Math 555: Differential Equations

3.3 Complex Roots of the
Characteristic Equation

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3.3.1 Complex Conjugate Roots

In this section, we continue our study of the differential equation

$$ ay'' + by' + cy = 0, $$
where $a$, $b$, and $c$ are real-valued constants. During Section 3.1 , we discovered that solutions for this equation of the form $y = e^{rt}$ could be found for values of $r$ satisfying the characteristic equation

$$ ar^2 + br + c = 0. $$
We showed that if the roots $r_1$ and $r_2$ are real and distinct (which happens when the discriminant $b^2 - 4ac\gt 0$), then the general solution of the ODE has the form

$$ y(t) = c_1 e^{r_1 t} + c_2 e^{r_2 t}. $$
Suppose now that the discriminant $b^2 - 4ac\lt 0$, then the roots of the characteristic equation are complex-valued

$$ r_1 = \lambda + i\mu,\qquad r_2 = \lambda - i\mu, $$
for real-valued constants $\lambda$ and $\mu$ (the real part and imaginary part of the complex number, respectively) and the imaginary unit $i = \sqrt{-1}$.

If these roots are used in the form $y = e^{rt}$, then our solution candidates for the ODE look like

$$ y_1(t) = e^{(\lambda + i\mu)t},\qquad y_2(t) = e^{(\lambda - i\mu)t}, $$
or
$$ y_1(t) = \exp\Big((\lambda + i\mu)t\Big),\qquad y_2(t) = \exp\Big((\lambda - i\mu)t\Big), $$
which is the exponential function $e^t$ being raised to a complex-valued power. What does it mean to take the exponential function of a complex power? We need to understand this before moving forward.

3.3.2 Euler's Formula

Euler's formula (yep, same guy as before) is an important relationship between two sets of elementary functions that involves complex numbers. Our goal here is to demonstrate that relationship, while giving meaning to the complex exponential function. We will do the following:

1. Establish a definition of the complex exponential function
2. Ensure that definition is compatible with the real-valued exponential function
3. Use that definition to establish Euler's formula

To begin, let's consider the Taylor series expansion of $e^t$.

$$ e^t = 1 + t + \dfrac{t^2}{2!} + \ldots + \dfrac{t^n}{n!} + \dots = \sum_{n=0}^\infty \dfrac{t^n}{n!}, $$
which converges for all $t\in(-\infty,\infty)$. Now, let's replace $t$ with $it$ and examine the series

$$ e^{it} = \sum_{n=0}^\infty \dfrac{(it)^n}{n!} $$
To simplify this expression, we write $(it)^n$ as $i^n t^n$ and use the identities $i^2 = -1$, $i^3 = -i$, and $i^4 = 1$, etc., noting that for even powers $n = 2k$ of $i$, we have that $i^{2k} = (-1)^k$ and for odd powers $n=2k+1$, $i^{2k+1} = i(-1)^k$. This indicates that we should separate the series into its even and odd terms:

$$ e^{it} = \sum_{k=0}^\infty \dfrac{(-1)^k t^{2k}}{(2k)!} + i\sum_{k=0}^\infty \dfrac{(-1)^k t^{2k+1}}{(2k+1)!}. $$
These are the series for $\cos(t)$ and $\sin(t)$, which leads to the famous Euler's formula

$$ e^{it} = \cos(t) + i\sin(t). $$
We will use this equation as the definition of $e^{it}$, so complex exponentials will always be interpreted using this formula.

In addition, it is useful to establish some basic variations of Euler's formula, like

$$ e^{-it} = \cos(t) - i\sin(t) $$
since $\cos(-t) = \cos(t)$ and $\sin(-t) = -\sin(t)$ and

$$ e^{i\mu t} = \cos(\mu t) + i\sin(\mu t), $$
which follows from the work with Taylor series above.

Now we are able to write the expression

$$ \begin{aligned} e^{(\lambda + i\mu)t} &= e^{\lambda t}e^{i\mu t} \\ \\ &= e^{\lambda t}\Big(\cos(\mu t) + i\sin(\mu t)\Big) \\ \\ &= e^{\lambda t}\cos(\mu t) + ie^{\lambda t}\sin(\mu t), \end{aligned} $$
which we will take as the definition of $e^{(\lambda + i\mu)t}$.

