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Section 10.3 Undetermined Coefficients Method

In the last sections, we learned how to choose a form for the particular solution \(y_p\) by matching it to the forcing function \(f(x)\text{,}\) and how to modify \(y_p\) when it overlaps with the homogeneous solution \(y_h\text{.}\)
Now, we’ll finish the process by finding the exact values of the unknown coefficients in \(y_p\text{.}\) This final step is called the Method of Undetermined Coefficients, and it uses substitution and simplification to solve for the unknowns.

Subsection Determining the Coefficients

With \(y_p\) properly chosen and modified, the final step is to determine the values of the unknown coefficients. This step is purely algebraic: we plug \(y_p\) into the differential equation, simplify, and match coefficients to solve for the unknowns.
We will walk through this process using a concrete example. To solve
\begin{equation*} y'' - 2y' + y = 2x^2 + 3x \end{equation*}
we first find \(y_h\) by solving the homogeneous equation
\begin{equation*} y'' - 2y' + y = 0\text{.} \end{equation*}
The characteristic equation is
\begin{equation*} r^2 - 2r + 1 = 0, \end{equation*}
which factors as \((r - 1)^2 = 0\text{.}\) This gives a repeated root \(r = 1\text{,}\) so
\begin{equation*} y_h = c_1 e^x + c_2 x e^x. \end{equation*}
Now, pick \(y_p\) to match \(f(x) = 2x^2 + 3x\text{.}\) Since \(f(x)\) is a quadratic polynomial, we guess
\begin{equation*} y_p = A x^2 + B x + C. \end{equation*}
Since \(y_p\) has no terms in common with \(y_h\text{,}\) no modification is needed.
Substituting \(y_p\) and its derivatives,
\begin{equation*} y_p' = 2Ax + B,\quad y_p'' = 2A\text{,} \end{equation*}
into the equation and collecting like-terms gives
\begin{gather*} (2A) - 2(2Ax + B) + (Ax^2 + Bx + C) = 2x^2 + 3x \\ 2A - 4Ax - 2B + Ax^2 + Bx + C = 2x^2 + 3x \\ \us{\ds\text{1️⃣}}{\ul{A}}x^2 + \us{\ds\text{2️⃣}}{\ul{(-4A + B)}}x + \us{\ds\text{3️⃣}}{\ul{(2A - 2B + C)}} = \us{\ds\text{1️⃣}}{\ul{2}}x^2 + \us{\ds\text{2️⃣}}{\ul{3}}x + \us{\ds\text{3️⃣}}{\ul{0}} \text{.} \end{gather*}
Matching the underlined coefficients leads to the following system of equations and solution for \(A\text{,}\) \(B\text{,}\) and \(C\text{:}\)
\begin{align*} \amp\ \text{1️⃣}\ (x^2\ \text{terms}) \\ \amp\ \text{2️⃣}\ (x\ \text{terms}) \\ \amp\ \text{3️⃣}\ (\text{free terms}) \end{align*}
\begin{align*} A \amp = 2 \\ -4A + B \amp = 3 \\ 2A - 2B + C \amp = 0 \end{align*}
\begin{gather*} \rightarrow \\ \rightarrow \\ \rightarrow \end{gather*}
\begin{align*} \amp\ A = 2 \\ \amp\ B = 11 \\ \amp\ C = 18 \end{align*}
Therefore, \(y_p = 2x^2 + 11x + 18\) and we can now write the general solution as
\begin{equation*} y = y_h + y_p = c_1 e^x + c_2 x e^x + 2x^2 + 11x + 18\text{.} \end{equation*}
Assuming the forcing function is of the type found in Table 169, this process should always lead to a single solution. If not, then double-check your work and make sure the form of \(y_p\) was selected and modified correctly.

Checkpoint 182. 📖❓ Not A Step in Finding \(y_p\).

Which of the following is NOT a step in finding the undetermined coefficients of the particular solution?
  • Applying the initial conditions.
  • Correct! The initial conditions are used to solve for the unknown constants in the general solution, not the particular solution.
  • Plugging \(y_p \) into the differential equation.
  • Incorrect. This is the first step in determining the coefficients.
  • Collecting and matching like-terms.
  • Incorrect. This is a crucial step for setting up the system of equations used to find the undetermined coefficients: \(A\text{,}\) \(B\text{,}\) ...
  • Solving a system of equations.
  • Incorrect. This is the final step in finding the coefficients.

