Second Order Homogenous Differential Equations

# Second Order Homogenous Differential Equations

We will now begin to look at methods for solving some types of second order differential equations. For the time being, we will investigate second order linear homogenous differential equations. Before we do so though, we should become a little more acquainted with what a general second order differential equation is.

 Definition: A Second Order Differential Equation is a differential equation containing a second derivative and can be written in the form $\frac{d^2 y}{dt^2} = f \left (t, y, \frac{dy}{dt} \right )$.

We will now look at an important type of second order linear differential equation (one in which, but before we do, recall that for $P$, $Q$, $R$, and $G$ as functions of $t$, a second order differential equation is of the form:

(1)
\begin{align} \quad P(t) \frac{d^2 y}{dt^2} + Q(t) \frac{dy}{dt} + R(t) y = G(t) \end{align}

We are now ready to look at what are known as homogenous differential equations.

 Definition: A second order linear differential equation $P(t) \frac{d^2 y}{dt^2} + Q(t) \frac{dy}{dt} + R(t) y = G(t)$ is said to be Homogenous if $G(t) = 0$, that is, $P(t) \frac{d^2 y}{dt^2} + Q(t) \frac{dy}{dt} + R(t) y = 0$.

For example, the differential equation $2t y'' + t y' + (e^t - 2)y = 0$ is a linear homogenous second order differential equation.

## Second Order Linear Homogenous Differential Equations with Constant Coefficients

Now suppose that $P(t) = a$, $Q(t) = b$, and $R(t) = c$ where $a, b, c \in \mathbb{R}$, that is, the functions $P$, $Q$, and $R$ are all constants. Then the differential equation $P(t) \frac{d^2 y}{dt^2} + Q(t) \frac{dy}{dt} + R(t) y = 0$ can be rewritten as:

(2)
\begin{align} \quad a \frac{d^2 y}{dt^2} + b \frac{dy}{dt} + c y = 0 \end{align}

From the differential equation above, we suspect that the term $re^t$ where $r \in \mathbb{R}$ will appear in our general solution since $\frac{d}{dt} (e^{rt}) = re^{rt}$ and $\frac{d^2}{dt^2} (e^{rt}) = r^2 e^{rt}$. If indeed $y = e^{rt}$ is a solution to the differential equation above, then:

(3)
\begin{align} \quad ar^2 e^{rt} + bre^{rt} + ce^{rt} = 0 \\ \quad (ar^2 + br + c) e^{rt} = 0 \end{align}

The equation above implies that either $ar^2 + br + c = 0$ or $e^{rt} = 0$. Note though that $e^{rt} > 0$ for all $t \in \mathbb{R}$ though, and thus only $ar^2 + br + c = 0$. The equation $ar^2 + br + c = 0$ is important and we define it below.

 Definition: The Characteristic Equation for the second order linear homogenous differential equation $a \frac{d^2 y}{dt^2} + b \frac{dy}{dt} + c y = 0$ with constant coefficients $a, b, c \in \mathbb{R}$ is $ar^2 + br + c = 0$.

Some people use the term "Auxiliary Equation" in place of "Characteristic Equation" to mean the same thing.

We should note that the characteristic equation is a quadratic polynomial, and so there are precisely three different possibilities for the forms of the two roots $r_1$ and $r_2$ of this polynomial

• The roots may be both real and distinct
• The roots may be real but not distinct (that is, one of the roots has multiplicity $2$),
• The roots may be complex conjugates.

We note that the characteristic equation cannot have one real root and one complex root since the $a, b, c \in \mathbb{R}$ and a polynomial with real coefficients has the property such that complex roots come in pairs. More precisely, if $z = \lambda + \mu t$ is a complex root of a polynomial with real coefficients, then the complex conjugate $\bar{z} = \lambda - \mu t$ is also a root of this polynomial. This will be important later.