Under this definition, the derivative of $e^{rt}$ works just like it does for real values

$$ \dfrac{d}{dt}e^{rt} = re^{rt}. $$

3.3.3 Linear Independence of Solutions

We have now developed a formula for $e^{rt}$ for complex values of $r$, and we know that solutions of a quadratic equation with real coefficients will always produce complex roots in conjugate pairs $r = \lambda \pm i\mu$, and hence solutions to our ODE of the form

$$ y_{1,2}(t) = e^{(\lambda \pm i\mu)t} = e^{\lambda t}\cos(\mu t) \pm ie^{\lambda t}\sin(\mu t). $$
Taking a linear combination of these solutions yields

$$ \begin{aligned} y(t) &= c_1 y_1 (t) + c_2 y_2 (t) \\ \\ &= (c_1+c_2) e^{\lambda t}\cos(\mu t) + i(c_1 - c_2) e^{\lambda t}\sin(\mu t). \end{aligned} $$
This can be thought of as just a linear combination of $ e^{\lambda t}\cos(\mu t) $ and $ ie^{\lambda t}\sin(\mu t) $, so we will use these instead to keep things simpler. In the previous section, the Wronskian was established as a way to test for linear independence. Computing the Wronskian of this pair of solutions,

$$ \begin{aligned} W &\left[ e^{\lambda t}\cos(\mu t), ie^{\lambda t}\sin(\mu t)\right](t) = \begin{vmatrix} e^{\lambda t}\cos(\mu t) & ie^{\lambda t}\sin(\mu t) \\ \\ \lambda e^{\lambda t}\cos(\mu t) - \mu e^{\lambda t}\sin(\mu t) & i\lambda e^{\lambda t}\sin(\mu t) + i\mu e^{\lambda t}\cos(\mu t) \end{vmatrix} \\ \\ &= ie^{\lambda t}\cos(\mu t)\left(\lambda e^{\lambda t}\sin(\mu t) + \mu e^{\lambda t}\cos(\mu t) \right) - i \mu e^{\lambda t}\sin(\mu t) \left(\lambda e^{\lambda t}\cos(\mu t) - \mu e^{\lambda t}\sin(\mu t) \right) \\ \\ &= i e^{\lambda t} \left(\mu\cos^2(\mu t) + \lambda\cos(\mu t)\sin(\mu t) - \lambda\sin(\mu t)\cos(\mu t) + \mu\sin^2(\mu t) \right) \\ \\ &= i \mu e^{\lambda t}, \end{aligned} $$
which is zero only when $\mu = 0$, and hence our solutions are linearly independent. Furthermore, the $i$ being multiplied by $\sin(\mu t)$ in Euler's formula is just a constant, so we typically omit writing it and express the general solution of the homogeneous equation with constant coefficients in the case of complex conjugate roots as


$$ y(t) = c_1 e^{\lambda t}\cos(\mu t) + c_2 e^{\lambda t}\sin(\mu t). $$
In general, it is possible for the constants $c_1$ and $c_2$ to be complex-valued, but since our original ODE has real coefficients $a$, $b$, and $c$ these constants will always end up being real.

If $\mu$ is zero, then the roots of the characteristic equation are real (and repeated), which we will handle next section .

3.3.4 Examples

Example 3.3.1

Consider the initial value problem

$$ y'' + y' + 6.5y = 0,\qquad y(0) = 3,\ y'(0) = -2. $$
The characteristic equation is

$$ r^2 + r + 6.5 = 0, $$
whose roots are

$$ r_1 = -\dfrac{1}{2} + \dfrac{5}{2}i,\qquad r_2 = -\dfrac{1}{2} - \dfrac{5}{2}i. $$
And so the general solution of the ODE is given by

$$ y(t) = c_1 e^{-\frac{t}{2}}\cos\left(\frac{5}{2}t\right) + c_2 e^{-\frac{t}{2}}\sin\left(\frac{5}{2}t\right).$$
Now we may apply the initial conditions. Direct evaluation is sufficient here (instead of Cramer's rule), since the first equation simplifies to

