeneous basis already found, but replaces constant coefficients by functions chosen through a small linear system.