When must the modification rule be applied?

When the trial particular solution duplicates a term in the homogeneous solution

Method of Undetermined Coefficients

§ 01The Non-Homogeneous Problem

In the previous topic we solved ay'' + by' + cy = 0, the homogeneous equation. Now we add a forcing function g(x) on the right and ask: what particular function y satisfies the full equation?

Non-Homogeneous Second-Order Linear ODE

ay'' + by' + cy = g(x) a, b, c ∈ ℝ with a ≠ 0 | g(x) ≠ 0 is the forcing function (or non-homogeneous term)

The method of undetermined coefficients applies when g(x) is one of the following types, or a sum or product of them:

A polynomial: g(x) = xⁿ + … (any degree)
An exponential: g(x) = eᵅˣ
A sine or cosine: g(x) = sin(βx) or cos(βx)
Any product of the above: g(x) = x²e³ˣ, eˣ sin(2x), etc.

If g(x) is something else (like ln x, tan x, or 1/x) the method does not apply and you must use variation of parameters instead.

§ 02Superposition: y = y_h + y_p

The complete general solution to the non-homogeneous equation is the sum of two parts: the homogeneous solution and a particular solution. This is the superposition principle.

General Solution

y = y_h + y_p

Homogeneous Solution

y_h — solves ay''+by'+cy = 0
Contains C₁, C₂

Particular Solution

y_p — solves ay''+by'+cy = g(x)
No arbitrary constants

Why does this work? If y_h satisfies the homogeneous equation and y_p satisfies the full non-homogeneous equation, then substituting y = y_h + y_p into the left-hand side gives:

Proof of Superposition

a(y_h+y_p)'' + b(y_h+y_p)' + c(y_h+y_p) = [ay_h''+by_h'+cy_h] + [ay_p''+by_p'+cy_p] = 0 + g(x) = g(x) ✓

The arbitrary constants C₁ and C₂ live entirely in y_h. The particular solution y_p has no free constants, it is a single, specific function. Initial conditions are applied to the complete solution y = y_h + y_p at the end.

The Five-Step Method

Solve the homogeneous equation ay''+by'+cy = 0 using the characteristic equation. Write down y_h.
Identify the form of g(x) and choose the appropriate trial particular solution y_p from the table in § 03.
Check for overlap between y_p and y_h. If any term of the trial y_p appears in y_h, apply the modification rule (multiply by x, or x² if needed).
Substitute the trial y_p (and its derivatives) into the full ODE. Collect like terms and solve for the undetermined coefficients.
Write the complete general solution y = y_h + y_p, then apply initial conditions if given.

§ 03The Trial Solution Table

The trial particular solution y_p must be chosen to match the family of functions that g(x) belongs to. This means including all members of that family, not just the exact terms that appear.

Forcing Function g(x)	Trial Particular Solution y_p	Notes
c (constant)	A	Single undetermined constant
bx + c (degree 1)	Ax + B	Include all terms down to degree 0
ax² + bx + c (degree 2)	Ax² + Bx + C	Full quadratic — all three terms required
Pₙ(x) (degree n poly)	Aₙxⁿ + … + A₁x + A₀	n+1 undetermined coefficients
e^αx	Ae^αx	One constant; α must not be a root of char. eq.
sin(βx) or cos(βx)	A cos(βx) + B sin(βx)	Always include BOTH sin and cos, even if g has only one
e^αx sin(βx) or e^αx cos(βx)	e^αx(A cos βx + B sin βx)	Exponential envelope × both trig terms
Pₙ(x)·e^αx	(Aₙxⁿ+…+A₀)e^αx	Polynomial of same degree × exponential
Pₙ(x)·sin(βx) or Pₙ(x)·cos(βx)	(Aₙxⁿ+…+A₀)cos βx + (Bₙxⁿ+…+B₀)sin βx	Two full polynomials — one for cos, one for sin

The "include the whole family" rule If g(x) = 3x², you might be tempted to try y_p = Ax². But differentiating Ax² twice produces a constant, which means the y and y' terms need x² and x as well. Always use the full polynomial Ax² + Bx + C for any degree-2 forcing. Similarly, if g(x) = 5 sin(3x), always use A cos(3x) + B sin(3x) — differentiation mixes sin and cos, so you need both.