$$ y(0) = 3 = c_1, $$
and
$$ y'(t) = -\frac{1}{2} c_1 e^{-\frac{t}{2}}\cos\left(\frac{5}{2}t\right) - \frac{5}{2}c_1 e^{-\frac{t}{2}}\sin\left(\frac{5}{2}t\right) - \frac{1}{2} c_2 e^{-\frac{t}{2}}\sin\left(\frac{5}{2}t\right) + \frac{5}{2} c_2 e^{-\frac{t}{2}}\cos\left(\frac{5}{2}t\right) $$
implies that

$$ y'(0) = -2 = -\frac{1}{2}c_1 + \frac{5}{2}c_2, $$
so $c_2 = -\frac{1}{5}$. Hence the solution to the initial value problem is

$$ y(t) = 3 e^{-\frac{t}{2}}\cos\left(\frac{5}{2}t\right) - \frac{1}{5} e^{-\frac{t}{2}}\sin\left(\frac{5}{2}t\right). $$

Notice that as $t$ increases, the oscillations shrink in size, with the solution tending asymptotically towards $y = 0$ as $t\rightarrow\infty.$

Example 3.3.2

Solve the initial value problem

$$ 25y'' - 10y' + 26y = 0,\qquad y(0) = -2,\ y'(0) = 2 $$
The characteristic equation for this ODE is

$$ 25r^2 - 10r + 26 = 0 $$
with roots

$$ r = \dfrac{-(-10) \pm \sqrt{(-10)^2 - 4(25)(26)}}{2(25)} = \frac{1}{5} \pm 1i. $$
The general solution is

$$ y(t) = c_1 e^{\frac{t}{5}}\cos(t) + c_2 e^{\frac{t}{5}}\sin(t). $$
To solve the initial value problem, we take the derivative of this function

$$ y'(t) = \frac{1}{5}c_1 e^{\frac{t}{5}}\cos(t) - \frac{1}{5}c_1 e^{\frac{t}{5}}\sin(t) + c_2 e^{\frac{t}{5}}\sin(t) + c_2 e^{\frac{t}{5}}\cos(t) $$
and evaluate these at the initial point, which yields the matrix equation

$$ \begin{bmatrix} 1 & 0 \\ \frac{1}{5} & 1 \end{bmatrix} \begin{bmatrix} c_1 \\ c_2 \end{bmatrix} = \begin{bmatrix} -2 \\ 2 \end{bmatrix}. $$
An application of Cramer's rule gives the coefficients $c_1$ and $c_2$,

$$ c_1 = \dfrac{\begin{vmatrix} \color{#307fe2}{-2} & 0 \\ \color{#307fe2}{2} & 1 \end{vmatrix}}{\begin{vmatrix} 1 & 0 \\ \frac{1}{5} & 1 \end{vmatrix}} = \dfrac{-2}{1}=-2, \qquad c_2 = \dfrac{\begin{vmatrix} 1 & \color{#307fe2}{-2} \\ \frac{1}{5} & \color{#307fe2}{2} \end{vmatrix}}{\begin{vmatrix} 1 & 0 \\ \frac{1}{5} & 1 \end{vmatrix}} = \dfrac{\frac{12}{5}}{1}=\frac{12}{5}. $$

The solution to the initial value problem is

$$ y(t) = -2 e^{\frac{t}{5}}\cos(t) + \frac{15}{2} e^{\frac{t}{5}}\sin(t), $$
which unlike the previous example does not vanish as $t$ increases.

We see that the oscillations grow with $t$. This indicates that the solution has fundamentally different properties than the previous example, something that will be explored more deeply later this chapter.

Example 3.3.3

Solve the initial value problem

$$ 9y'' + 4y = 0,\qquad y(0) = -3,\ y'(0) = 1 . $$

Check Your Work

$$ y(t) = -3 \cos\left(\frac{2}{3}t\right) + \frac{3}{2} \sin\left(\frac{2}{3}t\right) $$

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