DE-BOOK Higher-Order Equations

Section Higher-Order Equations

Like second-order linear homogeneous constant coefficient (LHCC) equations, we solve higher-order equations by finding the roots of a characteristic equation. However, with higher-order comes higher-degree characteristic polynomials and an increased number of roots. While this results in additional terms, determining which ones go into in the general solution is similar to the second-order case.

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Checkpoint 145. 📖❓ Just One Root.

Suppose the characteristic equation of an LHCC equation has a single root, \(r = -2\text{.}\) Select the correct statement.

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The equation is \(1\)st order with general solution \(\ ce^{-2x}\text{.}\)
Correct! The order of the equation is \(1\) because there is only one root, and the general solution is given by the exponential function with that root.
The equation is \(1\)st order with general solution \(\ ce^{2x}\text{.}\)
Incorrect. The root is negative, so the general solution should involve \(e^{-2x}\text{,}\) not \(e^{2x}\text{.}\)
The equation is \(2\)nd order with general solution \(\ c_1 e^{-2x} + c_2 x e^{-2x}\text{.}\)
Incorrect. The order of the equation is determined by the number of roots, which is \(1\) in this case, not \(2\text{.}\)
The equation can have any order with general solution \(\ ce^{-2x}\text{.}\)
Incorrect. The order of the equation is determined by the number of roots, which is \(1\) in this case, not \(2\text{.}\)

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Subsection The Characteristic Equation

The standard \(n\)th-order LHCC equation has the form:

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\begin{equation*} a_n y^{(n)} + a_{n-1} y^{(n-1)} + \cdots + a_1 y' + a_0 y = 0 \end{equation*}

To find solutions, we assume \(y = e^{rx}\text{.}\) Substituting this into the equation yields:

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\begin{align*} a_n r^n e^{rx} + a_{n-1} r^{n-1} e^{rx} + \cdots + a_1 r e^{rx} + a_0 e^{rx} \amp = 0 \\ (a_n r^n + a_{n-1} r^{n-1} + \cdots + a_1 r + a_0) e^{rx} \amp = 0 \end{align*}

Since \(e^{rx} \ne 0\text{,}\) we obtain the characteristic equation:

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\begin{equation*} a_n r^n + a_{n-1} r^{n-1} + \cdots + a_1 r + a_0 = 0 \end{equation*}

This is a polynomial of degree \(n\text{,}\) and solving it gives us the roots that define the structure of the general solution. Similar to second-order equations, each root contributes a different term in the general solution, as detailed next.

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Checkpoint 146. 📖❓ Degree of the Characteristic Equation.

The order of an LHCC differential equation is equal to the degree of its characteristic equation

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True.
This is true, since the order of each derivative in the differential equation translates directly to a corresponding power of \(r\) in the characteristic equation.
False.
This is true, since the order of each derivative in the differential equation translates directly to a corresponding power of \(r\) in the characteristic equation.

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Subsection Real and Distinct Roots

Every real root \(r\) that is distinct from others contributes a single term of the form

\begin{equation*} c e^{r x} \end{equation*}

to the general solution, where \(c\) is a constant.

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For example, if the characteristic equation has roots \(r = 2, -1\text{,}\) & \(4.5\) that only appear once, then part of the general solution is:

\begin{equation*} c_1 e^{2x} + c_2 e^{-x} + c_3 e^{4.5x}\text{.} \end{equation*}

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Subsection Repeated Roots

Every root \(r\) that repeats \(k\) times (multiplicity \(k\)) contributes

\begin{equation} c_1 e^{r x} + c_2 x e^{r x} + c_3 x^2 e^{r x} + \dots + c_k x^{k-1} e^{r x}\tag{37} \end{equation}

to the general solution.

