Undetermined Coefficients and Resonance

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The idea

For a linear equation with constant coefficients, polynomial, exponential and sinusoidal forcing admit finite trial families. A particular solution accounts for the forcing; the homogeneous solution supplies the freedom needed for initial conditions.

Visual guide

VISUAL GUIDEResonant forcing builds an increasing envelope

Resonant response
Positive envelope
Negative envelope

For y″ + y = cos t with zero initial state, y = (t/2) sin t. The dashed lines ±t/2 bound its oscillation. This unbounded growth belongs to the ideal undamped model under sustained resonant forcing.

Method and assumptions

Choose a trial closed under differentiation: a polynomial of the same degree, an exponential times such a polynomial, or both sine and cosine. If the trial overlaps a homogeneous mode, multiply it by $t$ enough times to restore independence. Substitute to determine coefficients.

Worked example: nonresonant forcing

For $y''+y=\cos2t$ , try $A\cos2t+B\sin2t$ . Substitution gives $-3A=1,-3B=0$ , so $y_p=-\cos2t/3$ . Add $c_1\cos t+c_2\sin t$ before imposing data.

PAUSE & THINKA quick check, not a grade

Try it yourself.

Choose an answer and explain your reasoning to yourself. Use a hint if you get stuck.

For y″+y=cos t, which particular-solution trial handles resonance?

t(A cos t+B sin t)

A cos t+B sin t

A eᵗ

Hint 1 · Find a starting point

The forcing already belongs to the homogeneous solution family.

Hint 2 · Take the next step

Multiply the overlapping sinusoidal trial by t.

Show the reasoning

Answer: t(A cos t+B sin t)

The extra factor removes the overlap; a valid particular solution is (t/2)sin t.

Before moving on: what would make one of the other answers wrong? Saying why is part of understanding.

Worked example: resonance

For $y''+y=\cos t$ , the ordinary sinusoidal trial is already homogeneous. A working choice is $y_p=(t/2)\sin t$ . Differentiating twice gives $y_p''+y_p=\cos t$ . With zero initial displacement and velocity, this is the full solution, whose envelope grows linearly.

Interpreting the result

Resonance in this ideal undamped model produces unbounded growth under sustained forcing. Real damping changes the response. A large finite peak in a damped system is different from the exact secular growth shown here.

Practice

Choose a trial for $y''+y=t^2$ .
Find a particular solution for $y''-y=e^{2t}$ .
Why does $A\cos t$ fail for resonant forcing?

Show worked solutions

Use $At^2+Bt+C$ ; matching gives $y_p=t^2-2$ .
$y_p=e^{2t}/3$ because $4-1=3$ .
The operator sends it to zero, so no value of $A$ can produce $\cos t$ .

Further study

MIT OpenCourseWare: Differential Equations provides a full university course with additional lectures and exercises.

MAKE IT YOURS

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