DE-BOOK Higher-Order Equations

Section 6.5 Higher-Order Equations

Solving higher-order linear homogeneous constant coefficient (LHCC) equations is an extension of the methods used for second-order equations. The basic strategy remains the same: solve the characteristic equation, then construct the general solution. However, for higher-order equations, the characteristic equation is a higher-degree polynomial, which can make finding the solutions more challenging.

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Definition 7. LHCC Characteristic Equation.

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The characteristic equation of an

n

-th order LHCC equation

a_{n} y^{(n)} + a_{n - 1} y^{(n - 1)} + \dots + a_{1} y^{'} + a_{0} y = 0

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is the

n

-th degree polynomial

⁷

Click here for an explanation.

a_{n} r^{n} + a_{n - 1} r^{n - 1} + \dots + a_{1} r + a_{0} = 0 .

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The solutions to this polynomial,

r_{1}, r_{2}, \dots r_{n},

correspond to the roots that determine the form of the general solution.

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In general, the characteristic equation will have

n

roots, which may be real or complex and may also include repeated roots. Each root of the characteristic equation leads to a term in the general solution of the LHCC. Examples in the table below illustrate the process of constructing these general solutions based on the different types of roots.

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Table 8. Examples of LHCC General Solutions

Characteristic Equation Solutions	General Solution
$r = 3, - 3, 5.3$ (3rd order)	$c_{1} e^{3 t} + c_{2} e^{- 3 t} + c_{3} e^{5.3 t}$
$r = 6 \pm i \sqrt{7.7}, 0$ (3rd order)	$e^{6 t} (c_{1} \sin (\sqrt{7.7} t) + c_{2} \cos (\sqrt{7.7} t)) + c_{3}$
$r = - 4 (triple), 5.3$ (4th order)	$(c_{1} t^{2} + c_{2} t + c_{3}) e^{- 4 t} + c_{4} e^{5.3 t}$
$r = \pm \frac{i}{2}, 2 \pm i$ (4th order)	$c_{1} \sin (\frac{t}{2}) + c_{2} \cos (\frac{t}{2}) + e^{2 t} (c_{3} \sin t + c_{4} \cos t)$
$r = 0 (double), 3 (5-repeats)$ (7th order)	$c_{1} t + c_{2} + (c_{3} t^{4} + c_{4} t^{3} + c_{5} t^{2} + c_{6} t + c_{7}) e^{3 t}$
$r = \pm i, π (double), 5$ (5th order)	$c_{1} \sin t + c_{2} \cos t + (c_{3} t + c_{4}) e^{π t} + c_{5} e^{5 t}$

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As the table demonstrates, constructing the general solution of an LHCC equation is systematic once you have the roots of the characteristic equation. However, solving higher-order polynomial equations by hand can be difficult. The next example shows a few algebra techniques you can use to handle higher degree equations.

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Example 9.

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Solve the following characteristic equations.

$4 r^{2} - 9 = 0$
$r^{3} + 3 r^{2} - 4 r = 0$
$r^{5} + 10 r^{4} = 0$
$r^{5} - 4 r^{3} = 0$
$r^{4} - 25 = 0$
$r^{4} - 5 r^{2} + 4 = 0$

Solution. Solution

Note: DoS = Difference of Squares.

$4 r^{2} - 9 = 0$ $\leftarrow$ DoS

$(2 r + 3) (2 r - 3) = 0$

$2 r + 3 = 0, 2 r - 3 = 0$

$r_{1} = - \frac{3}{2}, r_{2} = \frac{3}{2}$
$r^{3} + 3 r^{2} - 4 r = 0$ $\leftarrow r$ common

$r (r^{2} + 3 r - 4) = 0$

$r (r + 4) (r - 1) = 0$

$r_{1} = 0, r_{2} = - 4, r_{3} = 1$
$r^{5} + 10 r^{4} = 0$ $\leftarrow r^{4}$ common

$r^{4} (r + 10) = 0$

$r_{1} = 0 (4 repeats), r_{2} = - 10$

Technically, $r^{4} = (r - 0)^{4}$ and represents 4 repeated factors.
$r^{5} - 4 r^{3} = 0$ $\leftarrow$ common

$r^{3} (r^{2} - 4) = 0$ $\leftarrow$ DoS

$r^{3} (r + 2) (r - 2) = 0$

$r_{1} = 0 (3 repeats), r_{2} = - 2, r_{3} = 2$
$r^{4} - 25 = 0$ $\leftarrow$ DoS

