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Section 9.7 Constant Coefficients, Homogeneous

Form of the equation.

Consider an \(n\)th order linear ODE of the form

\begin{equation} \frac{d^ny}{dx^n} + a_{n-1} \frac{d^{n-1}y}{dx^{n-1}} + ... + a_0 y = 0\label{linearconsthomo}\tag{9.7.1} \end{equation}

where the coefficients \(a_i\) are constant. This very special case of the general \(n\)th order linear ODE, for which all of the \(a_i\)'s are constant, comes up in physics incredibly often. Especially important is the case of small (damped, for \(b\ne 0\)) oscillations, such as for a pendulum, which are described by a second order linear ODE (\(n=2\)), using the independent variable \(t\text{:}\)

\begin{equation} \frac{d^2 \theta}{dt^2} +2b\,\omega_0 \frac{d\theta}{dt} + \omega_0^2 \theta = 0\tag{9.7.2} \end{equation}

It will be important that you recognize this equation whenever it comes up, regardless of the algebraic letters that are use. Look for second order, linear, homogeneous, and CONSTANT coefficients. Reread this section 9.1of the book if you need to review the meaning of any of these words.

Method.

Often, when we are solving differential equations, we make an Ansatz, i.e. an initial guess for the basic form of the solution with some unknown parameters or functions. We then plug the guess into the differential equation to get a simpler equation for the unknown parameters. The differential equation eqn. (9.7.1) can be solved by making the Ansatz

\begin{equation} y = e^{k x}\label{kalpha}\tag{9.7.3} \end{equation}

where \(k\) is an unknown constant. We will find the value of the unknown \(k\) by substituting this expression back into the original equation. This Ansatz is a reasonable guess because the derivative of \(e^{kx}\) is just a multiple of itself and higher order derivatives

\begin{equation} \frac{d^me^{k x}}{dx^m} = k^m e^{k x}\tag{9.7.4} \end{equation}

are also multiples of the original function. Each term of equation (9.7.1) after substitution will have a factor of \(e^{kx}\text{,}\) i.e. we obtain

\begin{equation} \left( k^n + a_{n-1} k^{n-1} + ... a_0 \right) e^{k x} = 0\tag{9.7.5} \end{equation}

After we factor out \(e^{kx}\text{,}\) the differential equation has been reduced to the characteristic polynomial equation

\begin{equation} k^n + a_{n-1} k^{n-1} + ... a_0 = 0\label{characteristic}\tag{9.7.6} \end{equation}

for the unknown parameter \(k\text{.}\) This is a huge simplification! The ODE (9.7.1) has become a polynomial equation (9.7.6), which is much easier to solve. Notice that the order of the original ODE, namely \(n\text{,}\) is also the order of the characteristic polynomial. Over the complex numbers, \(n\)th order polynomials have exactly \(n\) roots which may be degenerate (i.e there may be repeated roots). If there is no degeneracy, then we obtain \(n\) linearly-independent solutions of the \(n\)th-order differential equation by plugging in each of the different values of \(k\) into the Ansatz. If there is degeneracy, we need to do something a little trickier to get \(n\) linearly independent solutions. See Section 9.7.2, below.

There is a theorem, which we quote without proof, that says that the trick below always works and we can obtain \(n\) solutions in both the non-degenerate and degenerate cases.

Activity 9.7.1. Generic Example.

Find the general solution of:

\begin{equation} \frac{d^2y}{dx^2} + \frac{dy}{dx} - 6 y = 0\text{.}\label{ExH1}\tag{9.7.7} \end{equation}
Answer
\begin{equation} y = C_1 e^{-3x} + C_2 e^{2x}\tag{9.7.8} \end{equation}
Solution

Substituting (9.7.3) into (9.7.7) leads to

\begin{equation} 0 = k^2 + k - 6 = (k+3)(k-2)\tag{9.7.9} \end{equation}

so that \(k=-3\) or \(k=2\text{.}\) You should check that both \(e^{-3x}\) and \(e^{2x}\) do indeed satisfy (9.7.7); the general solution of this homogeneous ODE is given by

\begin{equation} y = C_1 e^{-3x} + C_2 e^{2x}\tag{9.7.10} \end{equation}

for arbitrary constants \(C_i\text{.}\)

Activity 9.7.2. Special Case: Repeated Roots.

Find the general solution of:

\begin{equation} \frac{d^2y}{dx^2} - 2\frac{dy}{dx} + y = 0 ?\label{ExH2}\tag{9.7.11} \end{equation}
Answer
\begin{equation} y=C_1 e^x + C_2 x e^x\tag{9.7.12} \end{equation}
Solution

In this case, the characteristic polynomial is

\begin{equation} 0 = k^2 - 2k + 1 = (k-1)^2\tag{9.7.13} \end{equation}

and indeed \(e^x\) does satisfy (9.7.11). But wait a minute—that's only one solution! This is a second order equation, so the theorem above guarantees a second solution. How do we find it?

There's a trick in the case of repeated roots (degeneracy), which we will state without proof. The extra solutions are of the form \(x^m e^{\alpha x}\text{.}\) In this case, the second solution is \(x\,e^{\alpha x}\text{;}\) in general, the other linearly independent solutions include terms with \(m=0,1,...,r-1\) for a root of multiplicity \(r\text{.}\)

For example (9.7.11), the general solution is:

\begin{equation} y=C_1 e^x + C_2 x e^x\tag{9.7.14} \end{equation}
Activity 9.7.3. Special Case: Complex Roots.
Find the general solution of:
\begin{equation} \frac{d^2y}{dx^2} + 4y = 0\label{ExH3}\tag{9.7.15} \end{equation}
Answer

We can write the general solution of (9.7.15) in two ways:

\begin{align} y\amp =C_1 e^{2ix} + C_2 e^{-2ix}\tag{9.7.16}\\ \amp =D_1 \cos{2x} + D_2 \sin{2x}\tag{9.7.17} \end{align}
Solution

Even if the coefficients of the ODE are real, some of the roots of the characteristic polynomial could be complex. In this case, the independent solutions are \(y=e^{\pm2ix}\text{.}\) Since such complex solutions always come in pairs if the coefficients are real (Why?), we can use Euler's formula to replace the complex exponentials by trig functions, which in this case are

\begin{align} \cos(2x) \amp = \frac12 (e^{2ix}+e^{-2ix}) ,\tag{9.7.18}\\ \sin(2x) \amp = \frac1{2i} (e^{2ix}-e^{-2ix})\text{.}\tag{9.7.19} \end{align}

We can write the general solution of (9.7.15) in two ways:

\begin{align} y\amp =C_1 e^{2ix} + C_2 e^{-2ix}\notag\\ \amp =C_1 \left(\cos{2x} + i\sin{2x}\right) + C_2 \left(\cos{2x} -i\sin{2x}\right)\notag\\ \amp =D_1 \cos{2x} + D_2 \sin{2x}\label{ExH3Sol}\tag{9.7.20} \end{align}

There is, of course, a simple relationship between the pairs of coefficients \(C_1\text{,}\) \(C_2\) and \(D_1\text{,}\) \(D_2\text{,}\) but since these coefficients are unknown until we specify the boundary conditions or initial conditions, we rarely need to solve for this relationship. Once you understand the relationship between these two solutions, you can of course always just write down the second solution instead of the first, without going through the algebraic steps in (9.7.20).

In the language of vector spaces (see Chapter 4), the algebraic steps in (9.7.20) are just a change of basis in the two-dimensional vector space of solutions.