Finding a particular solution for constant forcing resonance

Key takeaway: This problem asks for a particular solution to a second-order linear differential equation with constant forcing, using the method of undetermined coefficients and the idea of resonance.

Recognize the differential equation structure

We are given

$\frac{d^2y}{dt^2}+9y=f(t),$

with homogeneous solution

$y_h(t)=A\cos(3t)+B\sin(3t).$

The coefficient 9 tells us the natural frequency is 3. To find a particular solution for a constant force $f(t)=a_0$, we use the fact that a constant forcing term usually suggests trying a constant trial solution.

Try a constant particular solution

Assume

$y_p(t)=C.$

Then

$y_p'(t)=0,\qquad y_p''(t)=0.$

Substitute into the differential equation:

$0+9C=a_0.$

So

$C=\frac{a_0}{9}.$

Therefore the particular solution is

$\boxed{y_p(t)=\frac{a_0}{9}}.$

Why this works and what it means physically

A constant forcing function acts like a steady displacement. Since the system is not being driven at its natural frequency, there is no resonance issue. The particular solution is just a constant offset from equilibrium.

You can verify it directly: if $y(t)=\frac{a_0}{9}$, then $y''=0$ and $9y=a_0$, so the equation is satisfied exactly.

Common method and resonance check

The method of undetermined coefficients asks you to match the forcing shape with a trial form. For a polynomial of degree 0, the trial is a constant, unless that constant duplicates a homogeneous solution. Here, constants are not in $y_h$, so no modification is needed.

The complete solution would be

$y(t)=A\cos(3t)+B\sin(3t)+\frac{a_0}{9}.$

That is the standard forced oscillator solution with constant input.

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Pitfalls the pros know 👇 A common error is to guess $y_p(t)=a_0$ instead of solving for the actual constant. Another mistake is to think any forcing term automatically requires multiplication by $t$. That only happens when the trial function overlaps with the homogeneous solution. Since $1$ is not part of $\cos(3t)$ or $\sin(3t)$, the constant trial works without adjustment. Also, be careful not to carry over the symbol $x$ from the homogeneous solution in the prompt; the independent variable here is $t$, so the final answer should be written in $t$ consistently.

What if the problem changes? If the forcing were linear, say $f(t)=a_0+a_1t$, then the appropriate trial would be a linear polynomial $y_p(t)=At+B$, provided it does not overlap with the homogeneous solution. If the equation were changed to $y''+9y=9a_0$, the same method would give $y_p=a_0$. If the coefficient of $y$ were changed to $4$ instead of $9$, the constant-forcing approach would still work, but the denominator in the final constant would change from 9 to 4. These variants show that the trial shape follows the forcing, while the coefficient in the differential equation determines the scaling.

Tags: undetermined coefficients, forced oscillator, homogeneous solution