DE-BOOK Undetermined Coefficients Method

Section 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 \(y_h\text{.}\)

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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.

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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.

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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*}

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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*}

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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.

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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*}

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Matching the underlined coefficients leads to the following system of equations and solution for \(A\text{,}\) \(B\text{,}\) and \(C\text{:}\)

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\begin{align*} \amp\ \text{1️⃣}\ (x^2\ \text{terms}) \\ \amp\ \text{2️⃣}\ (x\ \text{terms}) \\ \amp\ \text{3️⃣}\ (\text{free terms}) \end{align*}

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\begin{align*} A \amp = 2 \\ -4A + B \amp = 3 \\ 2A - 2B + C \amp = 0 \end{align*}

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\begin{gather*} \rightarrow \\ \rightarrow \\ \rightarrow \end{gather*}

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\begin{align*} \amp\ A = 2 \\ \amp\ B = 11 \\ \amp\ C = 18 \end{align*}

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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*}

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Assuming the forcing function is of the type found in Table 159, 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.

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Checkpoint 172. 📖❓ Not A Step in Finding \(y_p\).

Which of the following is NOT a step in finding the undetermined coefficients of the particular solution?

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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.

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Subsection Undetermined Coefficient Method 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 \(y_p\text{.}\)

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✍🏻 Method 5. 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{42} \end{equation}

is determined from the following steps:

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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*}

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Step 2: Select the form of \(y_p\) 🔗: Use Table 159 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.
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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.
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Step 4: Find the Undetermined Coefficients of \(y_p\) 🔗: Substitute \(y_p\) and its derivatives into (42). Simplify and match coefficients of like terms to set up a system of equations. Solving this system gives the coefficients, \(A\text{,}\) \(B\text{,}\) \(C\text{,}\) ....
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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.

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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.

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Checkpoint 173.

(a) 📖❓ Not A Step in the Undetermined Coefficient Method.

What are the main steps for finding \(y_p\) using the method of undetermined coefficients?

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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 \(y_p\text{.}\)

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(b) 📖❓ Final Step of the Method.

In the final step of the method, after substituting \(y_p\) and simplifying, what do you do next?

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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.

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Subsection Worked Examples

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

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🌌 Example 174. 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*}

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Solution.

Step 1: Solve the homogeneous equation to get \(y_h\)

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characteristic equation:

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\(\ds r - 3 = 0 \quad\Rightarrow\quad r = 3 \)

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homogeneous solution:

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\(\ds y_h = c_1 e^{3x} \)

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Step 2: Select the form of \(y_p\)

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initial \(y_p\text{:}\)

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\(\ds y_p = A x^2 + B x + C \)

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Step 3: Modify \(y_p\) if it conflicts with \(y_h\) ➜ No modification needed.

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Step 4: Find the Undetermined Coefficients of \(y_p\)

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compute \(y_p'\text{:}\)

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\(y_p' = 2A x + B \)

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substitute \(y_p\text{,}\) \(y_p'\text{:}\)

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\((2A x + B) - 3 (A x^2 + B x + C) = 9x^2 + x \)

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group powers of \(x\text{:}\)

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\((-3A)\ x^2 + (2A - 3B)\ x + (B - 3C) = 9x^2 + x \)

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Match coefficients:

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\begin{align*} -3A \amp = 9 \\ 2A - 3B \amp = 1 \\ B - 3C \amp = 0 \end{align*}

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\begin{gather*} \Rightarrow \\ \Rightarrow \\ \Rightarrow \end{gather*}

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\begin{align*} A = -3 \amp \\ 2(-3)-3B \amp = -2 \\ -\sfrac34 - 3C \amp = 0 \end{align*}

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\begin{gather*} \\ \Rightarrow \\ \Rightarrow \end{gather*}

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\begin{align*} \\ B \amp = -\sfrac{3}{4} \\ C \amp = -\sfrac{1}{4} \end{align*}

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particular solution:

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\(\ds y_p = -3 x^2 - \frac34 x - \frac14 \)

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Step 5: Give the Solution

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general solution:

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\(\ds y = c_1 e^{3x} - 3 x^2 - \frac34 x - \frac14 \)

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apply \(y(0) = 5\text{:}\)

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\(\ds 5 = y(0) = c_1 - \frac14 \quad\Rightarrow c_1 = \frac{21}{4} \)

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final solution:

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\(\ds y = \frac{21}{4} e^{3x} - 3 x^2 - \frac34 x - \frac14 \)

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🌌 Example 175. Exponential Forcing Function.