Subsection Method of Undetermined Coefficients Steps

Let’s summarize the entire process of using the Method of Undetermined Coefficients. Your goal is to find the full general solution \(y = y_h + y_p\) by first solving the homogeneous equation and then selecting, modifying, and finding \(y_p\text{.}\)

✍🏻 Method 183. Undetermined Coefficients.

The general solution for the linear nonhomogeneous constant-coefficient equation,
\begin{equation} a_n y^{(n)} + a_{n-1} y^{(n-1)} + \dots + a_1 y' + a_0 y = f(x)\text{,}\tag{10.2} \end{equation}
is determined from the following steps:
Step 1: Solve the homogeneous equation to get \(y_h\)
Find the homogeneous solution \(y_h\) by solving the characteristic equation from
\begin{equation*} a_n y^{(n)} + a_{n-1} y^{(n-1)} + \cdots + a_1 y' + a_0 y = 0\text{.} \end{equation*}
Step 2: Select the form of \(y_p\)
Use Table 169 to select a starting form for \(y_p\) based on \(f(x)\text{.}\) The constants \(A\text{,}\) \(B\text{,}\) \(C\text{,}\) ... are the undetermined coefficients.
Step 3: Modify \(y_p\) if it conflicts with \(y_h\)
Check \(y_p\) for conflicting terms already in \(y_h\text{.}\) If present, multiply these terms by \(x\text{.}\) Repeat until no terms are shared.
Step 4: Find the Undetermined Coefficients of \(y_p\)
Substitute \(y_p\) and its derivatives into (10.2). Simplify and match coefficients of like terms to set up a system of equations. Solving this system gives the coefficients, \(A\text{,}\) \(B\text{,}\) \(C\text{,}\) ....
Step 5: Give the Solution
Write the general solution as
\begin{equation*} y = y_h + y_p. \end{equation*}
and apply any initial conditions, if given.
You’ll follow these same five steps for any forcing function that fits the method. Now let’s apply this process to some worked examples to see how it all comes together.

Checkpoint 184.

(a) 📖❓ Steps for Finding \(y_p\).

What are the main steps for finding \(y_p\) using the method of undetermined coefficients?
  • Select a \(y_p\) form ➜ modify it, if necessary ➜ find its coefficients.
  • Correct! The process involves making an educated guess, substituting it, and solving for unknown coefficients.
  • Select a \(y_p\) form ➜ integrate it ➜ find its coefficients.
  • Incorrect. Integration is not a part of the method of undetermined coefficients.
  • Select a \(y_p\) form ➜ find its coefficients ➜ differentiate it.
  • Incorrect. The focus is on selecting and solving for \(y_p\text{,}\) not guessing the homogeneous solution.
  • Select a \(y_p\) form ➜ differentiate it ➜ verify the solution.
  • Incorrect. Integration is not a step in this method; it’s about solving for coefficients of \(y_p\text{.}\)

(b) 📖❓ Final Step of the Method.

In the final step of the method, after substituting \(y_p\) and simplifying, what do you do next?
  • Graph \(y_h\) and \(y_p\) to verify behavior.
  • Use initial conditions to determine integration constants.
  • Match coefficients of like terms and solve for unknown constants in \(y_p\text{.}\)
  • Integrate \(y_p\) to eliminate forcing terms.

Subsection Worked Examples

Apply the method of undetermined coefficients to find the general solution of the following linear nonhomogeneous differential equations.

🌌 Example 185. First-Order LNCC with an Initial Condition.

