# Higher Order Homogenous Differential Equations – Complex Roots of The Characteristic Equation

Recall from the Higher Order Homogenous Differential Equations – Constant Coefficients page that if we have an $n^{\mathrm{th}}$ order linear homogenous differential equation with constant coefficients $a_0, a_1, …, a_n \in \mathbb{R}$, that is, $a_0 \frac{d^{n}y}{dt^{n}} + a_1 \frac{d^{n-1}y}{dt^{n-1}} + … + a_{n-1} \frac{dy}{dt} + a_n y = 0$, then if the roots $r_1$, $r_2$, …, $r_n$ of the characteristic equation $a_0r^n + a_1r^{n-1} + … + a_{n-1}r + a_n = 0$ are all real and distinct, then the general solution to this differential equation is given by:

(1)

We will now look at the case in which some of the roots of the characteristic equation are complex.

Recall that when we were dealing with second order linear homogenous differential equations with constant coefficients, say $a \frac{d^2y}{dt^2} + b \frac{dy}{dt} + cy = 0$ where $a, b, c \in \mathbb{R}$, then if the roots of the characteristic equation $ar^2 + br + c = 0$ where complex numbers, then they were complex conjugates of each other and for $\lambda, \mu \in \mathbb{R}$ we had that $r_1 = \lambda + \mu i$ and $r_2 = \lambda – \mu i$. Also recall that the general solution to this differential equation was $y = Ce^{\lambda t} \cos (\mu t) + De^{\lambda t} \sin (\mu t)$.

Now going back to our $n^{\mathrm{th}}$ order linear homogenous differential equation with constant coefficients, we note that since $a_0, a_1, …, a_n \in \mathbb{R}$ then the characteristic equation $a_0r^n + a_1r^{n-1} + … + a_{n-1}r + a_n = 0$ has real coefficients and so once again, it is a property of polynomials with real coefficients that any complex roots come in pairs, namely, if $\lambda + \mu i$ (for $\lambda, \mu \in \mathbb{R}$) is a complex root of the characteristic equation, then so is $\lambda – \mu i$. Assuming that none of these complex roots are repeated, then we have that $e^{\lambda t} \cos (\mu t)$ and $e^{\lambda t} \sin (\mu t)$ will still be solutions to our differential equation.

Thus if we have $n$ distinct roots $r_1$, $r_2$, …, $r_n$ of our characteristic equation, then let $p_1$, $p_2$, …, $p_l$ be the distinct complex roots of the characteristic equation ($l$ is even), and let $q_1$, $q_2$, …, $q_m$ be distinct real roots of the characteristic equation. Thus $n = l + m$. Now we will have $\frac{l}{2}$ pairs of complex roots, say they’re $\lambda_1 \pm \mu_1 i$, $\lambda_2 \pm \mu_2 i$, …, $\lambda_{l/2} \pm \mu_{l/2} i$. Thus for some constants $C_1, C_2, …, C_{l/m}, D_1, D_2, …, D_{l/m}, K_1, K_2, …, K_m$, the general solution to our differential equation is:

(2)

(3)

## Example 1

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

We first note that the characteristic equation for this differential equation is:

(4)

The roots of this polynomial are $r_1 = i$, $r_2 = -i$, and $r_3 = 2$. For the roots $r_1$ and $r_2$, we have that $\lambda = 0$ and $\mu = 1$. Thus the general solution to our differential equation is:

(5)

## 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:

(6)

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

(7)

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

(8)

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