§ 04Polynomial Forcing

When g(x) is a polynomial, the trial solution is a polynomial of the same degree. Substitute and match coefficients of equal powers of x.

Example 1 — Constant Forcing

Solve: y'' + 3y' + 2y = 10.

Homogeneous solution. Characteristic equation: r² + 3r + 2 = (r+1)(r+2) = 0. Roots r = −1, −2.y_h = C₁e^−x + C₂e^−2x

Trial solution. g(x) = 10 (constant), so try y_p = A. Then y_p' = 0, y_p'' = 0.

Substitute. 0 + 3(0) + 2A = 10 → 2A = 10 → A = 5.

y = C₁e^−x + C₂e^−2x + 5

Example 2 — Quadratic Forcing

Find a particular solution of y'' − y = x² − 1.

Trial solution. g(x) = x² − 1 has degree 2, so try y_p = Ax² + Bx + C.

Compute derivatives. y_p' = 2Ax + B, y_p'' = 2A.

Substitute into y'' − y = x² − 1:2A − (Ax² + Bx + C) = x² − 1−Ax² − Bx + (2A − C) = x² − 1

Match coefficients by power of x:
x²: −A = 1 → A = −1
x¹: −B = 0 → B = 0
x⁰: 2A − C = −1 → −2 − C = −1 → C = −1

y_p = −x² − 1

Example 3 — Linear Forcing with Full IVP

Solve: y'' + y' = 2x + 3, y(0) = 1, y'(0) = 0.

Homogeneous solution. r² + r = r(r+1) = 0. Roots r = 0, r = −1.y_h = C₁ + C₂e^−x

Trial solution. g(x) = 2x + 3 suggests y_p = Ax + B. But C₁ (a constant) is already in y_h — does this overlap? The constant term B in y_p overlaps with C₁·1. Apply the modification rule: multiply by x.y_p = x(Ax + B) = Ax² + Bx

Derivatives. y_p' = 2Ax + B, y_p'' = 2A.

Substitute: 2A + (2Ax + B) = 2x + 3.
x¹: 2A = 2 → A = 1
x⁰: 2A + B = 3 → 2 + B = 3 → B = 1

Complete solution: y = C₁ + C₂e^−x + x² + x.
Apply y(0)=1: C₁ + C₂ = 1.
y' = −C₂e^−x + 2x + 1. Apply y'(0)=0: −C₂ + 1 = 0 → C₂ = 1 → C₁ = 0.

y = e^−x + x² + x

§ 05Exponential Forcing

When g(x) = Ke^αx, try y_p = Ae^αx, unless α is a root of the characteristic equation, in which case the modification rule applies.

Example 4 — Standard Exponential

Solve: y'' − 3y' + 2y = 4e^3x.

Homogeneous solution. r² − 3r + 2 = (r−1)(r−2) = 0. Roots r = 1, 2.y_h = C₁e^x + C₂e^2x

Trial solution. g(x) = 4e^3x. Check: is α = 3 a root? No (roots are 1 and 2). So try y_p = Ae^3x.

Compute derivatives. y_p' = 3Ae^3x, y_p'' = 9Ae^3x.

Substitute: 9Ae^3x − 3(3Ae^3x) + 2(Ae^3x) = 4e^3x.Ae^3x(9 − 9 + 2) = 4e^3x → 2A = 4 → A = 2

y = C₁e^x + C₂e^2x + 2e^3x

Example 5 — Modification Rule (α is a simple root) Modification Rule

Solve: y'' − 3y' + 2y = 5e^x.

Homogeneous solution. From Example 4: y_h = C₁e^x + C₂e^2x.

