Method of Undetermined Coefficients
Differential Equations · Section 3.5 · generated practice set
Pick the right trial form, repair it when it collides with the homogeneous solution, then solve for the constants.
Practice this set → Fresh numbers on every attempt. No account needed.
Method
For a non-homogeneous equation $ay'' + by' + cy = g(t)$ the general solution is always $$y = y_h + Y_p,$$ the homogeneous solution plus any one particular solution. Undetermined coefficients finds $Y_p$ by guessing its shape and solving for the constants.
Always solve the homogeneous problem first
Not because you need it for the final answer — you do — but because the correct guess depends on it. A guess that already solves the homogeneous equation contributes nothing and produces $0 = g(t)$.
The guess table
- $g = Ae^{kt}$ → $Y_p = Ce^{kt}$
- $g$ = polynomial of degree $n$ → $Y_p$ = full polynomial of degree $n$ (every lower term included)
- $g = Acosomega t$ or $Asinomega t$ → $Y_p = C_1cosomega t + C_2sinomega t$ (always both)
- products multiply the guesses: $g = Ate^{kt}$ → $Y_p = (C_1t + C_0)e^{kt}$
The repair rule
If any term of your guess solves the homogeneous equation, multiply the whole guess by $t$. Repeat if it still does. A double root can need $t^2$.
Superposition
If $g = g_1 + g_2$, find a particular solution for each piece separately and add them. Do not try to guess one shape for the sum.
Definitions and theorems
Worked example
Solve $y'' - 3y' - 4y = 2e^{-t}$.
- Homogeneous: $r^2 - 3r - 4 = (r-4)(r+1) = 0$, so $y_h = c_1e^{4t} + c_2e^{-t}$.
- The forcing exponent is $-1$, which is a root. The plain guess $Ce^{-t}$ would give $0 = 2e^{-t}$.
- Multiply by $t$: $Y_p = Cte^{-t}$. Then $Y_p' = C(1-t)e^{-t}$ and $Y_p'' = C(t-2)e^{-t}$.
- Substitute: $Cleft[(t-2) - 3(1-t) - 4t ight]e^{-t} = C(-5)e^{-t} = 2e^{-t}$, so $C = - frac25$.
- $y = c_1e^{4t} + c_2e^{-t} - frac25 te^{-t}$.
Common mistakes
- Guessing before solving the homogeneous equation. You cannot know whether you need the factor of $t$ until you have the roots.
- Dropping lower-order terms in a polynomial guess. For $g = 3t$ the guess is $pt + q$, not $pt$ — the constant is almost always non-zero.
- Guessing only cosine for cosine forcing. Differentiating cosine produces sine, so both are needed unless there is no $y'$ term.
- Multiplying only part of the guess by $t$. The repair rule multiplies the entire trial form.
- Forgetting $y_h$ in the final answer. $Y_p$ alone cannot satisfy initial conditions. Apply initial conditions to $y_h + Y_p$, never to $y_h$ first.
Practice it
Reading the method is not the same as being able to run it under time pressure. This set generates a new problem with new coefficients every attempt, marks each part separately, and shows the full worked solution.