Nonhomogeneous equations and particular integrals

A homogeneous linear ODE only ever settles to zero on its right-hand side. The moment a real driving term $f(x)$ — a polynomial force, an exponential input, a sinusoidal forcing — sits on the right, the complementary function $y_c$ alone can no longer satisfy the equation, and a second ingredient, the particular integral (PI) $y_p$ , must be added on top of it. This article picks up exactly where second-order constant-coefficient ODEs leave off. You already know how to find $y_c$ from the auxiliary equation. Here the tool is the method of undetermined coefficients: guess a trial form matching $f(x)$ , substitute it into the ODE, and match coefficients — with one crucial twist, the resonance modification, whenever the trial form collides with $y_c$ .

The core insight

For $a_ny^{(n)}+\cdots+a_0y=f(x)$ , guess a trial $y_p$ matching the type of $f(x)$ , multiply it by $x$ (or $x^2$ ) if it duplicates a term of $y_c$ , then substitute and match coefficients to pin down $y_p$ , so that $y=y_c+y_p$ solves the full equation.

The core idea

For a linear constant-coefficient ODE

a_n y^{(n)} + \cdots + a_1 y' + a_0 y = f(x),

the general solution splits into two pieces added together,

y = y_c + y_p,

where: $y_c$ is the complementary function, the general solution of the associated homogeneous equation $L[y]=0$ , and $y_p$ is any single particular integral, one solution of the full nonhomogeneous equation $L[y]=f(x)$ . This split works because the operator $L[y]=a_ny^{(n)}+\cdots+a_0y$ is linear, so $L[y_c+y_p]=L[y_c]+L[y_p]=0+f(x)=f(x)$ .

Choosing the trial form. For $f(x)$ built from polynomials, exponentials $e^{kx}$ , or $\sin kx$ / $\cos kx$ (and products of these), the trial $y_p$ has the same "shape" as $f(x)$ , with unknown coefficients to be found:

polynomial of degree $k$ $\to$ trial polynomial of degree $k$ (every lower-degree term included, even if $f(x)$ is a pure power)
$e^{kx}$ $\to$ trial $Ae^{kx}$
$\sin kx$ or $\cos kx$ $\to$ trial $A\cos kx + B\sin kx$ (both terms, even if only one appears in $f(x)$ )
$e^{px}\sin qx$ or $e^{px}\cos qx$ $\to$ trial $e^{px}(A\cos qx + B\sin qx)$

The resonance modification. If any term of the naive trial already appears in $y_c$ — i.e. it duplicates a root of the auxiliary equation — substituting the naive trial gives an impossible equation ( $0$ equals a nonzero constant). The fix is to multiply the entire trial by $x$ (or by $x^2$ if a single multiplication still duplicates $y_c$ ) before solving for coefficients. This is the one place students lose the most marks, and it is the subject of the derivation below.

Figure

Flowchart from a driving term f of x through trial selection resonance check and coefficient matching to the particular integral

Derivation AWhy a duplicate trial term forces multiplication by x

The goal is to show, with nothing beyond substitution and algebra, why a trial $y_p=Ae^{kx}$ fails whenever $e^{kx}$ already solves the homogeneous equation, and why multiplying the trial by $x$ repairs it — this is the resonance modification used every time the driving term $f(x)$ shares a root with the auxiliary equation, and it is the step students forget most often.

Step 1. Set up the resonance case concretely. Take $L[y]=y''-4y=e^{2x}$ . The auxiliary equation is $m^2-4=0$ , giving roots $m=2,-2$ , so $y_c=C_1e^{2x}+C_2e^{-2x}$ . The driving term $e^{2x}$ matches the root $m=2$ exactly — this is what "duplicates $y_c$ " means.

Step 2. Try the naive trial and watch it collapse. If $y_p=Ae^{2x}$ , then $y_p''=4Ae^{2x}$ , so $L[y_p]=y_p''-4y_p=4Ae^{2x}-4Ae^{2x}=0$ for every value of $A$ . Substituting into $L[y_p]=e^{2x}$ demands $0=e^{2x}$ , an equation with no solution — the naive trial can never match the right-hand side, no matter how $A$ is chosen, because $Ae^{2x}$ already solves the homogeneous equation and $L$ annihilates it.

Step 3. Multiply the trial by $x$ and retry. Let $y_p=Axe^{2x}$ . Differentiating, $y_p'=Ae^{2x}+2Axe^{2x}$ and $y_p''=4Ae^{2x}+4Axe^{2x}$ . Then $L[y_p]=y_p''-4y_p=(4Ae^{2x}+4Axe^{2x})-4(Axe^{2x})=4Ae^{2x}$ . The $xe^{2x}$ terms have cancelled, leaving a clean nonzero multiple of $e^{2x}$ — exactly the shape needed to match the right-hand side.

Step 4. Match coefficients and read off the general pattern. Setting $4Ae^{2x}=e^{2x}$ gives $A=\tfrac14$ , so $y_p=\tfrac14 xe^{2x}$ solves the equation exactly. The mechanism is general: multiplying by $x$ shifts the trial out of the null space of $L$ (the space of solutions of $L[y]=0$ ), so $L$ no longer annihilates it, and a genuine coefficient equation appears. If a single multiplication by $x$ still lands back in that null space (a repeated root duplicated twice), multiply by $x^2$ instead, for the same reason.

