DE-BOOK Solving High-Order LHCC Equations

Section Solving High-Order LHCC Equations

As the previous section demonstrated, constructing the general solution of a LHCC equation is fairly straightforward given the roots of the characteristic equation. However, solving high-degree polynomial equations can be quite challenging. While there are strategies to tackle this by hand, sometimes computational tools are necessary to obtain the solutions. Below, we will outline some common approaches

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Subsection Factoring

Some high-degree characteristic polynomials can be factored by hand using classic algebraic techniques. Factoring is the fastest way to solve these problems when it works. We’ll start by reviewing common methods of factoring and apply them to characteristic equations that appear in LHCC differential equations.

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Here are a few algebraic strategies you may find helpful:

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🧮 Factoring Techniques.

\(\textbf{Common Factor}\) 🔗: \begin{gather*} ab + ac + ad = a(b+c+d) \end{gather*}

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\(\textbf{Grouping}\) 🔗: \begin{align*} ab + ac + bd + cd \amp = a(b+c) + d(b+c) \\ \amp = (a+d)(b+c) \end{align*}

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\(\textbf{Difference of Squares}\) 🔗: \begin{equation*} a^2 - b^2 = (a+b)(a-b) \end{equation*}

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\(\textbf{Difference of Cubes}\) 🔗: \begin{equation*} a^3 - b^3 = (a-b)(a^2+ab+b^2) \end{equation*}

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\(\textbf{Sum of Cubes}\) 🔗: \begin{equation*} a^3 + b^3 = (a+b)(a^2-ab+b^2) \end{equation*}

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Let’s now apply these tools to solve higher-order LHCC equations by factoring their characteristic equations.

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🌌 Example 150. Solving LHCC Equations by Factoring the Characteristic Equation.

These examples show how to find the general solution to higher-order LHCC equations using a variety of factoring techniques to find the roots of the characteristic equation.

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1. \(\ 4 y'' - 9 = 0\)

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Solution.

The characteristic equation is:

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\begin{equation*} 4 r^2 - 9 = 0 \end{equation*}

Applying the difference of squares:

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\begin{align*} \amp (2 r + 3)(2 r - 3) = 0 \\ \amp r = -\sfrac{3}{2},\ \sfrac{3}{2} \\ \amp \Rightarrow\quad y = c_1 e^{-\frac{3}{2}x} + c_2 e^{\frac{3}{2}x} \end{align*}

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2. \(\ y^{(5)} - 4 y''' = 0\)

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Solution.

The characteristic equation is:

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\begin{equation*} r^5 - 4 r^3 = 0 \end{equation*}

Factoring out the common term \(r^3\text{,}\) followed by a differences of squares gives:

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\begin{align*} r^3(r^2 - 4) \amp = 0\\ r^3(r + 2)(r - 2) \amp = 0 \end{align*}

\begin{equation*} r = 0\ (\text{triple}),\ -2,\ 2 \end{equation*}

\begin{equation*} \Rightarrow\quad y = c_1 + c_2 x + c_3 x^2 + c_4 e^{-2x} + c_5 e^{2x} \end{equation*}

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3. \(\ y^{(4)} - 25 = 0\)

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Solution.

\begin{equation*} r^4 - 25 = 0 \end{equation*}

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This factors as a difference of squares twice:

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\begin{align*} (r^2 + 5)(r^2 - 5) \amp = 0\\ (r^2 + 5)(r + \sqrt{5})(r - \sqrt{5}) \amp = 0 \end{align*}

\begin{equation*} r = \pm i\sqrt{5},\ -\sqrt{5},\ \sqrt{5} \end{equation*}

\begin{align*} \Rightarrow\quad y = c_1 \cos(\sqrt{5}x) \amp + c_2 \sin(\sqrt{5}x) \\ \amp + c_3 e^{\sqrt{5}x} + c_4 e^{-\sqrt{5}x} \end{align*}

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4. \(\ y^{(4)} - 5y'' + 4 = 0\)

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Solution.

