Recipe

Find a particular solution with undetermined coefficients

Use when: you have a nonhomogeneous linear ODE $ay''+by'+cy=g(x)$ where $g(x)$ is a polynomial, exponential, sine/cosine, or a sum/product of these

Goal: a particular solution $y_p$ to add to the homogeneous solution $y_h$

Taught in The method of undetermined coefficients · a later section extends this recipe with more cases

Quick recipe

Guess a trial form matched to $g(x)$ 's shape
Substitute the guess into the equation
Match coefficients term by term
Solve for the unknown coefficients
Add $y_p$ to the homogeneous solution

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 $y''-y=2x^2+3$ , the forcing $g(x)=2x^2+3$ is a degree-2 polynomial, so guess a full degree-2 polynomial back: every degree up through 2, plus the constant term: $y_p=Ax^2+Bx+C$
- Not yet: the natural guess duplicates a term already in $y_h$ — covered in The resonance modification rule
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, rather than needing a whole new method.
2
Compute $y_p'$ and $y_p''$ , then substitute all three into $ay''+by'+cy=g(x)$ .

Example

Continuing with $y_p=Ax^2+Bx+C$ : $y_p'=2Ax+B$ and $y_p''=2A$ . Substituting into $y''-y=2x^2+3$ : $2A-(Ax^2+Bx+C)=2x^2+3$ which rearranges to $-Ax^2-Bx+(2A-C)=2x^2+0x+3$ .

Why: this is the step that turns an unknown function into a system of unknown numbers.
3
Group the substituted equation by term type (the $x^2$ terms, the $x$ terms, the constant terms, and so on) and set each side's coefficients equal.

Example

From $-Ax^2-Bx+(2A-C)=2x^2+0x+3$ , matching term by term: $x^2:\ -A=2 \qquad x:\ -B=0 \qquad \text{constant:}\ 2A-C=3$

Why: two polynomials (or two exponential/trig combinations) are equal only if they agree term by term, which is what turns the guess into a solvable linear system for the unknown coefficients.
4
Solve the resulting system for $A$ , $B$ , and any other unknowns in the guess.

Example

From $-A=2$ : $A=-2$ . From $-B=0$ : $B=0$ . Substituting $A=-2$ into $2A-C=3$ : $-4-C=3$ , so $C=-7$ . That gives $y_p=-2x^2-7$

Why: this pins down the specific numbers that make the guessed form actually satisfy the equation.
5
Write the full general solution as $y=y_h+y_p$ .

Example

The homogeneous equation $y''-y=0$ has characteristic roots $r=\pm1$ (Chapter 7's recipe), so $y_h=C_1e^x+C_2e^{-x}$ . Adding the particular solution from Step 4: $y=C_1e^x+C_2e^{-x}-2x^2-7$

Why: a nonhomogeneous equation's general solution is always the homogeneous solution plus any one particular solution, a consequence of superposition, not a coincidence specific to this method.

Common mistakes

Guessing a trial form one degree too low, such as $y_p=A$ for a linear forcing $g(x)=3x$ , instead of matching every degree present.
Substituting $y_p$ into the equation but forgetting to also substitute $y_p'$ and $y_p''$ everywhere they appear.
Matching coefficients by eye instead of grouping systematically by term type, and missing one.
Reporting $y_p$ alone as "the solution" instead of adding it to $y_h$ .