Find the general solution of
\begin{equation*} y' - 3y = 9x^2 + x, \quad y(0) = 5 \end{equation*}
Solution.
Characteristic equation:
\(\ds r - 3 = 0 \quad\Rightarrow\quad r = 3 \)
Homogeneous solution:
\(\ds y_h = c_1 e^{3x} \)
Initial \(y_p\text{:}\)
\(\ds y_p = A x^2 + B x + C \)
Compute \(y_p'\text{:}\)
\(y_p' = 2A x + B \)
Substitute \(y_p\text{,}\) \(y_p'\text{:}\)
\((2A x + B) - 3 (A x^2 + B x + C) = 9x^2 + x \)
Group powers of \(x\text{:}\)
\((-3A)\ x^2 + (2A - 3B)\ x + (B - 3C) = 9x^2 + x \)
Match coefficients:
\begin{align*} -3A \amp = 9 \\ 2A - 3B \amp = 1 \\ B - 3C \amp = 0 \end{align*}
\begin{gather*} \\ \Rightarrow \\ \Rightarrow \end{gather*}
\begin{align*} \\ 2(-3) - 3B \amp = 1 \\ B - 3C \amp = 0 \end{align*}
\begin{gather*} \Rightarrow \\ \Rightarrow \\ \Rightarrow \end{gather*}
\begin{align*} A \amp = -3 \\ B \amp = -\sfrac{7}{3} \\ C \amp = -\sfrac{7}{9} \end{align*}
Particular solution:
\(\ds y_p = -3 x^2 - \frac{7}{3} x - \frac{7}{9} \)
General solution:
\(\ds y = c_1 e^{3x} - 3 x^2 - \frac{7}{3} x - \frac{7}{9} \)
Apply \(y(0) = 5\text{:}\)
\(\ds 5 = y(0) = c_1 - \frac{7}{9} \quad\Rightarrow c_1 = \frac{52}{9} \)
final solution:
\(\ds y = \frac{52}{9} e^{3x} - 3 x^2 - \frac{7}{3} x - \frac{7}{9} \)

🌌 Example 186. Exponential Forcing Function.

Find the general solution of
\begin{equation*} y'' - 3y' + 2y = 5e^{2x} \end{equation*}
Solution.
Characteristic equation:
\begin{align*} r^2 - 3r + 2 = 0 \amp\quad\Rightarrow\quad (r - 1)(r - 2) = 0 \\ \amp\quad\Rightarrow\quad r = 1, 2 \end{align*}
Homogeneous solution:
\(\ds y_h = c_1 e^x + c_2 e^{2x} \)
Initial \(y_p\text{:}\)
\(\ds y_p = A e^{2x} \)
Since \(Ae^{2x}\) and \(c_2e^{2x}\) are like terms, we modify \(y_p\text{:}\)
Modified \(y_p\text{:}\)
\(\ds y_p = Axe^{2x} \)
Compute \(y_p'\text{,}\) \(y_p''\text{:}\)
\begin{align*} y_p' \amp = A e^{2x} + 2Ax e^{2x}\\ y_p'' \amp = 2A e^{2x} + 2A e^{2x} + 4Ax e^{2x} = 4A e^{2x} + 4Ax e^{2x} \end{align*}
Substitute \(y_p\text{,}\) \(y_p'\text{,}\) \(y_p''\text{:}\)
\begin{equation*} (4A e^{2x} + 4Ax e^{2x}) - 3(A e^{2x} + 2Ax e^{2x}) + 2(A x e^{2x}) = 5e^{2x} \end{equation*}
Group like terms:
\begin{gather*} \ub{(2A - 3A + 2A)}_{\ds = A} e^{2x} + \ub{(4A - 6A + 2A)}_{\ds = 0} x e^{2x} = 5e^{2x}\\ A e^{2x} = 5e^{2x} \end{gather*}
Match coefficients:
\(A = 5 \)
Particular solution:
\(\ds y_p = 5 x e^{2x} \)
General solution:
\(\ds y = c_1 e^x + c_2 e^{2x} + 5 x e^{2x} \)

🌌 Example 187. Trigonometric Forcing Function.

Find the general solution of
\begin{equation*} y'' + y = \cos(2x). \end{equation*}
Solution.
Characteristic equation:
\(\ds r^2 + 1 = 0 \Rightarrow r = \pm i \)
Homogeneous solution:
\(\ds y_h = c_1 \cos x + c_2 \sin x \)
Initial \(y_p\text{:}\)
\(\ds A \cos 2x + B \sin 2x \)
\(\{\cos 2x, \sin 2x\}\) and \(\{\cos x, \sin x\}\) are not like terms ➜ No modification needed.
Compute \(y_p'\) and \(y_p''\text{:}\)
\begin{gather*} y_p' = -2A \sin 2x + 2B \cos 2x \\ y_p'' = -4A \cos 2x - 4B \sin 2x \end{gather*}
Substitute \(y_p\text{,}\) \(y_p''\text{:}\)
\begin{equation*} (-4A \cos 2x - 4B \sin 2x) + (A \cos 2x + B \sin 2x) = \cos 2x \end{equation*}
Group like-terms:
\(\ds (-3A) \cos 2x + (-3B) \sin 2x = \cos 2x \)
Match coefficients:
\begin{align*} -3A \amp = 1 \\ -3B \amp = 0 \end{align*}
\begin{gather*} \Rightarrow \\ \Rightarrow \end{gather*}
\begin{align*} A \amp = -\sfrac{1}{3} \\ B \amp = 0 \end{align*}
Particular solution:
\(\ds y_p = -\frac{1}{3} \cos 2x \)
\begin{equation*} y = c_1 \cos x + c_2 \sin x - \frac{1}{3} \cos 2x \end{equation*}