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For example, if \(r = 4\) is a triple root, then part of the general solution will be:

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\begin{equation*} c_1 e^{4x} + c_2 x e^{4x} + c_3 x^2 e^{4x} \end{equation*}

Note: If the repeated root is \(r = 0\text{,}\) then \(e^{0x} = 1\text{,}\) and (37) reduces to:

\begin{equation*} c_1 + c_2 x + \dots + c_k x^{k-1}\text{.} \end{equation*}

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Subsection Complex Roots

Every complex root, \(\alpha \pm i\beta\) (conjugate pair), contributes

\begin{equation*} e^{\alpha x} (c_1 \cos(\beta x) + c_2 \sin(\beta x)) \end{equation*}

to the general solution. There may be multiple pairs, each contributing their own term.

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Subsection From Characteristic Roots to General Solution

For a general \(n\)th-order LHCC equation, the full general solution includes exactly \(n\) terms—one for each root of the characteristic polynomial. The form of each term depends on the type of root (real or complex) and whether it repeats. These rules are summarized in Table 147.

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Table 147. Characteristic Root Contribution to the General Solution

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root type	contribution to general solution
real (multiplicity \(1\))	\(c e^{rx}\)
real (multiplicity \(k\))	\(c_1 e^{rx} + c_2 x e^{rx} + \cdots + c_k x^{k-1} e^{rx}\)
complex (\(\alpha \pm i\beta\))	\(e^{\alpha x} (c_1 \cos(\beta x) + c_2 \sin(\beta x))\)

Let’s practice building general solutions from the roots of the characteristic equation.

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🌌 Example 148. From Charateristic Roots to General Solution.

For each set of characteristic roots, write the general solution of the LHCC equation.

\(\displaystyle r = 3,\quad -3,\quad 5.3 \)

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\(\displaystyle r = 6 \pm i\sqrt{7.7},\quad 0\ (\text{double root}) \)

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\(\displaystyle r = -4\ (\text{triple root}),\quad 5.3 \)

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\(\displaystyle r = \pm \frac{i}{2},\quad 2 \pm i \)

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Solution.

Three different real roots:
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\begin{equation*} y = c_1 e^{3x} + c_2 e^{-3x} + c_3 e^{5.3x} \end{equation*}

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A complex conjugate pair and a double real root at \(r = 0\text{:}\)
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\begin{equation*} y = \ub{e^{6x} (c_1 \sin(\sqrt{7.7}x) + c_2 \cos(\sqrt{7.7}x))}_{\ds\text{from}\ 6 \pm i\sqrt{7.7}} + \ub{c_3\ +\ c_4 x}_{\ds\text{from}\ 0\ \text{(double)}} \end{equation*}

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A triple \(-4\) root and single \(5.3\) root:
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\begin{equation*} y = \ub{c_1 e^{-4x} + c_2 x e^{-4x} + c_3 x^2 e^{-4x}\!}_{\ds\text{from}\ -4\ \text{(triple)}} + c_4 e^{5.3x} \end{equation*}

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Two complex conjugate pairs:
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\begin{equation*} y = \ub{c_1 \sin\left(\frac{x}{2}\right) + c_2 \cos\left(\frac{x}{2}\right)}_{\ds\text{from}\ \pm \sfrac{i}{2}} + \ub{e^{2x} (c_3 \sin x + c_4 \cos x)}_{\ds\text{from}\ 2 \pm i} \end{equation*}

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Subsection 📤 Wrap-Up

🗝️ \(\textbf{Key Takeaways...}\)

Knowing that an LHCC equation has order \(n\) tells you that:
- The characteristic polynomial will have degree \(n\text{.}\)
  
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- There will be \(n\) characteristic roots.
  
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- There will be \(n\) terms to in general solution.
  
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The terms that are included in the general solution depend on the characteristic root type, as summarized in Table 147.

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Check Your Understanding.

Checkpoint 149. 🤔💭 Separation of Variables Reading Questions.

(a) 🤔💭3rd-Order General Solution.

Give the general solution for a 3rd-order LHCC equation whose characteristic equation has the solutions:

\begin{equation*} r = 0, 1, -8\text{.} \end{equation*}

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\(\quad y = c_1 + c_2 e^{x} + c_3 e^{-8x}\)
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Correct! This is the form when there are three distinct real roots.
\(\quad y = c_1 e^{x} + c_2 x e^{x} + c_3 e^{-8x}\)
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Incorrect.
\(\quad y = c_1 x e^{x} + c_2 e^{-8x}\)
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Incorrect.
\(\quad y = c_1 e^{x} + c_2 e^{-8x}\)
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Incorrect.

