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.

Definition7.LHCC Characteristic Equation.

The characteristic equation of an \(n\)-th order LHCC equation

The solutions to this polynomial, \(r_1, r_2, \dots r_n\text{,}\) correspond to the roots that determine the form of the general solution.

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.

\(r = \pm i, \pi \text{ (double)}, 5\) (5th order)

\begin{equation*}
c_1\sin t + c_2\cos t + (c_3t + c_4)e^{\pi t} + c_5e^{5t}
\end{equation*}

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.

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.

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

Reading QuestionsCheck-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.

\(y'' + 3y' + 2y = 0 \)

\(r^2 + 3r + 2 = 0 \)

\(y''' - y' + y = 0 \)

\(r^3 - r + 1 = 0 \)

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

\(r^4 - 4r^2 + 4 = 0 \)

\(y'' - y = 0 \)

\(r^2 - 1 = 0 \)

\(y'' + y = 0 \)

\(r^2 + 1 = 0 \)

\(y' + 3y = 0 \)

\(r + 3 = 0 \)

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

Give the general solution for a 3rd-order LHCC equation whose characteristic equation has the solution: \(0, 1, -8\text{.}\)

\(y = c_1 + c_2 e^{x} + c_3 e^{-8x}\)

Correct! This is the form when there are three distinct real roots.

\(y = (c_1 + c_2 x) e^{x} + c_3 e^{-8x}\)

Incorrect.

\(y = c_1 x e^{x} + c_2 e^{-8x}\)

Incorrect.

\(y = c_1 e^{x} + c_2 e^{-8x}\)

Incorrect.

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

Give the general solution for a 4th-order LHCC equation whose characteristic equation has the solutions: \(r = 1 \pm 2i, -3 \pm 4i\text{.}\)