🌌 Example 188. Second-Order LNCC with \(y_p\) Modification.

Find the general solution of
\begin{equation*} y'' + 3y' - 28y = 14t - 1 + e^{4t} \end{equation*}
Solution.
Characteristic equation:
\begin{align*} r^2 + 3r - 28 = 0 \amp\quad\Rightarrow\quad (r - 4)(r + 7) = 0 \\ \amp\quad\Rightarrow\quad r = 4, -7 \end{align*}
Homogeneous solution:
\(\ds y_h = c_1 e^{4t} + c_2 e^{-7t} \)
Initial \(y_p\text{:}\)
\(\ds y_p = At + B + Ce^{4t} \)
Since \(y_p\) and \(y_h\) share an \(e^{4t}\) term, we modify this term in \(y_p\text{:}\)
Modified \(y_p\text{:}\)
\(\ds y_p = At + B + Ct\,e^{4t} \)
Compute \(y_p'\text{,}\) \(y_p''\text{:}\)
\begin{align*} y_p' \amp = A + C e^{4t} + 4Ct\, e^{4t}\\ y_p'' \amp = 8C e^{4t} + 16Ct\, e^{4t} \end{align*}
Substitute \(y_p\text{,}\) \(y_p'\text{,}\) \(y_p''\text{:}\)
\begin{align*} 8C\ e^{4t} + 16Ct\, e^{4t} + 3(A + C e^{4t} + 4Ct\, e^{4t}) - 28(At \amp + B + Ct\, e^{4t})\\ \amp = 14t - 1 + e^{4t} \end{align*}
The \(te^{4t}\) terms cancel, but you can group the \(e^{4t}\text{,}\) \(t\text{,}\) and free terms:
\begin{align*} \us{\ds\text{1️⃣}}{\ul{11C}} e^{4t} - \us{\ds\text{2️⃣}}{\ul{28A}} t \amp + \us{\ds\text{3️⃣}}{\ul{3A - 28B}} = \us{\ds\text{2️⃣}}{\ul{14}} t + \us{\ds\text{3️⃣}}{\ul{-1}} + \us{\ds\text{1️⃣}}{\ul{1}} e^{4t} \end{align*}
Match coefficients:
\begin{align*} \amp\ \text{1️⃣} \\ \amp\ \text{2️⃣} \\ \amp\ \text{3️⃣} \end{align*}
\begin{align*} 11C \amp = 1 \\ -28A \amp = 14 \\ 3A - 28B \amp = -1 \end{align*}
\begin{gather*} \rightarrow \\ \rightarrow \\ \rightarrow \end{gather*}
\begin{align*} \amp\ C = 1/11 \\ \amp\ A = -1/2 \\ \amp\ B = -1/56 \end{align*}
Particular solution:
\(\ds y_p = -\frac{1}{2} t - \frac{1}{56} + \frac{1}{11} t e^{4t} \)
General solution:
\(\ds y = c_1 e^{4t} + c_2 e^{-7t} - \frac{1}{2} t - \frac{1}{56} + \frac{1}{11} t e^{4t} \)

🌌 Example 189. Forcing Function as a Product.