Check for overlap. g(x) = 5eˣ suggests trial Aeˣ. But eˣ is already in y_h (the C₁eˣ term). The trial would satisfy the homogeneous equation and give 0 = 5eˣ — impossible. Apply the modification rule and multiply by x:y_p = Axe^x

Compute derivatives (product rule).
y_p' = Ae^x + Axe^x = A(1+x)e^x
y_p'' = Ae^x + A(1+x)e^x = A(2+x)e^x

Substitute: A(2+x)eˣ − 3A(1+x)eˣ + 2Axeˣ = 5eˣ.eˣ · A[(2+x) − 3(1+x) + 2x] = 5eˣeˣ · A[2+x−3−3x+2x] = eˣ · A(−1) = 5eˣ−A = 5 → A = −5

y = C₁e^x + C₂e^2x − 5xe^x

Example 6 — Modification Rule (α is a double root) Modification Rule ×2

Find a particular solution of y'' − 2y' + y = e^x.

Homogeneous solution. r² − 2r + 1 = (r−1)² = 0. Double root r = 1.y_h = (C₁ + C₂x)e^x

Check overlap. Trial Aeˣ overlaps with C₁eˣ. Multiply by x: Axeˣ overlaps with C₂xeˣ. Must multiply by x again:y_p = Ax²e^x

Derivatives.
y_p' = (2Ax + Ax²)e^x = Ax(2+x)e^x
y_p'' = A(2+x)e^x + Ax(2+x)e^x + Axe^x = A(2 + 4x + x²)e^x

Substitute: A(2+4x+x²)eˣ − 2Ax(2+x)eˣ + Ax²eˣ = eˣ.A[(2+4x+x²) − (4x+2x²) + x²]eˣ = eˣ → 2Aeˣ = eˣ → A = ½

y_p = ½x²e^x

§ 06Trigonometric Forcing

When g(x) involves sin(βx) or cos(βx), the trial solution must include both A cos(βx) and B sin(βx), because differentiation converts one into the other. Never use only one of the pair.

Example 7 — Sinusoidal Forcing

Find a particular solution of y'' + 4y = 3 sin(x).

Homogeneous solution. r² + 4 = 0 → r = ±2i. Roots are ±2i (not ±i), so β = 1 does not overlap.y_h = C₁cos(2x) + C₂sin(2x)

Trial solution. y_p = A cos(x) + B sin(x).

Derivatives. y_p'' = −A cos(x) − B sin(x).

Substitute into y'' + 4y:(−Acos x − Bsin x) + 4(Acos x + Bsin x) = 3sin x3Acos x + 3Bsin x = 3sin x

Match coefficients: cos x: 3A = 0 → A = 0. sin x: 3B = 3 → B = 1.

y_p = sin(x) Full solution: y = C₁cos(2x) + C₂sin(2x) + sin(x)

Example 8 — Resonance Modification Rule

Find a particular solution of y'' + 4y = 3 cos(2x).

Homogeneous solution. y_h = C₁cos(2x) + C₂sin(2x) (same as Example 7).

Check overlap. Trial A cos(2x) + B sin(2x) exactly duplicates y_h. This is resonance — the forcing frequency matches the natural frequency. Apply the modification rule: multiply by x.y_p = x(A cos(2x) + B sin(2x))

Derivatives (product rule).
y_p' = (A cos 2x + B sin 2x) + x(−2A sin 2x + 2B cos 2x)
y_p'' = (−2A sin 2x + 2B cos 2x) + (−2A sin 2x + 2B cos 2x) + x(−4A cos 2x − 4B sin 2x)
y_p'' = (−4A sin 2x + 4B cos 2x) − 4x(A cos 2x + B sin 2x)

Substitute into y'' + 4y:(−4A sin 2x + 4B cos 2x) − 4x(A cos 2x + B sin 2x) + 4x(A cos 2x + B sin 2x) = 3cos 2x−4A sin 2x + 4B cos 2x = 3 cos 2x

Match: sin 2x: −4A = 0 → A = 0. cos 2x: 4B = 3 → B = ¾.

y_p = ¾ x sin(2x) Full solution: y = C₁cos(2x) + C₂sin(2x) + ¾x sin(2x)

§ 07Combined and Product Forcing

When g(x) is a sum of terms, the particular solution is the sum of the particular solutions for each term separately. When g(x) is a product (like e^αx sin βx), treat the whole product as one unit.