Find the general solution of

\begin{equation*} y'' - 3y' + 2y = 5e^{2x} \end{equation*}

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Solution.

Step 1: Solve the homogeneous equation to get \(y_h\)

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characteristic equation:

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\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*}

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homogeneous solution:

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\(\ds y_h = c_1 e^x + c_2 e^{2x} \)

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Step 2: Select the form of \(y_p\)

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initial \(y_p\text{:}\)

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\(\ds y_p = A e^{2x} \)

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Step 3: Modify \(y_p\) if it conflicts with \(y_h\)

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Since \(Ae^{2x}\) and \(c_2e^{2x}\) are like terms, we modify \(y_p\text{:}\)

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modified \(y_p\text{:}\)

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\(\ds y_p = Axe^{2x} \)

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Step 4: Find the Undetermined Coefficients of \(y_p\)

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compute \(y_p'\text{,}\) \(y_p''\text{:}\)

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\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*}

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substitute \(y_p\text{,}\) \(y_p'\text{,}\) \(y_p''\text{:}\)

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\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:

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\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:

\(\ds y_p = 5 x e^{2x} \)

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Step 5: Give the Solution

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general solution:

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\(\ds y = c_1 e^x + c_2 e^{2x} + 5 x e^{2x} \)

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🌌 Example 176. Trigonometric Forcing Function.

Find the general solution of

\begin{equation*} y'' + y = \cos(2x). \end{equation*}

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Solution.

Step 1: Solve the homogeneous equation to get \(y_h\)

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characteristic equation:

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\(\ds r^2 + 1 = 0 \Rightarrow r = \pm i \)

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homogeneous solution:

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\(\ds y_h = c_1 \cos x + c_2 \sin x \)

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Step 2: Select the form of \(y_p\)

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initial \(y_p\text{:}\)

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\(\ds A \cos 2x + B \sin 2x \)

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Step 3: Modify \(y_p\) if it conflicts with \(y_h\)

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\(\{\cos 2x, \sin 2x\}\) and \(\{\cos x, \sin x\}\) are not like terms ➜ No modification needed.

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Step 4: Find the Undetermined Coefficients of \(y_p\)

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compute \(y_p'\) and \(y_p''\text{:}\)

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\begin{gather*} y_p' = -2A \sin 2x + 2B \cos 2x \\ y_p'' = -4A \cos 2x - 4B \sin 2x \end{gather*}

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substitute \(y_p\text{,}\) \(y_p'\text{:}\)

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\begin{equation*} (-4A \cos 2x - 4B \sin 2x) + (-2A \sin 2x + 2B \cos 2x) = \cos 2x \end{equation*}

group like-terms:

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\(\ds (-4A + 2B) \cos 2x + (-2A - 4B) \sin 2x = \cos 2x \)

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Match coefficients:

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\begin{align*} -4A + 2B \amp = 1 \\ -2A - 4B \amp = 0 \end{align*}

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\begin{gather*} \Rightarrow \end{gather*}

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\begin{align*} -4A + 2B \amp = 1 \\ 4A + 8B \amp = 0 \end{align*}

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\begin{gather*} \Rightarrow \end{gather*}

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\begin{align*} A \amp = \sfrac{2}{10} \\ B \amp = \sfrac{1}{10} \end{align*}

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particular solution:

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\(\ds y_p = \frac{2}{10} \sin 2x + \frac{1}{10} \cos 2x \)

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Step 5: Give the Solution

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\begin{equation*} y = c_1 \cos x + c_2 \sin x + \frac{2}{10} \sin 2x + \frac{1}{10} \cos 2x \end{equation*}

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🌌 Example 177. Second-Order LNCC with \(y_p\) Modification.