Find the general solution of
\begin{equation*} y'' - 10y' = x \sin x \end{equation*}
Solution.
Characteristic equation:
\(\ds r^2 - 10r = 0, \ r(r - 10) = 0, \ r = 0, 10 \)
Homogeneous solution:
\(\ds y_h = c_1 + c_2 e^{10x} \)
Initial \(y_p\text{:}\)
\(\ds y_p = (A x + B) \cos x + (C x + D) \sin x \)
Note: This part can be a bit intense. The product rule is used multiple times, and after each step, we regroup the \(\cos x\text{,}\) \(\sin x\text{,}\) \(x\cos x\text{,}\) and \(x\sin x\) terms.
Compute \(y_p'\) and \(y_p''\text{:}\)
\begin{align*} y_p' \amp = A \cos x - (Ax + B) \sin x + C \sin x + (Cx + D) \cos x \\ \amp = (A + D)\cos x + (-B + C)\sin x + Cx\cos x - Ax\sin x \\ y_p'' \amp = -(A + D)\sin x + (-B + C)\cos x \\ \amp \hspace{4cm} + C\cos x - Cx\sin x - A\sin x - Ax\cos x \\ \amp = (-B + 2C)\cos x + (-2A - D)\sin x - Ax\cos x - Cx\sin x \end{align*}
Substitute \(y_p''\text{,}\) \(y_p'\text{:}\)
\begin{alignat*}{5} (-B + 2C) \amp {}\cos x{} {}+{} \amp (-2A + D) \amp {}\sin x{} {}+{} \amp (-A) \amp {}x\cos x{} {}+{} \amp (-C) \amp {}x\sin x{} \amp\\ -10((A + D) \amp {}\cos x{} {}+{} \amp (-B + C) \amp {}\sin x{} {}+{} \amp (C) \amp {}x\cos x{} {}+{} \amp (-A) \amp {}x\sin x{}) \amp\\ \amp \amp \amp \amp \amp \amp {}={} \amp x \sin x \amp \end{alignat*}
Group like-terms:
\begin{align*} \amp (-B + 2C - 10A - 10D)\cos x + (10B - 2A - D - 10C)\sin x \\ \amp + (-A - 10C)\ x\cos x + {\DLO (10A - C)}\ x\sin x = {\DLO 1} x \sin x \end{align*}
Match coefficients:
\begin{alignat*}{4} -10{}A{}\amp {}-{}\amp {}B{}\amp {}+{}\amp 2{}C{}\amp {}-{}\amp 10 {}D{}\amp {}={} 0 \\ -2{}A{}\amp {}+{}\amp 10{}B{}\amp {}+{}\amp {}C{}\amp {}-{}\amp {}D{}\amp {}={} 0 \\ - {}A{}\amp {} {}\amp {} {}\amp {}-{}\amp 10{}C{}\amp {} {}\amp {} {}\amp {}={} 0 \\ 10{}A{}\amp {} {}\amp {} {}\amp {}-{}\amp {}C{}\amp {} {} \amp {} {}\amp {}={} 1 \end{alignat*}
\begin{gather*} \\ \\ \Rightarrow \\ \Rightarrow \end{gather*}
\begin{align*} \\ \\ A \amp = \sfrac{10}{101} \\ C \amp = -\sfrac{1}{101} \end{align*}
Substituting \(A\) and \(C\) into the first two equations gives:
\begin{alignat*}{2} - {}B{}\amp {}-{}\amp 10 {}D{}\amp {}={} \sfrac{102}{101} \\ 10 {}B{}\amp {}-{}\amp {}D{}\amp {}={} \sfrac{10}{101} \end{alignat*}
\begin{gather*} \Rightarrow\\ \Rightarrow \end{gather*}
\begin{alignat*}{2} -10{}B{}\amp {}-{}\amp 100 {}D{}\amp {}={} \sfrac{1020}{101} \\ 10{}B{}\amp {}-{}\amp {}D{}\amp {}={} \sfrac{10}{101} \end{alignat*}
Adding the two equations gives: \(B = -\sfrac{2}{10201}\) and \(D = -\sfrac{1030}{10201}\)
Particular solution:
\begin{equation*} y_p = \left(\frac{10}{101}x - \frac{2}{10201}\right) \cos x + \left(-\frac{1}{101}x - \frac{1030}{10201}\right) \sin x \end{equation*}
General solution:
\begin{equation*} y = c_1 + c_2 e^{10x} + \left(\frac{10}{101}x - \frac{2}{10201}\right) \cos x + \left(-\frac{1}{101}x - \frac{1030}{10201}\right) \sin x \end{equation*}
These examples illustrate the systematic nature of this method. No matter how complex the forcing function, the steps remain consistent: identify the homogeneous solution, propose a form for the particular solution, modify it if necessary, compute derivatives, substitute into the differential equation, and match coefficients to solve for unknowns.