Example 9 — Sum of Forcing Terms

Find a particular solution of y'' + y' − 2y = 2e^x + 3.

Homogeneous solution. r² + r − 2 = (r+2)(r−1) = 0. Roots r = 1, −2.y_h = C₁e^x + C₂e^−2x

Split: find y_p1 for 2eˣ. Trial Aeˣ overlaps with C₁eˣ in y_h. Apply modification rule: y_p1 = Axeˣ.
Derivatives: y_p1' = A(1+x)eˣ, y_p1'' = A(2+x)eˣ.
Substitute: A(2+x)eˣ + A(1+x)eˣ − 2Axeˣ = 2eˣ → A(3)eˣ = 2eˣ → A = 2/3.

Find y_p2 for 3 (constant). Trial y_p2 = B. Substitute: 0 + 0 − 2B = 3 → B = −3/2.

y_p = ⅔xe^x − 3/2 y = C₁e^x + C₂e^−2x + ⅔xe^x − 3/2

Example 10 — Exponential × Trig Forcing

Find a particular solution of y'' + 2y' + 2y = e^−x cos(x).

Homogeneous solution. r² + 2r + 2 = 0. r = [−2 ± √(4−8)]/2 = −1 ± i.y_h = e^−x(C₁cos x + C₂sin x)

Check overlap. Trial e^−x(Acos x + Bsin x) is exactly y_h. Apply modification rule:y_p = xe^−x(Acos x + Bsin x)

Compute y_p' and y_p'' (product rule, collecting terms):
y_p' = e^−x[(A+B)cos x + (B−A)sin x] + xe^−x[(B−A)cos x − (A+B)sin x]

After computing y_p'' and substituting into y''+2y'+2y, the x-terms cancel (as expected from the modification rule) and the remaining terms give:2B cos x − 2A sin x = cos x

Match: cos x: 2B = 1 → B = ½. sin x: −2A = 0 → A = 0.

y_p = ½ xe^−x sin(x)

Example 11 — Polynomial × Exponential Forcing

Find a particular solution of y'' − y' = 2xe^x.

Homogeneous solution. r² − r = r(r−1) = 0. Roots r = 0, 1.y_h = C₁ + C₂e^x

Trial solution. g(x) = 2xeˣ — a degree-1 polynomial times eˣ. Standard trial: (Ax + B)eˣ. Check overlap: Beˣ overlaps with C₂eˣ. Multiply the whole trial by x:y_p = x(Ax + B)e^x = (Ax² + Bx)e^x

Derivatives.
y_p' = (2Ax + B)eˣ + (Ax²+Bx)eˣ = (Ax² + (2A+B)x + B)eˣ
y_p'' = (Ax² + (4A+B)x + (2A+2B))eˣ

Substitute into y'' − y':[Ax²+(4A+B)x+(2A+2B)]eˣ − [Ax²+(2A+B)x+B]eˣ = 2xeˣ[2Ax + (2A+B)]eˣ = 2xeˣ

Match: x¹: 2A = 2 → A = 1. x⁰: 2A + B = 0 → B = −2.

y_p = (x² − 2x)e^x

Example 12 — Complete IVP

Solve: y'' − 2y' + y = eˣ + 4, y(0) = 2, y'(0) = 3.

Homogeneous. (r−1)² = 0. Double root r = 1.y_h = (C₁ + C₂x)e^x

Particular for eˣ. From Example 6 (same equation): y_p1 = ½x²eˣ.

Particular for 4 (constant). Trial y_p2 = A. Substitute: 0 − 0 + A = 4 → A = 4.

