Method of Undetermined Coefficients

What is undetermined coefficients for?

For a non-homogeneous linear ODE $a y'' + b y' + c y = g(x),$ the general solution has the form $y = y_h + y_p$ where

• $y_h$ is the homogeneous solution (general solution of the ODE with $g$ replaced by $0$), and
• $y_p$ is any particular solution of the full non-homogeneous equation.

Why this works: if $L$ denotes the linear operator on the left ($L y = a y'' + b y' + c y$), then $L y_h = 0$ and $L y_p = g$, so $L(y_h + y_p) = 0 + g = g$. Any particular $y_p$ does the trick — we just need one.

The method of undetermined coefficients is the fastest way to find $y_p$ when $g(x)$ is one of a few "nice" forms — polynomials, exponentials, sines/cosines, and products of these. You guess a function of the same form as $g$ (with unknown coefficients) and plug in to determine the coefficients.

Limitation: only works for those special right-hand sides. For arbitrary $g$, use variation of parameters (see #187).

The given equation

$y'' + y = \sin(2 x)$

Step 1 — Find the homogeneous solution

$y'' + y = 0$: characteristic equation $r^{2} + 1 = 0 \implies r = \pm i$ (see #184). $y_h(x) = C_1 \cos x + C_2 \sin x.$

Step 2 — Guess the particular solution

The right side is $\sin 2x$. The derivatives of $\sin 2x$ and $\cos 2x$ stay inside the 2D span $\{\sin 2x, \cos 2x\}$, so the right guess is $y_p = A \sin 2x + B \cos 2x$ (even though the RHS only has $\sin 2x$, the derivatives will bring in $\cos 2x$ and we need both coefficients to balance).

Important check — resonance. Before committing to this guess, check that it's not already a homogeneous solution. Here $y_h$ has frequency $1$ ($\cos x, \sin x$) while the guess has frequency $2$ — no overlap. ✓ If there were overlap, we'd multiply by $x$ (or higher powers) to escape. More on this below.

Step 3 — Compute $y_p'$ and $y_p''$

$y_p' = 2 A \cos 2x - 2 B \sin 2x$ $y_p'' = -4 A \sin 2x - 4 B \cos 2x$

Step 4 — Substitute into the ODE and match coefficients

$y_p'' + y_p = (-4 A \sin 2x - 4 B \cos 2x) + (A \sin 2x + B \cos 2x)$ $= -3 A \sin 2x - 3 B \cos 2x$

Set equal to the right side $\sin 2x = 1 \cdot \sin 2x + 0 \cdot \cos 2x$:

• $\sin 2x$ coefficient: $-3 A = 1 \implies A = -\dfrac{1}{3}$.
• $\cos 2x$ coefficient: $-3 B = 0 \implies B = 0$.

Step 5 — Assemble particular solution. $y_p(x) = -\frac{1}{3} \sin 2x.$

Step 6 — General solution

$\boxed{\, y(x) = C_1 \cos x + C_2 \sin x - \tfrac{1}{3} \sin 2x \,}$

Verification

$y_p = -\tfrac{1}{3} \sin 2x$, so $y_p'' = \tfrac{4}{3} \sin 2x$. $y_p'' + y_p = \tfrac{4}{3} \sin 2x - \tfrac{1}{3} \sin 2x = \sin 2x \; \checkmark$

The guess-table for undetermined coefficients

Match the right side $g(x)$ to the corresponding guess for $y_p$:

• $g =$ polynomial of degree $n$ → $y_p = A_n x^{n} + \ldots + A_1 x + A_0$ (full polynomial of same degree).
• $g = e^{\alpha x}$ → $y_p = A e^{\alpha x}$.
• $g = \sin \beta x$ or $\cos \beta x$ → $y_p = A \sin \beta x + B \cos \beta x$ (both!).
• $g = e^{\alpha x} \sin \beta x$ or $e^{\alpha x} \cos \beta x$ → $y_p = e^{\alpha x}(A \sin \beta x + B \cos \beta x)$.
• $g$ = product of the above → $y_p$ = product of the corresponding guesses.
• $g$ = sum of different forms → $y_p$ = sum of corresponding guesses (solve each part separately).

The resonance fix — multiply by $x^s$

If your guess overlaps with $y_h$, it fails (plugging in gives $0 = g$, a contradiction). Fix: multiply the guess by the smallest $x^s$ that eliminates the overlap.

Example. $y'' + y = \sin x$. Now the guess $A \sin x + B \cos x$ is part of $y_h$ — this is the resonant forcing. Upgrade: $y_p = x (A \sin x + B \cos x).$

Solving gives $A = 0$, $B = -1/2$, so $y_p = -\tfrac{1}{2} x \cos x$ — the particular solution grows linearly because we're driving the oscillator at its natural frequency. This is the archetype of physical resonance: amplitude builds indefinitely.

Initial value problem

For our problem with $y(0) = 0$, $y'(0) = 0$:

$y = C_1 \cos x + C_2 \sin x - \tfrac{1}{3} \sin 2x$ $y(0) = C_1 = 0$ $y' = -C_1 \sin x + C_2 \cos x - \tfrac{2}{3} \cos 2x$ $y'(0) = C_2 - \tfrac{2}{3} = 0 \implies C_2 = \tfrac{2}{3}$

$y(x) = \tfrac{2}{3} \sin x - \tfrac{1}{3} \sin 2x.$

Common mistakes

• Leaving out $\cos$ when forcing is $\sin$. The derivative of $\sin$ is $\cos$; the guess must be a full linear combination of both.
• Not checking resonance. Plug in the homogeneous solutions $y_1, y_2$ and see if your guess already solves $L y = 0$. If yes, bump up by $x$.
• Using the wrong multiplicity of $x$. A guess that matches a double homogeneous root needs $x^{2}$, not $x$.
• Forgetting to include $y_h$. The general solution is $y_h + y_p$; $y_p$ alone only works for zero initial conditions by coincidence.

Try it in the visualization

Sweep the forcing frequency and watch the amplitude of the particular solution. Cross the natural frequency of the oscillator — the amplitude blows up (mathematically reflected in the $x^s$ resonance fix producing $x \cos \omega_0 x$ / $x \sin \omega_0 x$ growth).