Checkpoint 190. 🤔💭 Reading Questions.

(a) 📖❓ Undetermined Coefficients and Nonlinear Equations.

    The method of undetermined coefficients can be applied to nonlinear differential equations.

  • True.

  • Incorrect. The method of undetermined coefficients applies only to linear nonhomogeneous equations with constant coefficients.

  • False.

  • Incorrect. The method of undetermined coefficients applies only to linear nonhomogeneous equations with constant coefficients.

(b) 📖❓ Sort the Steps in Order.

(c) 📖❓ Combining Constants.

Which of the following are valid ways to combine constants when finalizing your solution? Select ALL that apply.
  • \(\quad \left\{ \begin{align} |y| \amp = e^{2x + c_1}\\ y \amp = c_2 e^{2x} \end{align} \right.\qquad\) where \(c_2=\pm e^{c_1}\text{.}\)
  • Correct, since
    \begin{align*} |y| \amp = e^{2x + c_1} \\ y \amp = \pm e^{2x + c_1} \\ y \amp = \pm e^{2x} \cdot e^{c_1} \\ y \amp = c_2 e^{2x} \quad \text{where } c_2 = \pm e^{c_1} \end{align*}
  • \(\quad \left\{ \begin{align} \frac{y^2}{2} \amp = \sin(x) + c_1\\ y^2 \amp = 2\sin(x) + c_2 \end{align} \right.\qquad\) where \(c_2 = 2c_1\text{.}\)
  • Correct, since
    \begin{align*} \frac{y^2}{2} \amp = \sin(x) + c_1 \\ y^2 \amp = \sin(x) + 2c_1 \\ y^2 \amp = 2\sin(x) + c_2 \quad \text{where } c_2 = 2c_1 \end{align*}
  • \(\quad \left\{ \begin{align} y^2 \amp = 2\sin(x) + c_2\\ y \amp = \pm\sqrt{2\sin(x)} + c_3 \end{align} \right.\qquad\) where \(c_3=\pm \sqrt{c_2}\text{.}\)
  • Incorrect, you should not combine constants here since \(\quad\sqrt{A + B} \neq \sqrt{A} + \sqrt{B}\text{.}\)
    Your solution should be written as
    \begin{align*} y^2 \amp = 2\sin(x) + c_2 \\ y \amp = \pm\sqrt{2\sin(x) + c_2}\quad \leftarrow\text{ done} \end{align*}
  • \(\quad \left\{ \begin{align} e^y \amp = 3y + c_1\\ y \amp = \ln(3y) + c_2 \end{align} \right.\qquad\) where \(c_2= \ln(c_1)\text{.}\)
  • Incorrect; you should not combine constants here since \(\ln(A + B) \neq \ln(A) + \ln(B)\text{.}\)
    Your solution should be
    \begin{align*} e^y \amp = 3y + c_1 \\ y \amp = \ln(3y + c_1)\quad \leftarrow\text{ done} \end{align*}
  • \(\quad \left\{ \begin{align} xy \amp = e^x + c_1\\ y \amp = \frac{e^x}{x} + c_2 \end{align} \right.\qquad\) where \(\ds c_2=\frac{c_1}{x}\text{.}\)
  • Incorrect; you should not combine constants here since \(\frac{c_1}{x}\) is not constant. Instead, your solution should be left as
    \begin{align*} xy \amp = e^x + c_1 \\ y \amp = \frac{e^x + c_1}{x} \\ y \amp = \frac{e^x}{x} + \frac{c_1}{x}\quad \leftarrow\text{ done} \end{align*}
  • \(\quad \left\{ \begin{align} \sqrt{y} \amp = c_1e^{3x}\\ y \amp = c_2\, e^{6x} \end{align} \right.\qquad\) where \(c_2= c_1^2\text{.}\)
  • Correct, since
    \begin{align*} \sqrt{y} \amp = c_1e^{3x} \\ y \amp = (c_1e^{3x})^2 \\ y \amp = c_1^2(e^{3x})^2 \\ y \amp = c_2\, e^{6x}\quad \text{where } c_2 = c_1^2 \end{align*}
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