Worked example

Find the general solution of $y'' - y' - 2y = 4x^2$ , then use initial conditions $y(0)=0$ , $y'(0)=1$ to pin down a unique solution.

Step 1. Find the complementary function. The auxiliary equation is $m^2-m-2=0$ , which factors as $(m-2)(m+1)=0$ , giving $m=2,-1$ . So $y_c=C_1e^{2x}+C_2e^{-x}$ .

Step 2. Choose the trial PI. Since $f(x)=4x^2$ is a degree-2 polynomial, and no term of a polynomial trial duplicates $y_c$ (which is built from exponentials), try $y_p=Ax^2+Bx+C$ — all three terms, not just the leading one.

Step 3. Differentiate and substitute. With $y_p'=2Ax+B$ and $y_p''=2A$ ,

y_p''-y_p'-2y_p = 2A-(2Ax+B)-2(Ax^2+Bx+C) = -2Ax^2+(-2A-2B)x+(2A-B-2C).

Step 4. Match coefficients against $4x^2+0\cdot x+0$ . From $x^2$ : $-2A=4\Rightarrow A=-2$ . From $x^1$ : $-2A-2B=0\Rightarrow 4-2B=0\Rightarrow B=2$ . From $x^0$ : $2A-B-2C=0\Rightarrow -4-2-2C=0\Rightarrow C=-3$ . So $y_p=-2x^2+2x-3$ , and the general solution is

y = C_1e^{2x}+C_2e^{-x}-2x^2+2x-3.

Step 5. Apply the initial conditions to find $C_1,C_2$ . Differentiating, $y'=2C_1e^{2x}-C_2e^{-x}-4x+2$ . At $x=0$ : $y(0)=C_1+C_2-3=0\Rightarrow C_1+C_2=3$ , and $y'(0)=2C_1-C_2+2=1\Rightarrow 2C_1-C_2=-1$ . Adding the two equations eliminates $C_2$ : $3C_1=2\Rightarrow C_1=\tfrac23$ , so $C_2=3-\tfrac23=\tfrac73$ .

The particular solution is $y=\dfrac{2}{3}e^{2x}+\dfrac{7}{3}e^{-x}-2x^2+2x-3$ .

Competitive stretch

Check the answer quickly at $x=0$ without re-expanding: $y(0)=\tfrac23+\tfrac73-3=\tfrac93-3=0$ matches the given condition, confirming the algebra before trusting the full expression.

Common traps

Forgetting the resonance modification. If a trial term duplicates $y_c$ , undetermined coefficients gives an impossible equation ( $0=$ nonzero) unless the trial is multiplied by $x$ (or $x^2$ for a doubly repeated root).
Using an incomplete trial polynomial. For $f(x)$ a degree- $k$ polynomial, the trial must include every lower-degree term, $Ax^k+\cdots+C$ , not just the leading power.
Dropping sine or cosine. When $f(x)$ contains only $\sin kx$ (or only $\cos kx$ ), the trial must still carry both $A\cos kx$ and $B\sin kx$ , since differentiating one produces the other.
Losing $y_c$ at the end. The final general solution is always $y=y_c+y_p$ ; adding $y_c$ into the trial itself, or forgetting to re-attach it, throws away the arbitrary constants the problem needs.

Ask the tutor: Not sure which trial form fits a mixed right-hand side like $xe^{3x}\cos 2x$ , or want your resonance check verified before you substitute? Ask the AI tutor to walk through the trial selection with you.

Why does multiplying the trial by x fix resonance instead of, say, adding a constant?

Multiplying by $x$ works because the trouble is that the original trial already satisfies the homogeneous equation, so the differential operator sends it straight to zero and it can never match a nonzero right-hand side. Multiplying by $x$ moves the trial outside that set of homogeneous solutions while keeping the same exponential or trigonometric "flavour," so the operator applied to it no longer vanishes identically — it leaves behind exactly the kind of term needed to match $f(x)$ . Adding a constant would not fix this, since constants (or other unrelated terms) do not track the structure of $f(x)$ and would not survive coefficient matching correctly.

What if f(x) is a sum of two different types, like 4x squared plus e to the 2x?

Because the equation is linear, particular integrals add: find a particular integral $y_{p1}$ for the polynomial part and, separately, a particular integral $y_{p2}$ for the exponential part (checking each piece for resonance on its own), then take $y_p=y_{p1}+y_{p2}$ . This is the same superposition idea that lets $y=y_c+y_p$ work in the first place, just applied one more time, splitting $f(x)$ into pieces that are each easy to match with a trial.

Does it matter which particular solution I pick for y sub p, since the trial method could give different constants?

No — any correct $y_p$ that satisfies the full nonhomogeneous equation works, because two different particular integrals for the same $f(x)$ can only differ by a solution of the homogeneous equation, which is already absorbed into $y_c$ 's arbitrary constants. In practice the method of undetermined coefficients with a given trial form produces a unique $y_p$ once the coefficients are matched, so this ambiguity rarely shows up — but it is why you never need to worry about "the wrong" particular integral, only whether your trial form was complete and resonance-checked.

Nonhomogeneous equations and particular integrals

The core idea

Worked example

Competitive stretch

Common traps

Before this

Where this leads