Homogeneous solution

The homogeneous solution (also called the complementary solution) to eq. ¿eq:ioode? is the family of solutions¹ that satisfy eq. ¿eq:ioode? with the forcing function $f(t) = 0$, but is not restricted by the initial conditions.
The equation is

\[ \frac{d^n y}{d t^n} + a_{n-1} \frac{d^{n-1} y}{d t^{n-1}} + \cdots + a_1 \frac{d y}{d t} + a_0 y = 0. \qquad{(1)}\]

What function might solve this, the homogeneous equation?
The natural exponential with base $e$ is a good candidate, since it is its own derivative.
It turns out that a linear combination (weighted sum) of linearly independent exponentials² is the family of solutions we’re looking for.

Characteristic equation and its roots

It can be shown that, for an exponential function $C e^{\lambda t}$ with complex $C$ and $\lambda$, the latter must satisfy

\[ \lambda^n + a_{n-1} \lambda^{n-1} + \cdots + a_1 \lambda + a_0 = 0, \qquad{(2)}\]

called the characteristic equation.
It has $n$ solutions or roots $\lambda_i$.
When these roots are all distinct—meaning none is equal to another—the homogeneous solution $y_h$ to eq. ¿eq:ioode? is

\[ y_h(t) = \sum_{i=1}^n C_i e^{\lambda_i t}. \qquad{(3)}\]

Repeated roots

If a root is not distinct, it is said to have multiplicity $\mu$ equal to the number of its instances.
So a root that appears thrice has multiplicity three.
This multiplicity causes the linear combination of exponentials to be degenerate or linearly dependent.
This is easily remedied, however, by augmenting the solution of eq. 3 with a polynomial term in $t$, as follows.
Let there be $n'$ distinct roots, each with multiplicity $\mu_i$.
Then the solution is:³

\[ y_h(t) = \sum_{i=1}^{n'} \sum_{k=1}^{\mu_i} C_{ik} t^{(k-1)} e^{\lambda_{ik} t}. \qquad{(4)}\]

Note that, as we would expect, when all roots are distinct, the factor $t$ in each term has exponent zero, $t^0 = 1$, and we recover eq. 3.

What have we done?

We have found the homogeneous solution to eq. ¿eq:ioode?, which we know sums with another term to form its general solution.
We should probably consider what this homogeneous solution looks like.
It’s a sum of weighted complex exponential functions of time.
Complex solutions to the characteristic equation, eq. 2, always arise in complex conjugate pairs $\sigma \pm j \omega$, yielding terms like

\[ e^{(\sigma + j \omega) t} + e^{(\sigma - j \omega)t} = e^{\sigma t} \left( e^{j \omega t} + e^{- j \omega t} \right) = 2 e^{\sigma t} \cos(\omega t). \]

This last identity is from a form of Euler’s formula.
The result shows us the possible forms of terms in the homogeneous solution:

For a real root, $\omega=0$ and a real exponential results (note, also, that the $2$ goes away).
For imaginary roots, $\sigma=0$ and a sinusoid results.
For complex roots, a sinusoidal oscillation with an exponential envelope occurs.

Everything we know about the exponential also applies.
For instance, if $\sigma < 0$, an exponential decay results, whereas if $\sigma > 0$, we get an exponential growth.

Example 7.1

A homogeneous solution
Find the homogeneous solution for the equation

$$ \frac{d^5 y}{d t^5} + 14 \frac{d^4 y}{d t^4} + 81 \frac{d^3 y}{d t^3} + 248 \frac{d^2 y}{d t^2} + 408 \frac{d y}{d t} + 288 y = f(t). $$

First, we write the characteristic equation:

λ⁵ + 14λ⁴ + 81λ³ + 248λ² + 408λ + 288 = 0.

Your favorite root finder (e.g. Matlab, Wolfram Alpha, many calculators) will yield, in arbitrary order:

$$ \begin{aligned} \lambda_{11} &= -4, \\ \lambda_{21,22} &= -3, \\ \lambda_{31,41} &= -2 \pm j 2. \end{aligned} $$

So λ₁₁ is a real root, λ₂₁ and λ₂₂ are repeated real roots with multiplicity μ = 2, and λ₃₁ and λ₄₁ are complex conjugate roots.

So the homogeneous solution is, à la [@eq:homogeneous_solution]:

$$ \begin{aligned} y_h(t) &= \sum_{i=1}^{n'} \sum_{j=1}^{\mu_i} C_i t^{(j-1)} e^{\lambda_i t} \\ &= C_{11} e^{\lambda_{11} t} + C_{21} e^{\lambda_{21} t} + C_{22} t e^{\lambda_{22} t} + C_{31} e^{\lambda_{31} t} + C_{41} e^{\lambda_{41} t}. \end{aligned} $$

Now that we’ve applied our equation, we can drop the double-subscripting, if we like:

y_h(t) = C₁e^λ₁t + C₂e^λ₂t + C₃te^λ₃t + C₄e^λ₄t + C₅e^λ₅t,

where we map indices as follows:

$$ \begin{aligned} 1 &\rightarrow 11, \\ 2 &\rightarrow 21, \\ 3 &\rightarrow 21, \\ 4 &\rightarrow 31, \\ 5 &\rightarrow 41. \end{aligned} $$

Once we’re comfortable dealing with multiplicity, we can skip the strict application of [@eq:homogeneous_solution] and use single subscripts.

This family of solutions is actually the general solution of the homogeneous equation, eq. 1. Lest it be confused with the general solution of eq. ¿eq:ioode?, we avoid calling it this in the following.↩︎
The special case of repeated characteristic equation roots requires a slight modification of this statement, as we’ll see. In this case, some of the exponentials pick up extra factors.↩︎
We use a double subscript $ik$, here, meaning the $i$th distinct root and the $k$th copy. Don’t be alarmed: it’s just to make sure there are enough distinct constants. If one prefers, the constants can be numbered $1$ through $n$, but it’s difficult to write that in summation form.↩︎

Online Resources for Section 7.3

No online resources.