The characteristic equation is:

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\begin{equation*} r^4 - 5r^2 + 4 = 0 \end{equation*}

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Use the substitution, \(u = r^2\text{,}\) to reveal a quadratic in disguise:

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\begin{align*} u^2 - 5u + 4 \amp = 0\\ (u - 4)(u - 1) \amp = 0 \end{align*}

Reverse the substitution, \(u = r^2\text{,}\) back into \(r\text{:}\)

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\begin{align*} (r^2 - 4)(r^2 - 1) \amp = 0\\ (r - 2)(r + 2)(r - 1)(r + 1) \amp = 0 \end{align*}

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

\begin{equation*} \Rightarrow\quad y = c_1 e^{x} + c_2 e^{-x} + c_3 e^{2x} + c_4 e^{-2x} \end{equation*}

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5. \(\ y^{(4)} - 6y''' + 9y'' = 0\)

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Solution.

The characteristic equation is:

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\begin{equation*} r^4 - 6r^3 + 9r^2 = 0 \end{equation*}

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Take out \(r^2\) and use standard factoring:

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\begin{align*} r^4 - 6r^3 + 9r^2 \amp = 0\\ r^2(r^2 - 6r + 9) \amp = 0\\ r^2(r - 3)^2 \amp = 0 \end{align*}

\begin{equation*} r = 0\ (\text{double}),\ 3\ (\text{double}) \end{equation*}

\begin{equation*} \Rightarrow\quad y = c_1 + c_2 x + c_3 e^{3x} + c_4 x e^{3x} \end{equation*}

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6. \(\ y^{(5)} - 4 y''' + 8 y'' + 48 = 0\)

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Solution.

The characteristic equation is:

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\begin{equation*} r^5 - 4 r^3 + 8 r^2 + 48 = 0 \end{equation*}

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We can factor this by grouping and recognizing a sum of cubes:

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\begin{align*} r^5 - 4 r^3 + 8 r^2 + 48 \amp = 0\\ r^3(r^2 - 4) + 8(r^2 - 4) \amp = 0\\ (r^3 + 8)(r^2 - 4) \amp = 0\\ (r + 2)(r^2 - 2r + 4)(r + 2)(r - 2) \amp = 0 \end{align*}

\begin{equation*} r = -2\ (\text{double}),\ 2,\ 1 \pm i\sqrt{3} \end{equation*}

\begin{align*} \Rightarrow\quad y = c_1 e^{-2x} \amp + c_2 x e^{-2x} + c_3 e^{2x} \\ + \amp\ e^{x}(c_4 \cos(\sqrt{3}x) + c_5 \sin(\sqrt{3}x)) \end{align*}

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Subsection Rational Root Theorem

When the characteristic polynomial is hard to factor by inspection, the Rational Root Theorem provides a systematic way to search for rational (fractional) solutions. Once you find a root, you can use polynomial division to reduce the polynomial to a lower degree. From there, you can repeat the process or apply other factoring techniques.

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While this method is reliable, it can be tedious by hand and prone to errors. That’s why technology often comes in handy.

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Subsection Using Technology: Factoring High-Degree Polynomials

Not all characteristic polynomials can be factored easily — and some can’t be factored at all using simple algebra. In these cases, it’s perfectly reasonable to turn to technology.

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Computer algebra systems (CAS), online factoring tools, and symbolic solvers can quickly find the roots of even complicated polynomials. Once you have the roots, you can build the general solution using the framework you learned earlier.

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For example, suppose you want to solve the following LHCC equation:

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\begin{equation*} 2 y^{(4)} + 9 y''' + y'' - 21 y' + 9 y = 0\text{.} \end{equation*}

The first step is to write down the characteristic equation:

\begin{equation*} 2r^4 + 9r^3 + r^2 - 21r + 9 = 0 \end{equation*}

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This is challenging to solve by hand, but we can use the tool below to factor it.

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Type 2*r^4 + 9*r^3 + r^2 - 21*r + 9 into the text box and press enter.

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🎮 Interactive 151. Polynomial Factoring Calculator Directions.

Directions:

Enter a polynomial in the box below.

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Use r for the variable, * for multiplication, and ^ for exponents.

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Press enter to see the factored form and approximate values.

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Note: It only factors polynomials with fractional (rational) roots.