$(r^{2} + 5) (r^{2} - 5) = 0$ $\leftarrow$ DoS

$(r^{2} + 5) (r - \sqrt{5}) (r + \sqrt{5}) = 0$

$r^{2} + 5 = 0, r - \sqrt{5} = 0, r + \sqrt{5} = 0$

$r_{1} = - i \sqrt{5}, r_{2} = i \sqrt{5}, r_{3} = - \sqrt{5}, r_{4} = \sqrt{5}$
$\begin{matrix} r^{4} - 5 r^{2} + 4 = 0 . \end{matrix}$

Let $u = r^{2},$ then we can rewrite the equation as:

$\begin{matrix} u^{2} - 5 u + 4 = 0 . \end{matrix}$

Solving for $u,$ we get:

$\begin{matrix} u = 1, u = 4 . \end{matrix}$

Thus, $r^{2} = 1$ gives $r = \pm 1,$ and $r^{2} = 4$ gives $r = \pm 2 .$ The general solution is:

$y = c_{1} e^{x} + c_{2} e^{- x} + c_{3} e^{2 x} + c_{4} e^{- 2 x} .$

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These factoring techniques useful, but they only apply to specific forms of polynomials and it is unlikely that you will be lucky enough to encounter such forms in practice. Therefore, it is reasonable to use technology to aid you with these tyes of problems. Many software tools are available and you are encouraged to use any that you are familiar with.

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Example 10.

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Find the general solution to the LHCC equations

$y^{‴} - 6 y^{″} + 11 y^{'} - 6 y = 0$
$2 y^{(4)} + 9 y^{‴} + y^{″} - 21 y^{'} + 9 y = 0$

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using the Polynomial Factoring Calculator.

Solution. Solution

First, write down the characteristic equation:

$\begin{matrix} r^{3} - 6 r^{2} + 11 r - 6 = 0 . \end{matrix}$

Plugging this into to the factoring tool shows that it factors as $(r - 3) (r - 2) (r - 1) .$ Solving for $r,$ we get

$\begin{matrix} r = 1, r = 2, r = 3 . \end{matrix}$

Therefore, the general solution is:

$y = c_{1} e^{x} + c_{2} e^{2 x} + c_{3} e^{3 x} .$
The characteristic equation is:

$9 r^{5} - 4 r^{4} + r + 10 = 0$

which is challenging to solve by hand. However, typing it into factoring tool above shows that it can be factored as

$2 {(r + 3)}^{2} (r - 1) (r - \frac{1}{2}) = 0,$

which gives the solutions

$r = - 3, r = - 3, r = 1, r = \frac{1}{2} .$

Therefore, the general solution is:

$y = (c_{1} x + c_{2}) e^{- 3 x} + c_{3} e^{x} + c_{4} e^{\frac{1}{2} x} .$

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Besides factoring software, another options is to use numerical solvers. The next example illustrates this.

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Example 11.

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Consider the fifth-order LHCC equation:

y^{(5)} - 4 y^{‴} + 3 y^{″} - y = 0 .

Solution. Solution

The characteristic equation is:

r^{5} - 4 r^{3} + 3 r^{2} - 1 = 0 .

Solving this polynomial equation analytically is challenging, so we use a numerical solver (Wolfram Alpha

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www.wolframalpha.com/input?i=r%5E5-4*r%5E3%2B3*r%5E2-1

). This equation has five roots. Three are real:

r_{1} \approx - 2.29, r_{2} \approx - 0.46, r_{3} \approx 1.52,

and two are complex:

r_{4} \approx 0.61 + 0.5 i, r_{5} \approx 0.61 - 0.5 i .

The general solution, incorporating these roots, is:

\begin{aligned} y = c_{1} e^{- 2.29 x} + c_{2} e^{- 0.46 x} + & c_{3} e^{1.52 x} \\ + & e^{0.61 x} (c_{4} \cos (0.5 x) + c_{5} \sin (0.5 x)) . \end{aligned}

y = c_{1} e^{- 2.29 t} + c_{2} e^{- 0.46 t} + c_{3} e^{1.52 t} + e^{0.61 t} (c_{4} \cos (0.5 t) + c_{5} \sin (0.5 t)) .

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By recognizing the nature of the roots and utilizing appropriate technology for solving polynomial equations, we can tackle higher-order LHCC equations with confidence.

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Reading Questions Check-Point Questions

1. Match the Differential Equation to its Characteristic Equation.

Match each differential equation on the left to its characteristic equation on the right.