Complete general solution:y = (C₁ + C₂x)eˣ + ½x²eˣ + 4

Apply y(0) = 2: C₁ + 0 + 0 + 4 = 2 → C₁ = −2.

Differentiate: y' = C₂eˣ + (C₁+C₂x)eˣ + xeˣ + ½x²eˣ.
Apply y'(0) = 3: C₂ + C₁ + 0 + 0 = 3 → C₂ + (−2) = 3 → C₂ = 5.

y = (−2 + 5x)eˣ + ½x²eˣ + 4 = (−2 + 5x + ½x²)eˣ + 4

§ 08The Modification Rule — Full Summary

The modification rule is the single most important subtlety in the method of undetermined coefficients. Master it and you can handle every case.

When to apply the modification rule If any term in the initial trial y_p is a solution of the homogeneous equation (i.e., it appears in y_h), then that trial will produce 0 = g(x) upon substitution — useless. The fix: multiply the entire trial by xˢ, where s is the smallest positive integer that eliminates all overlaps. Typically s = 1 (simple root) or s = 2 (double root).

Situation	Action	Example
No overlap between trial y_p and y_h	Use trial as-is	g=e³ˣ, roots 1,2 → y_p=Ae³ˣ
Trial overlaps y_h (simple root)	Multiply trial by x	g=eˣ, root 1 → y_p=Axeˣ
Trial overlaps y_h after ×x (double root)	Multiply original trial by x²	g=eˣ, double root 1 → y_p=Ax²eˣ
Trig trial overlaps (resonance)	Multiply trig trial by x	g=cos 2x, y_h has cos 2x → y_p=x(Acos 2x+Bsin 2x)
Polynomial trial overlaps (root r=0)	Multiply polynomial trial by x	g=3x, r=0 is root → y_p=x(Ax+B)=Ax²+Bx

What "overlap" means precisely A term in the trial y_p overlaps with y_h if substituting that term alone into the left-hand side ay''+by'+cy would give zero. Concretely: eᵅˣ overlaps if α is a root; cos(βx) or sin(βx) overlaps if ±βi are roots; a constant overlaps if 0 is a root (i.e., c = 0 in the ODE).

§ 09Common Mistakes

Mistake 1: Only including the terms that actually appear in g(x) If g(x) = 5x, using trial y_p = Ax misses the constant term. Because y_p'' and y_p' generate constants, you must use y_p = Ax + B. Similarly for trig: g(x) = cos x requires y_p = A cos x + B sin x, not just A cos x.

Mistake 2: Forgetting to check for overlap before writing the trial Always write out y_h first, then check every term of your proposed trial against it. Skipping this step and discovering halfway through that your trial gives 0 = g(x) wastes enormous effort.

Mistake 3: Applying initial conditions before adding y_p to y_h Initial conditions must be applied to the complete solution y = y_h + y_p. Applying them to y_h alone gives wrong values of C₁ and C₂.

Mistake 4: Using the method when g(x) is not of the required type The method only works for polynomials, exponentials, sines, cosines, and their products. It does not work for g(x) = ln x, g(x) = tan x, g(x) = sec x, or g(x) = 1/x. Use variation of parameters for those.

Mistake 5: Incorrect differentiation in the substitution step Expressions like xe^αx or xe^−xcos(βx) require the product rule (sometimes applied twice). Differentiating them as if the x is just a constant is the most common algebraic error in this method.

Continue in Differential Equations

Undetermined coefficients handles a wide range of forcing functions, but for cases it cannot handle, and for a deeper theoretical understanding, explore these topics next:

§ 01The Non-Homogeneous Problem

§ 02Superposition: y = yh + yp

§ 03The Trial Solution Table

§ 04Polynomial Forcing

§ 05Exponential Forcing

§ 06Trigonometric Forcing

§ 07Combined and Product Forcing

§ 08The Modification Rule — Full Summary

§ 09Common Mistakes

§ 1010-Question Quiz

Continue in Differential Equations

§ 02Superposition: y = y_h + y_p