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After entering the polynomial, you should see the factored form:

\begin{equation*} 2\left(r - 1\right)\left(r - \frac12\right)\left(r + 3\right)^2 \end{equation*}

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From here the roots are easily identified as:

\begin{equation*} r = 1,\quad r = -3\ (\text{double root}),\quad r = \frac{1}{2} \end{equation*}

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Therefore, the general solution is:

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\begin{equation*} y = c_1 e^{x} + c_2 e^{-3x} + c_3 x e^{-3x} + c_4 e^{\frac{1}{2}x} \end{equation*}

Checkpoint 152. 📖❓ Solving a High-Order LHCC Equation With a Factoring Tool.

Find the general solution to the LHCC equation

\begin{equation*} y''' - 6y'' + 11y' - 6y = 0 \end{equation*}

by using the factoring tool to find the characteristic roots.

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Characteristic Roots: \(r =\)

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General Solution: \(y(t) =\)

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Use c1, c2, c3 as the arbitrary constants of integration.

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Solution.

First, write down the characteristic equation:

\begin{gather*} r^3 - 6r^2 + 11r - 6 = 0 \text{.} \end{gather*}

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

\begin{gather*} r = 1, \quad r = 2, \quad r = 3 \text{.} \end{gather*}

Therefore, the general solution is:

\begin{equation*} y = c_1 e^{x} + c_2 e^{2x} + c_3 e^{3x} \text{.} \end{equation*}

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Subsection Using Technology: Approximating Roots

Some characteristic equations are too messy to factor, even with software. When this happens, a numerical root finder can provide decimal approximations of the roots, allowing us to write down the structure of the general solution.

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Consider the fifth-order LHCC equation:

\begin{equation*} y^{(5)} - 4y''' + 3y'' - y = 0 \end{equation*}

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The characteristic equation is:

\begin{equation*} r^5 - 4r^3 + 3r^2 - 1 = 0 \end{equation*}

This polynomial has no obvious factoring pattern, and standard techniques fail. Fortunately, numerical approximation tools can help.

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Try entering r^5 - 4*r^3 + 3*r^2 - 1 into the same factoring tool above. You should get the following approximations for three real roots:

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\begin{equation*} r_1 \approx -2.29, \quad r_2 \approx -0.46, \quad r_3 \approx 1.52 \end{equation*}

and the complex pair:

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\begin{equation*} r_4 \approx 0.61 + 0.5i, \quad r_5 \approx 0.61 - 0.5i \end{equation*}

using the standard root-to-solution rules, the general solution is:

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\begin{equation*} y = c_1 e^{-2.29x} + c_2 e^{-0.46x} + c_3 e^{1.52x} + e^{0.61x} \left(c_4 \cos(0.5x) + c_5 \sin(0.5x)\right) \end{equation*}

Checkpoint 153. 📖❓ Decimal Roots in the General Solution.

Suppose you find the characteristic roots are approximately

\begin{equation*} r \approx -2.7,\quad 0.8,\quad 0.8 + 2i,\quad 0.8 - 2i\text{.} \end{equation*}

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Which of the following best describes the general solution?

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\(\quad y = c_1 e^{-2.7x} + c_2 e^{0.8x} + e^{0.8x}(c_3 \cos(2x) + c_4 \sin(2x))\)
Correct! The real root gives a standalone exponential, and the complex pair contributes an oscillatory term with the same \(\alpha\text{.}\)
\(\quad y = c_1 e^{-2.7x} + c_2 e^{0.8x} + c_3 \cos(2x) + c_4 \sin(2x)\)
Incorrect. The sine and cosine must be multiplied by \(e^{0.8x}\text{.}\)
\(\quad y = c_1 e^{-2.7x} + c_2 x e^{0.8x} + c_3 x^2 e^{0.8x}\)
Incorrect. This would be appropriate if \(r = 0.8\) were a triple root.
\(\quad y = c_1 e^{-2.7x} + c_2 e^{0.8x} + c_3 e^{-0.8x} + c_4 e^{2ix}\)
This mixes formats and includes a non-real exponential.

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Regardless of the method you choose to find the roots, the general solution is relatively easy to construct once you have them.

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Subsection 📤 Wrap-Up

🗝️ \(\textbf{Key Takeaways...}\)

If the characteristic polynomial factors easily, use algebraic techniques like grouping, factoring by common terms, or recognizing patterns such as difference of squares or cubes.

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When factoring is hard or impossible by hand, use a factoring tool or numerical solver to find the roots.

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Once the roots are known, constructing the full general solution is mechanical and follows a predictable structure.

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