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(b) 🤔💭Two Complex Pair Roots.

Give the general solution for a 4th-order LHCC equation whose characteristic equation has the solutions:

\begin{equation*} r = \pm 2i, -3 \pm 4i\text{.} \end{equation*}

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\(\quad y = c_1 \cos(2x) + c_2 \sin(2x) + e^{-3x} \left(c_3 \cos(4x) + c_4 \sin(4x)\right)\)
Correct! This is the form when the characteristic equation has complex roots.
\(\quad y = c_1 e^{2x} + c_2 e^{-2x} + e^{-3x} \left(c_3 \cos(4x) + c_4 \sin(4x)\right)\)
Incorrect.
\(\quad y = c_1 \left(\cos(2x) + \sin(2x)\right) + c_2 e^{-3x} \left(\cos(4x) + \sin(4x)\right)\)
Incorrect. Not enough constants.
\(\quad y = e^{2x} \left(c_1 \cos(x) + c_2 \sin(x)\right) + e^{4x} \left(c_3 \cos(3x) + c_4 \sin(3x)\right)\)
Incorrect.

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(c) 🤔💭All Three Root Types.

Suppose the characteristic equation of a fifth-order LHCC equation has the following roots:

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\begin{equation*} r = 1 \ (\text{triple}),\quad -2 \ (\text{single}),\quad 2 \pm i \end{equation*}

Which of the following correctly represents the structure of the general solution?

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\(\quad y = c_1 e^{x} + c_2 x e^{x} + c_3 x^2 e^{x} + c_4 e^{-2x} + e^{2x}(c_5 \cos x + c_6 \sin x)\)
Correct! The triple root at \(r = 1\) contributes three terms, \(e^{x},\ x e^{x},\ x^2 e^{x}\text{;}\) the single real root \(-2\) contributes \(e^{-2x}\text{;}\) and the complex conjugate pair \(2 \pm i\) gives rise to the sinusoidal term multiplied by \(e^{2x}\text{.}\)
\(\quad y = c_1 e^{x} + c_2 e^{-2x} + e^{2x}(c_3 \cos x + c_4 \sin x)\)
Incorrect. The triple root at \(r = 1\) should contribute three terms, not just one.
\(\quad y = c_1 e^{x} + c_2 x e^{x} + c_3 e^{-2x} + e^{2x}(c_4 \cos x + c_5 \sin x)\)
Incorrect. This only includes two of the three terms needed for the triple root at \(r = 1\text{.}\)
\(\quad y = c_1 e^{x} + c_2 x e^{x} + c_3 x^2 e^{x} + c_4 e^{-2x} + c_5 \cos(2x) + c_6 \sin(2x)\)
Incorrect. The complex roots \(2 \pm i\) must be expressed as \(e^{2x}(c_5 \cos x + c_6 \sin x)\text{,}\) not as standalone sine and cosine terms.

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(d) 🤔💭Multiple Repeating Roots.

Give the general solution for a 5th-order LHCC equation whose characteristic equation has the solutions:

\begin{equation*} r = 2, 2, 2, -1, -1\text{.} \end{equation*}

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\(\quad y = c_1 e^{2x} + c_2 e^{-x}\)
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Incorrect. This form does not account for the multiplicity of the roots.
\(\quad y = c_1 e^{2x} + c_2 x e^{2x} + c_3 e^{-x} + c_4 x e^{-x}\)
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Incorrect. The multiplicity should be reflected for each root.
\(\quad y = c_1 e^{2x} + c_2 x e^{2x} + c_3 x^2 e^{2x} + c_4 e^{-x} + c_5 x e^{-x}\)
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Correct! This is the form when there are repeated real roots with appropriate multiplicity.
\(\quad y = c_1 x^2 e^{2x} + c_2 x e^{-x}\)
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Incorrect. Make sure to account for the cubic multiplicity for \(r = 2\text{.}\)

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