Complex Roots of The Characteristic Equation

Consider the following second order linear homogenous differential equation $a \frac{d^2 y}{dt^2} + b \frac{dy}{dt} + cy = 0$ where $a$, $b$, and $c$ are constants. Recall that the characteristic equation for this differential equation is the quadratic polynomial $ar^2 + br + c = 0$ and that the roots of this polynomial, $r_1$ and $r_2$ are either both real and distinct, both complex and conjugates of each other, or are the same real root repeated twice.

In the case when $r_1$ and $r_2$ were two distinct roots, we saw that the general solution to the second order linear homogenous differential equation $a \frac{d^2 y}{dt^2} + b \frac{dy}{dt} + cy = 0$ was $y = Ce^{r_1 t} + De^{r_2 t}$ where $C$ and $D$ are constants. We will now look at the case where the roots $r_1$ and $r_2$ are complex numbers.

Suppose that $r_1$ and $r_2$ are both complex number roots of the characteristic equation $ar^2 + br + c = 0$. One important theorem regarding polynomials is that a polynomial with real coefficients has the property that if $r_1$ is a complex root of the polynomial, then the complex conjugate of $r_1$ is also a root of the polynomial. In this case, since $r_1$ is a complex root of the characteristic equation, we must have that $r_2$, the second root, must be the complex conjugate of $r_1$, that is $r_1 = \overline{r_2}$. We thus have that the roots $r_1$ and $r_2$ can both be written as $r_1 = \lambda + \mu i$ and $r_2 = \lambda - \mu i$ where $\lambda, \mu \in \mathbb{R}$.

Therefore we have that two solutions for our differential equation are:

(1)

\begin{align} \quad y_1 = e^{(\lambda + \mu i)t} \quad \quad y_2 = e^{(\lambda - \mu i)t} \end{align}

As of the moment, the solutions given above are not that useful to us, so we will make use of perhaps one of the most famous formulas in mathematics known as Euler's Formula which is:

(2)

\begin{align} \quad e^{it} = \cos t + i \sin t \end{align}

With this in hand, we see that the our solution $y_1 = e^{(\lambda + \mu i)t}$ can be rewritten as:

(3)

\begin{align} \quad y_1 = e^{(\lambda + \mu i )t} = e^{\lambda t} e^{\mu i t} = e^{\lambda t} \left ( \cos (\mu t) + i \sin (\mu t) \right ) \end{align}

Furthermore, noting that $\cos (-\mu t) = \cos (\mu t )$ and $\sin (- \mu t) = - \sin (\mu t)$ we have that our solution $y_2 = e^{(\lambda - \mu i)t}$ can be rewritten as:

(4)

\begin{align} \quad y_2 = e^{(\lambda + \mu i)t} = e^{\lambda t} e^{-\mu i t} = e^{\lambda t} \left ( \cos (-\mu t) + i \sin(-\mu t) \right ) = e^{\lambda t} \left ( \cos( \mu t) - i \sin (\mu t) \right ) \end{align}

Once again, the solutions above are nicer, though they still include the complex number $i$, while our original differential equation has only real coefficients. Fortunately, the following theorem will allow us to write our solutions in terms of real-valued functions.

Theorem 1: If $\frac{d^2 y}{dt^2} + p(t) \frac{dy}{dt} + q(t) y = 0$ is a second order linear homogenous differential equation where $p$ and $q$ are continuous real-valued functions and if $y = u(t) + v(t) i$ is a complex-valued solution to the differential equation above, then $y = u(t)$ and $y = v(t)$ are also solutions to this differential equation.

• Proof: Let $\frac{d^2 y}{dt^2} + p(t) \frac{dy}{dt} + q(t) y = 0$ be a second order linear homogenous differential equation where $p$ and $q$ are continuous, and suppose that $y = u(t) + v(t) i$ is a complex-valued solution. Plugging in $y = u(t) + v(t)i$ into our differential equation and we get that:

(5)

\begin{align} \quad 0 = \frac{d^2 }{dt^2} (u(t) + v(t)i) + p(t) \frac{d}{dt} (u(t) + v(t)i) + q(t) (u(t) + v(t)i) \\ \quad 0 = \left [ \frac{d^2 u}{dt^2} + \frac{d^2 v}{dt^2} i \right ] + \left [ p(t) \frac{du}{dt} + p(t) \frac{dv}{dt} i \right ]+ \left [ q(t)u(t) + q(t)v(t) i \right ] \\ \quad 0 = \left [ \frac{d^2 u}{dt^2} + p(t) \frac{du}{dt} + q(t) u(t) \right ] + \left [ \frac{d^2 v}{dt^2} + p(t) \frac{dv}{dt} + q(t) v(t) \right ] i \end{align}

• Note that the righthand side of the equation above is a complex number for each $t$. However, a complex number $z$ is equal to $0$ if and only if $\mathrm{Re} (z) = 0$ and $\mathrm{Im} (z) = 0$. From above, this implies that both:

(6)

\begin{align} \quad \frac{d^2 u}{dt^2} + p(t) \frac{du}{dt} + q(t) u(t) = 0 \quad \mathrm{and} \quad \frac{d^2 v}{dt^2} + p(t) \frac{dv}{dt} + q(t) v(t) = 0 \end{align}

• Therefore $y = u(t)$ and $y = v(t)$ are both solutions to our second order linear homogenous differential equation $\frac{d^2 y}{dt^2} + p(t) \frac{dy}{dt} + q(t) y = 0$. $\blacksquare$

From Theorem 1 above, we thus see that $y_1 = e^{\lambda t} \cos (\mu t)$ and $y_2 = e^{\lambda t} \sin (\mu t)$ are both solutions to our second order linear homogenous differential equation. Therefore if the roots of the characteristic equation $ar^2 + br + c = 0$, then for $C$ and $D$ as constants, the general solution to our second order linear homogenous differential equation is given by:

(7)

\begin{align} \quad y = Ce^{\lambda t} \cos (\mu t) + De^{\lambda t} \sin (\mu t) \end{align}