3. Homogeneous solutions¶

The linear second-order problem \(\opA[x]=f\) has a nice structure, but only a few such problems are straightforward to solve. We will now, and for the rest of the chapter, limit ourselves to the constant-coefficient problem

(3.3)¶\[x'' + bx' + k x = f(t),\]

in which \(b\) and \(k\) are constant, and \(f(t)\) is a forcing function.

Note

In (3.3) the coefficient \(b\) has units \(1/T\), the coefficient \(k\) has units \(1/T^2\), and \(f(t)\) has units \(X/T^2\), where \(X\) and \(T\) are the units of \(x\) and \(t\), respectively.

We start with solutions of the homogeneous equation

\[ x'' + b x' + k x = 0. \]

We will use \(x_h\) to name the general solution, as a reminder that the ODE is homogeneous.

All the solutions to first-order linear ODEs were exponential functions. The same is true for second-order linear problems, though with a significant caveat.

3.1. Characteristic polynomial¶

The king of ODE solution methods is to guess the answer and then show that it works. In that tradition, we try

\[ x_h(t) = e^{\lambda t} \]

for a to-be-determined value of \(\lambda\). We substitute it into the ODE and get

\[\begin{align*} 0 &= \lambda^2 e^{\lambda t} + b \bigl(\lambda e^{\lambda t}\bigr) + k \bigl(e^{\lambda t}\bigr)\\ &= e^{\lambda t} \bigl(\lambda^2 + b \lambda + k \bigr). \end{align*}\]

We therefore know that \(x_h\) is a homogeneous solution provided that \(\lambda^2 + b \lambda + k =0\).

Definition 3.5 (Characteristic polynomial of a linear ODE)

The characteristic polynomial of the linear operator \(\opA[x]=x''+bx'+kx\) is

\[ p(s) = s^2 + b s + k. \]

Its roots are called the characteristic values of \(\opA\). For brevity, we will often refer to these as simply the roots of \(\opA\), although this usage is not standard.

Note

We earlier used the term characteristic polynomial with matrices. The connection goes way beyond the name, as we will see in the future.

Theorem 3.6 (Homogeneous solution, distinct roots)

Let \(\lambda_1,\lambda_2\) be the characteristic values of \(\opA\), where \(\opA[x]=x''+bx'+kx.\) If \(\lambda_1\neq \lambda_2\), then the general solution of \(\opA[x]= 0\) is

(3.4)¶\[x_h(t) = c_1 e^{\lambda_1 t} + c_2 e^{\lambda_2 t}.\]

By analogy with linear algebra, we may say that \(e^{\lambda_1 t}\) and \(e^{\lambda_2 t}\) form a basis for the homogeneous solution, as all such solutions are a linear combination of them and neither can be omitted.

We have to handle the case \(\lambda_1=\lambda_2\) separately. (You might recall that a double eigenvalue was a hassle for matrices as well.) Without fuss, we jump to the result, which is easy to derive via reduction of order.

Theorem 3.7 (Homogeneous solution, repeated root)

Let \(\lambda\) be a double root of \(\opA\), where \(\opA[x]=x''+bx'+kx.\) The general solution of \(\opA[x]= 0\) is

(3.5)¶\[x_h(t) = e^{\lambda t} \bigl( c_1 t + c_2 \bigr).\]

Example

We could think of the trivial problem \(x''=0\) as being linear with \(b=k=0\). The characteristic polynomial is \(s^2\), which makes zero a double root. So the general solution is

\[ x_h(t) = e^{0t} (c_1 t + c_2) = c_1 t + c_2, \]

which is obvious in retrospect.

Example

Find the general solution of \(x''-x'-2x=0\).

Example

Solve the IVP \(x'' - 5 x = 0\), \(x(0)=6\), \(x'(0)=0\).

3.2. Complex characteristic values¶

All seems well, until you remember that the roots of a real quadratic polynomial may come as a complex conjugate pair. What then are we to make of (3.4) or (3.5)? That’s what makes second-order problems really different from first-order ones, and we tackle it next.

Notes on linear algebra and ODEs

3. Homogeneous solutions¶

3.1. Characteristic polynomial¶

3.2. Complex characteristic values¶