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Lesson 3: Undetermined Coefficients

Estimated time: 40-50 minutes

Learning Objectives

By the end of this lesson, you will be able to:

State the structure: general solution = y_h + y_p
Choose the correct guess form for polynomial, exponential, and trigonometric forcing
Apply the modification rule when the guess overlaps with y_h
Solve complete nonhomogeneous IVPs
Handle combinations and sums of forcing terms

The Big Picture

For ay'' + by' + cy = g(t), the general solution is:

y = y_h + y_p

where y_h is the general solution to the homogeneous equation (from Lessons 1-2) and y_p is any particular solution to the full nonhomogeneous equation.

Method of Undetermined Coefficients: For certain types of g(t) (polynomials, exponentials, sines/cosines, and their products), we guess the form of y_p with unknown coefficients, substitute into the ODE, and solve for those coefficients.

The Guessing Rules

g(t)	Guess for y_p
a_ntⁿ + ... + a₀	A_ntⁿ + ... + A₁t + A₀
e^αt	Ae^αt
cos βt or sin βt	A cos βt + B sin βt (always include BOTH)
e^αt cos βt	e^αt(A cos βt + B sin βt)
tⁿe^αt	(A_ntⁿ + ... + A₀)e^αt

Basic Examples

Example 1: Polynomial Forcing

Solve y'' - 3y' + 2y = 4t.

Step 1 (y_h): r² - 3r + 2 = (r-1)(r-2) = 0. y_h = c₁e^t + c₂e^2t.

Step 2 (Guess): g(t) = 4t, so guess y_p = At + B.

Step 3 (Substitute): y_p' = A, y_p'' = 0. Plug in: 0 - 3A + 2(At + B) = 4t.

2At + (-3A + 2B) = 4t + 0.

Match coefficients: 2A = 4 ⇒ A = 2. -3(2) + 2B = 0 ⇒ B = 3.

Answer: y = c₁e^t + c₂e^2t + 2t + 3.

Example 2: Exponential Forcing

Solve y'' + y' - 2y = 3e^4t.

y_h: r² + r - 2 = (r+2)(r-1) = 0. y_h = c₁e^-2t + c₂e^t.

Guess: y_p = Ae^4t. (4 is not a root, so no modification needed.)

Substitute: 16Ae^4t + 4Ae^4t - 2Ae^4t = 3e^4t ⇒ 18A = 3 ⇒ A = 1/6.

Answer: y = c₁e^-2t + c₂e^t + (1/6)e^4t.

Example 3: Trigonometric Forcing

Solve y'' + 4y = sin(3t).

y_h: r² + 4 = 0 ⇒ r = ±2i. y_h = c₁cos(2t) + c₂sin(2t).

Guess: y_p = A cos(3t) + B sin(3t).

Substitute: y_p'' = -9A cos(3t) - 9B sin(3t).

(-9A + 4A) cos(3t) + (-9B + 4B) sin(3t) = sin(3t).

-5A cos(3t) - 5B sin(3t) = sin(3t).

A = 0, B = -1/5.

Answer: y = c₁cos(2t) + c₂sin(2t) - (1/5)sin(3t).

The Modification Rule

Modification Rule: If any term in your initial guess is already a solution to the homogeneous equation, multiply the entire guess by t. If it still overlaps, multiply by t².

Example 4: Modification Required

Solve y'' - 2y' + y = e^t.

y_h: (r-1)² = 0 ⇒ r = 1 (repeated). y_h = (c₁ + c₂t)e^t.

Initial guess: Ae^t -- but e^t is in y_h! Try Ate^t -- also in y_h!

Modified guess: y_p = At²e^t.

Substitute: y_p' = A(2t + t²)e^t, y_p'' = A(2 + 4t + t²)e^t.

A(2 + 4t + t²)e^t - 2A(2t + t²)e^t + At²e^t = e^t.

A(2)e^t = e^t ⇒ A = 1/2.

Answer: y = (c₁ + c₂t)e^t + (1/2)t²e^t.

Check Your Understanding

1. What form would you guess for y_p if g(t) = 5e^-3t and the characteristic roots are r = 1, 2?

Answer: Guess y_p = Ae^-3t. No modification needed since -3 is not a characteristic root.

2. What form would you guess if g(t) = 4 cos(2t) and the characteristic roots are r = ±2i?

Answer: Initial guess A cos(2t) + B sin(2t) overlaps with y_h. Modify: y_p = t[A cos(2t) + B sin(2t)].

3. Solve y'' + y = 2t + 1.

Answer: y_h = c₁cos t + c₂sin t. Guess y_p = At + B. Substitute: 0 + At + B = 2t + 1. So A = 2, B = 1. y = c₁cos t + c₂sin t + 2t + 1.

Key Takeaways

General solution = y_h + y_p (homogeneous + particular).
Guess the form based on g(t): polynomials, exponentials, trig (always include both sin and cos).
The modification rule: if the guess overlaps y_h, multiply by t (or t² if needed).
Works only for specific forms of g(t). For anything else, use variation of parameters (Lesson 4).
Apply IVP conditions to the complete general solution y = y_h + y_p, never to y_h alone.

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In Lesson 4, learn variation of parameters -- a method that works for any forcing function.

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