$y^{″} + 3 y^{'} + 2 y = 0$
$r^{2} + 3 r + 2 = 0$
$y^{‴} - y^{'} + y = 0$
$r^{3} - r + 1 = 0$
$y^{(4)} - 4 y^{″} + 4 y = 0$
$r^{4} - 4 r^{2} + 4 = 0$
$y^{″} - y = 0$
$r^{2} - 1 = 0$
$y^{″} + y = 0$
$r^{2} + 1 = 0$
$y^{'} + 3 y = 0$
$r + 3 = 0$

2. Give the general solution for a 3rd-order LHCC equation whose...

0, 1, - 8 .

$y = c_{1} + c_{2} e^{x} + c_{3} e^{- 8 x}$
Correct! This is the form when there are three distinct real roots.
$y = (c_{1} + c_{2} x) e^{x} + c_{3} e^{- 8 x}$
Incorrect.
$y = c_{1} x e^{x} + c_{2} e^{- 8 x}$
Incorrect.
$y = c_{1} e^{x} + c_{2} e^{- 8 x}$
Incorrect.

3. Give the general solution for a 4th-order LHCC equation whose...

r = 1 \pm 2 i, - 3 \pm 4 i .

$y = c_{1} e^{x} + c_{2} e^{- x} + c_{3} e^{2 i x} + c_{4} e^{- 2 i x}$
Incorrect.
$y = c_{1} e^{x} (\cos (2 x) + \sin (2 x)) + c_{2} e^{- 3 x} (\cos (4 x) + \sin (4 x))$
Incorrect. Not enough constants.
$y = e^{2 x} (c_{1} \cos (x) + c_{2} \sin (x)) + e^{4 x} (c_{3} \cos (3 x) + c_{4} \sin (3 x))$
Incorrect.
$y = e^{x} (c_{1} \cos (2 x) + c_{2} \sin (2 x)) + e^{- 3 x} (c_{3} \cos (4 x) + c_{4} \sin (4 x))$
Correct! This is the form when the characteristic equation has complex roots.

4. Give the general solution for a 5th-order LHCC equation whose...

= 2, 2, 2, - 1, - 1 .

$y = c_{1} e^{2 x} + c_{2} e^{- x}$
Incorrect. This form does not account for the multiplicity of the roots.
$y = c_{1} e^{2 x} + c_{2} x e^{2 x} + c_{3} e^{- x} + c_{4} x e^{- x}$
Incorrect. The multiplicity should be reflected for each root.
$y = (c_{1} + c_{2} x + c_{3} x^{2}) e^{2 x} + (c_{4} + c_{5} x) e^{- x}$
Correct! This is the form when there are repeated real roots with appropriate multiplicity.
$y = (c_{1} + c_{2} x) e^{2 x} + (c_{3} + c_{4} x) e^{- x}$
Incorrect. Make sure to account for the cubic multiplicity for $r = 2 .$

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$4 r^{2} - 9 = 0$	$\leftarrow$ DoS
$(2 r + 3) (2 r - 3) = 0$
$2 r + 3 = 0, 2 r - 3 = 0$
$r_{1} = - \frac{3}{2}, r_{2} = \frac{3}{2}$

$r^{3} + 3 r^{2} - 4 r = 0$	$\leftarrow r$ common
$r (r^{2} + 3 r - 4) = 0$
$r (r + 4) (r - 1) = 0$
$r_{1} = 0, r_{2} = - 4, r_{3} = 1$

$r^{5} + 10 r^{4} = 0$	$\leftarrow r^{4}$ common
$r^{4} (r + 10) = 0$
$r_{1} = 0 (4 repeats), r_{2} = - 10$

$r^{5} - 4 r^{3} = 0$	$\leftarrow$ common
$r^{3} (r^{2} - 4) = 0$	$\leftarrow$ DoS
$r^{3} (r + 2) (r - 2) = 0$
$r_{1} = 0 (3 repeats), r_{2} = - 2, r_{3} = 2$

$r^{4} - 25 = 0$	$\leftarrow$ DoS
$(r^{2} + 5) (r^{2} - 5) = 0$	$\leftarrow$ DoS
$(r^{2} + 5) (r - \sqrt{5}) (r + \sqrt{5}) = 0$
$r^{2} + 5 = 0, r - \sqrt{5} = 0, r + \sqrt{5} = 0$
$r_{1} = - i \sqrt{5}, r_{2} = i \sqrt{5}, r_{3} = - \sqrt{5}, r_{4} = \sqrt{5}$