Find the general solution of

\begin{equation*} y'' + 3y' - 28y = 14t - 1 + e^{4t} \end{equation*}

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Solution.

Step 1: Solve the homogeneous equation to get \(y_h\)

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characteristic equation:

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\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*}

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homogeneous solution:

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\(\ds y_h = c_1 e^{4t} + c_2 e^{-7t} \)

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Step 2: Select the form of \(y_p\)

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initial \(y_p\text{:}\)

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\(\ds y_p = At + B + Ce^{4t} \)

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Step 3: Modify \(y_p\) if it conflicts with \(y_h\)

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Since \(y_p\) and \(y_h\) share an \(e^{4t}\) term, we modify this term in \(y_p\text{:}\)

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modified \(y_p\text{:}\)

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\(\ds y_p = At + B + Ct\,e^{4t} \)

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Step 4: Find the Undetermined Coefficients of \(y_p\)

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compute \(y_p'\text{,}\) \(y_p''\text{:}\)

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\begin{align*} y_p' \amp = A + C e^{4t} + 4Ct\, e^{4t}\\ y_p'' \amp = 8C e^{4t} + 16Ct\, e^{4t} \end{align*}

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substitute \(y_p\text{,}\) \(y_p'\text{,}\) \(y_p''\text{:}\)

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\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*}

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The \(te^{4t}\) terms cancel, but you can group the \(e^{4t}\text{,}\) \(t\text{,}\) and free terms:

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\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*}

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match coefficients:

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\begin{align*} \amp\ \text{1️⃣} \\ \amp\ \text{2️⃣} \\ \amp\ \text{3️⃣} \end{align*}

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\begin{align*} 11C \amp = 1 \\ -28A \amp = 14 \\ 3A - 28B \amp = -1 \end{align*}

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\begin{gather*} \rightarrow \\ \rightarrow \\ \rightarrow \end{gather*}

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\begin{align*} \amp\ C = 1/11 \\ \amp\ A = -1/2 \\ \amp\ B = -1/56 \end{align*}

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particular solution:

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\(\ds y_p = -\frac{1}{2} t - \frac{1}{56} + \frac{1}{11} t e^{4t} \)

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Step 5: Give the Solution

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general solution:

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\(\ds y = c_1 e^{4t} + c_2 e^{-7t} - \frac{1}{2} t - \frac{1}{56} + \frac{1}{11} t e^{4t} \)

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🌌 Example 178. Forcing Function as a Product.

Find the general solution of

\begin{equation*} y'' - 10y' = x \sin x \end{equation*}

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Solution.

Step 1: Solve the homogeneous equation to get \(y_h\)

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characteristic equation:

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\(\ds r^2 - 10r = 0, \ r(r - 10) = 0, \ r = 0, 10 \)

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homogeneous solution:

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\(\ds y_h = c_1 + c_2 e^{10x} \)

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Step 2: Select the form of \(y_p\)

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initial \(y_p\text{:}\)

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\(\ds y_p = (A x + B) \cos x + (C x + D) \sin x \)

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Step 3: Modify \(y_p\) if it conflicts with \(y_h\) ➜ No modification needed.

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Step 4: Find the Undetermined Coefficients of \(y_p\)

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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.

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compute \(y_p'\) and \(y_p''\text{:}\)

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\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 \(y_p''\text{,}\) \(y_p'\text{:}\)

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\begin{align*} ((-B \amp + 2C)\cos x + (-2A + D)\sin x - Ax\cos x - Cx\sin x) \\ \amp - 10((A + D)\cos x + (-B + C)\sin x + Cx\cos x - Ax\sin x) = x \sin x \end{align*}

group like-terms:

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\begin{align*} \amp (-B + 2C - 10A - 10D)\cos x \\ \amp +\ (-2A + D + 10B - 10C)\sin x \\ \amp +\ (-A + C)\ x\cos x + {\DLO (-C - A)}\ x\sin x = {\DLO 1} x \sin x \end{align*}

Match coefficients:

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\begin{align*} -B + 2C - 10A - 10D \amp = 0 \\ -2A + D + 10B - 10C \amp = 0 \end{align*}

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\begin{align*} -A + C \amp = 0 \\ {\DLO -C - A} \amp = {\DLO 1} \end{align*}

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\begin{gather*} \Rightarrow \end{gather*}

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\begin{align*} A \amp = \sfrac12 \\ C \amp = \sfrac12 \end{align*}

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Substituting \(A = \sfrac12\) and \(C = \sfrac12\) into the first two equations gives:

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\begin{align*} -B - 10D \amp = 4 \\ 10B + D \amp = 6 \end{align*}

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\begin{gather*} \Rightarrow \end{gather*}

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\begin{align*} -10B - 100D \amp = 40 \\ 10B + D \amp = 6 \end{align*}

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\begin{gather*} \Rightarrow \end{gather*}

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\begin{align*} B \amp = \sfrac{64}{99} \\ D \amp = \sfrac{46}{99} \end{align*}

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particular solution:

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\begin{equation*} y_p = \left(\frac{1}{2} + \frac{64}{99}x\right) \cos x + \left(\frac{1}{2} + \frac{46}{99}x\right) \sin x \end{equation*}

Step 5: Give the Solution

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General solution:

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\begin{equation*} y = c_1 \cos x + c_2 \sin x + \left(\frac{1}{2} + \frac{64}{99}x\right) \cos x + \left(\frac{1}{2} + \frac{46}{99}x\right) \sin x \end{equation*}

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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.

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Subsection 📤 Wrap-Up

🗝️ \(\textbf{Key Takeaways...}\)

The general solution to a nonhomogeneous differential equation is \(y = y_h + y_p\text{,}\) where \(y_h\) solves the homogeneous equation and \(y_p\) accounts for the forcing function.

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The Method of Undetermined Coefficients provides a systematic way to find \(y_p\) when \(f(x)\) consists of polynomials, exponentials, sines, cosines, or their combinations.

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Choose \(y_p\) by matching the form of \(f(x)\) and modify it by multiplying by \(x\) when necessary to avoid overlap with \(y_h\text{.}\)

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Substitute \(y_p\) and its derivatives into the differential equation, simplify, and match like terms to solve for unknown coefficients.

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Check Your Understanding.

Checkpoint 179. 🤔💭 Separation of Variables Reading Questions.

(a) 🤔💭 SOV and Nonlinear Equations.

The separation of variables method cannot be applied to non-linear differential equations.

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True.
Being able to solve non-linear differential equations is one of the strengths of the separation of variables method.
False.
Being able to solve non-linear differential equations is one of the strengths of the separation of variables method.

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(b) 🤔💭 Sort the Steps in Order.

CONVERT TO PARSONS.. Suppose you want to use separation of variables method to determine the general solution of the differential equation

\begin{equation} \ds \frac{dq}{dt} - 3qt^2 = 2q\text{.}\tag{43} \end{equation}

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Match the scrambled tasks (on the left) to the task order (on the right) that it should be performed to use this method.

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Review the SOV Method.

Equation is 1st-order ✅
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\(\text{Task 1}\)
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\(\ds \frac{dq}{dt} = q(2 + 3t^2)\)
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\(\text{Task 2}\)
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\(\ds \frac{1}{q}\ dq = (2 + 3t^2)\ dt\)
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\(\text{Task 3}\)
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\(\ds \int \frac{1}{q} dq = \int (2 + 3t^2) dt\)
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\(\text{Task 4}\)
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\(\ds \ln|q| = 2t^2 + t^3 + c\)
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\(\text{Task 5}\)
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\(\ds q = c e\vphantom{|}^{\large 2t^2 + t^3}\)
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\(\text{Task 6}\)
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(c) 🤔💭 Combining Constants.

Which of the following are valid ways to combine constants when finalizing your solution? Select ALL that apply.

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\(\quad \left\{ \begin{align} |y| \amp = e^{2x + c_1}\\ y \amp = c_2 e^{2x} \end{align} \right.\qquad\) 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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