Recipe

Solve a forced constant-coefficient ODE for its steady state in frequency space

Use when: you have a constant-coefficient linear differential equation forced by a sinusoid, and only need the steady-state response

Goal: the steady-state $y(t)$ , found by dividing one complex number by another instead of solving a differential equation directly

Taught in Constant-coefficient differential equations become algebra in frequency space

Quick recipe

Write the equation as $Ly=f$ and build $L(i\omega)$
Rewrite the forcing term as a pure exponential
Evaluate $L(i\omega)$ and divide
Convert the complex answer back to a real solution

1
Collect every derivative into one operator $L$ built from $D$ and constants, then replace each $D$ with $i\omega$ and each $I$ with $1$ .

Example

$L=D^2+3D+2I$ becomes the polynomial $L(i\omega)=(i\omega)^2+3(i\omega)+2$ .

Why: Section 1 showed $e^{i\omega t}$ is an eigenfunction of any polynomial in $D$ , with eigenvalue exactly $L(i\omega)$ .
2
If the forcing is $\cos$ or $\sin$ , use Euler's formula to write it as the real (or imaginary) part of $e^{i\omega t}$ , and note which $\omega$ that is.

Example

$\cos(2t)=\operatorname{Re}\big[e^{i2t}\big]$ , so $\omega=2$ .

Why: the division trick in the next step applies directly only to a pure exponential mode, so a real sinusoid needs translating first.
3
Plug $\omega$ into $L(i\omega)$ as a complex number, then set $y(t)=\dfrac{1}{L(i\omega)}e^{i\omega t}$ .

Example

$L(2i)=(2i)^2+3(2i)+2=-2+6i$ , so $y(t)=\dfrac{1}{-2+6i}e^{i2t}$ .
- $L(i\omega)\ne0$ — the division is well-defined, giving a genuine bounded steady state.
- $L(i\omega)=0$ — this frequency is a resonance of the unforced equation, and no bounded steady state exists there.
Why: this step is pure algebra, since Step 1 already converted differentiation into multiplication by $i\omega$ .
4
Rationalize the fraction, convert it to polar form, and take the real part if the original forcing was $\cos$ or $\sin$ .

Example

$\dfrac{1}{-2+6i}=\dfrac{-2-6i}{40}\approx0.158\,e^{-i1.89}$ , so taking the real part gives $y(t)\approx0.158\cos(2t-1.89)$ .

Why: the physical forcing was real, so only the real part of the complex solution is an actual, physical signal.

Common mistakes

Dividing by $L(\omega)$ instead of $L(i\omega)$ ; the polynomial must be evaluated at the imaginary number $i\omega$ .
Forgetting to take the real part at the end, when the original forcing was a real cosine or sine.
Treating $L(i\omega)=0$ as an error to fix, rather than the genuine resonance it signals.
Reusing one frequency's answer for a different forcing frequency without recomputing $L(i\omega)$ there.