Complex Roots of The Characteristic Equation Examples 1

Recall from the Complex Roots of The Characteristic Equation page that if we have a second order linear homogenous differential equation with constant coefficients, $a \frac{d^2y}{dt^2} + b \frac{dy}{dt} + c y = 0$ where $a, b, c \in \mathbb{R}$, then if the roots $r_1$ and $r_2$ of the characteristic polynomial $ar^2 + br + c = 0$ are complex conjugates, then $r_1 = \lambda + \mu i$ and $r_2 = \lambda - \mu i$ for some $\lambda, \mu \in \mathbb{R}$ and the general solution to this differential equation for $C$ and $D$ as constants is given as:

(1)

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

We will now look at some examples of finding the general solutions to a differential equation of this type.

Example 1

Find the general solution to the differential equation $\frac{d^2y}{dt^2} + 6 \frac{dy}{dt} 13y = 0$. What is the behaviour of the general solutions at $t \to \infty$?

The characteristic equation for this differential equation is $r^2 + 6r + 13 = 0$. Applying the quadratic formula (or by completing the square) we can find the roots to this characteristic equation.

(2)

\begin{align} \quad r = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} = \frac{-6 \pm \sqrt{(-6)^2 - 4(1)(13)}}{2(1)} = \frac{-6 \sqrt{36 - 52}}{2} = -3 \pm \frac{\sqrt{-16}}{2} = -3 \pm 2i \end{align}

Therefore the roots to the characteristic equation are $r_1 = -3 + 2i$ and $r_1 = -3 - 2i$. Thus $\lambda = -3$ and $\mu = 2$ and so the general solution to our differential equation is:

(3)

\begin{align} \quad y = Ce^{-3t} \cos (2t) + De^{-3t} \sin(2t) \end{align}

Note that $\cos (2t)$ and $\sin (2t)$ are both bounded functions. More precisely, $-1 ≤ \cos (2t) ≤ 1$ and $-1 ≤ \sin (2t) ≤ 1$ for all $t \in \mathbb{R}$. Furthermore, note that $\lim_{t \to \infty} e^{-3t} = 0$. It's not hard to see that as $t \to \infty$, our solutions approach $0$, that is $\lim_{t \to \infty} \left ( Ce^{-3t} \cos (2t) + De^{-3t} \sin(2t) \right ) = 0$.

The image below is a graph of some solutions to our differential equation for which $C$ and $D$ are equal and positive.

Example 2

Find the general solution to the differential equation $\frac{d^2y}{dt^2} + \frac{dy}{dt} + \frac{5}{4}y = 0$. What is the behaviour of the general solutions at $t \to \infty$?

The characteristic equation to this differential equation is $r^2 + r + \frac{5}{4} = 0$. Applying the quadratic formula and we have that:

(4)

\begin{align} \quad r = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} = \frac{-1 \pm \sqrt{(-1)^2 - 4(1)\left ( \frac{5}{4} \right )}}{2(1)} = \frac{-1 \pm \sqrt{-4}}{2} = \frac{-1 \pm 2i}{2} = -\frac{1}{2} \pm i \end{align}

Therefore the roots to our characteristic equation are $r_1 = -\frac{1}{2} + i$ and $r_2 = -\frac{1}{2} - i$. Thus we see that $\lambda = -\frac{1}{2}$ and $\mu = 1$, and so the general solution to our differential equation is:

(5)

\begin{align} \quad y = Ce^{-t/2} \cos (t) + De^{-t/2} \sin (t) \end{align}

Once again the sine and cosine functions are both bounded between $-1$ and $1$. Once again we note that, $\lim_{t \to \infty} e^{-t/2} = 0$. Therefore, it's not hard to see that $\lim_{t \to \infty} \left ( Ce^{-t/2} \cos (t) + De^{-t/2} \sin (t) \right ) = 0$. The image below is a graph of some solutions to our differential equation for which $C$ and $D$ are equal and positive.