Recipe

Find a particular solution with undetermined coefficients

Use when: you have a nonhomogeneous linear ODE $ay''+by'+cy=g(x)$ , including the case where the naive trial function duplicates a term already in $y_c$

Goal: a particular solution $y_p$ that actually works, even when the naive guess collides with the homogeneous solution

Taught in The resonance modification rule · extends Find a particular solution with undetermined coefficients

Quick recipe

Guess a trial form matched to $g(x)$ 's shape
Check the guess against the homogeneous solution
Substitute the (possibly modified) guess into the equation
Match coefficients term by term
Solve for the coefficients and add $y_p$ to $y_c$

1
Write $y_p$ with the same kind of terms as $g(x)$ : a polynomial for a polynomial, an exponential for an exponential, a sine-and-cosine pair for a sine or cosine forcing, each carrying an unknown coefficient.

Example

For $y''-3y'+2y=e^x$ , the forcing $g(x)=e^x$ is a pure exponential, so the naive guess is $y_p=Ae^x$

Why: differentiating a polynomial, exponential, or sine/cosine stays inside that same family of functions, and that closure is exactly what makes the coefficients solvable by matching.
2
Solve the homogeneous equation first (Chapter 7's recipe) and compare its solution $y_c$ against the naive trial.

Example

The characteristic equation $r^2-3r+2=0$ factors as $(r-1)(r-2)=0$ , so $r=1$ and $r=2$ , giving $y_c=C_1e^x+C_2e^{2x}$ . The naive trial $Ae^x$ duplicates the $C_1e^x$ term already in $y_c$ ; the matching root $r=1$ is simple, so $s=1$ . Multiplying the entire trial by $x^1$ : $y_p=Axe^x$
- the naive trial shares no term with $y_c$ — use it as written, and continue to Step 4.
- the naive trial duplicates a term already in $y_c$ — multiply the ENTIRE trial by $x^s$ , where $s$ is the matching characteristic root's multiplicity ( $s=1$ for a simple root, $s=2$ for a repeated one), then continue.
Why: a trial that already satisfies the homogeneous equation substitutes to $0=g(x)$ , unsolvable for any coefficients, because it can't tell itself apart from the forcing; multiplying by $x^s$ breaks that tie while staying closed under differentiation.
3
Compute $y_p'$ and $y_p''$ for whichever trial Step 2 left you with, then substitute all three into $ay''+by'+cy=g(x)$ .

Example

For $y_p=Axe^x$ : $y_p'=Ae^x(1+x)$ and $y_p''=Ae^x(2+x)$ . Substituting into $y''-3y'+2y=e^x$ : $Ae^x(2+x)-3Ae^x(1+x)+2Axe^x=e^x$

Why: this is the step that turns an unknown function into a system of unknown numbers, exactly as it would for a trial with no resonance to worry about.
4
Group the substituted equation by term type and set each side's coefficients equal.

Example

Expanding the left side of Step 3's equation: $Ae^x[(2+x)-3(1+x)+2x]=Ae^x[-1+0x]=-Ae^x$ . So the equation reduces to $-Ae^x=e^x$ matching the (only) $e^x$ -coefficient on each side: $-A=1$ .

Why: two polynomials (or exponential/trig combinations) are equal only if they agree term by term.
5
Solve the resulting system, then write $y=y_c+y_p$ .

Example

From $-A=1$ : $A=-1$ , so $y_p=-xe^x$ . Adding to $y_c=C_1e^x+C_2e^{2x}$ from Step 2: $y=C_1e^x+C_2e^{2x}-xe^x$

Why: a nonhomogeneous equation's general solution is always the homogeneous solution plus any one particular solution, whether or not that particular solution needed the resonance fix.

Common mistakes

Writing the naive trial and substituting without ever checking it against $y_c$ first, which is exactly what produces the nonsense equation $0=g(x)$ .
Multiplying only PART of the trial by $x^s$ instead of the entire trial function.
Using $s=1$ when the matching root is actually repeated ( $s$ should be $2$ ).
Forgetting that a modified trial is still checked against $y_c$ the same way: for a doubly-repeated collision, $x^s$ may itself still overlap and need a second bump.