Higher Order O.D.E.'s Complex Roots Examples 1

Higher Order O.D.E.'s Complex Roots of The Characteristic Equation Examples 1

Consider the following $n^{\mathrm{th}}$ order linear homogenous differential equation:

(1)

\begin{align} \quad \quad a_0 \frac{d^{n}y}{dt^{n}} + a_1 \frac{d^{n-1}y}{dt^{n-1}} + ... + a_{n-1} \frac{dy}{dt} + a_n y = 0 \end{align}

Recall from the Higher Order Homogenous Differential Equations - Complex Roots of The Characteristic Equation page that sometimes the characteristic equation $a_0r^n + a_1r^{n-1} + ... + a_{n-1}r + a_{n-1}r + a_n = 0$ does not solely have distinct real roots. Instead, there many be some roots of the characteristic equation that are complex.

We note that we can only have an even number of complex roots (since complex roots come in conjugate pairs). Recall that for each pair of complex roots we have that they can be written in the form $\lambda + \mu i$ where $\lambda, \mu \in \mathbb{R}$. If these complex roots are distinct, then we will have that $e^{\lambda} (\cos (\mu t) + \sin (\mu t))$ will be a part of our general solution.

We will now look at some examples of higher order linear homogenous differential equations of this form.

Example 1

Find the general solution to the differential equation $\frac{d^3y}{dt^3} + 2 \frac{d^2y}{dt^2} + x +2$.

The characteristic equation for this differential equation is $r^3 + 2r^2 + r + 2 = 0$. By trial and error, we can immediately see that $r_1 = -2$ is a solution to this differential equation, and by applying long division we have that:

(2)

\begin{align} \quad (r + 2)(r^2 + 1) = 0 \end{align}

Note that the factor $r^2 + 1$ is irreducible as factors of real numbers. If we apply the quadratic formula, we'll see that this factor has two complex roots, namely $r_2 = i$ and $r_3 = -i$. For this pair of complex roots we have that $\lambda = 0$ and $\mu = 1$, and so the general solution to our differential equation is:

(3)

\begin{align} \quad y = C_1e^{-2t} + C_2\cos t + C_3\sin t \end{align}

Example 2

Find the general solution of the differential equation $\frac{d^3y}{dt^3} + 5 \frac{d^2y}{dt^2} + 17 \frac{dy}{dt} +13 = 0$.

The characteristic equation for this differential equation is $r^3 + 5r^2 + 17r + 13 = 0$. By trial and error we see that $r_1 = -1$ is a root to the characteristic equation, and upon factoring this root out by long division we get that:

(4)

\begin{align} \quad (r + 1)(r^2 +4r + 13) = 0 \end{align}

We will now use the quadratic formula on the term $r^2 +4r + 13$:

(5)

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

Therefore we have that $r_2 = -2 + 3i$ and $r_3 = -2 - 3i$ are both roots to our characteristic equation. So $\lambda = -2$ and $\mu = 3$. Thus we have that the general solution to our differential equation is

(6)

\begin{align} \quad y = C_1e^{-t} + C_2 e^{-2t} \cos (3t) + C_3e^{-2t} \sin (3t) \end{align}