Real and Distinct Roots of The Characteristic Equation

# Real and Distinct Roots of The Characteristic Equation

Recall from the Second Order Homogenous Differential Equations page that if have the second order homogenous linear differential equation $a \frac{d^2}{dt^2} + b \frac{dy}{dt} + c y = 0$ where $a, b, c \in \mathbb{R}$, then the characteristic equation for this differential equation is $ar^2 + br + c = 0$.

We will first look at the case where the roots of the characteristic equation, call them $r_1$ and $r_2$ are distinct and real. In that case, we have that $y = e^{r_1t}$ and $y = e^{r_2t}$ are both solutions to $a \frac{d^2 y}{dt^2} + b \frac{dy}{dt} + c y = 0$.

The following proposition tells us that any linear combination of the solutions $y = e^{r_1t}$ and $y = e^{r_2t}$ will also be a solution to our differential equation.

 Proposition 1: Let $a \frac{d^2 y}{dt^2} + b \frac{dy}{dt} + c y = 0$ be a second order linear homogenous differential equation and let $ar^2 + br + c = 0$ be the characteristic equation of this differential equation whose roots $r_1$ and $r_2$ are real and distinct. Then for $C, D \in \mathbb{R}$ we have that $y = Ce^{r_1t} + De^{r_2t}$ is a solution to this differential equation.
• Proof: Let $y = Ce^{r_1t} + De^{r_2t}$. Then $\frac{dy}{dt} = Cr_1e^{r_1t} + Dr_2e^{r_2t}$ and $\frac{d^2 y}{dt^2} = Cr_1^2e^{r_1t} + Dr_2^2e^{r_2t}$. Plugging this into the differential equation $a \frac{d^2 y}{dt^2} + b \frac{dy}{dt} + c y = 0$ and we have that:
(1)
\begin{align} \quad a [Cr_1^2e^{r_1t} + Dr_2^2e^{r_2t}] + b[Cr_1e^{r_1t} + Dr_2e^{r_2t}] + c[Ce^{r_1t} + De^{r_2t} ] = Ce^{r_1t}\underbrace{[ar_1^2 + br_1 + c]}_{=0} + De^{r_2t}\underbrace{[ar_2^2 + br_2 + c]}_{=0} = 0 \end{align}
• Thus $y = Ce^{r_1t} + De^{r_2t}$ is a solution to our differential equation. $\blacksquare$

The next question that we might ask ourselves is whether or not all initial value problems have a solution. Fortunately, if the roots of the characteristic equation are distinct real numbers, then the answer is simple - yes, as verified with the following proposition.

 Proposition 2: Let $a \frac{d^2 y}{dt^2} + b \frac{dy}{dt} + c y = 0$ be a second order linear homogenous differential equation and let $ar^2 + br + c = 0$ be the characteristic equation of this differential equation whose roots $r_1$ and $r_2$ are real and distinct. Then for any choice of initial conditions $y(t_0) = y_0$ and $y'(t_0) = y'_0$, a solution exists.
• Proof: The general solution to this differential equation is $y = Ce^{r_1t} + De^{r_2t}$ where $C, D \in \mathbb{R}$. Applying the initial condition $y(t_0) = y_0$ to our solution and we get that:
(2)
\begin{align} \quad y_0 = Ce^{r_1t_0} + De^{r_2t_0} \: (*) \end{align}
• Now the derivative of our solution is $y' = Cr_1e^{r_1t} + Dr_2e^{r_2t}$. Applying the second initial condition $y'(t_0) = y'_0$ to the derivative of our solution and we get that:
(3)
\begin{align} \quad y'_0 = Cr_1e^{r_1t_0} + Dr_2e^{r_2t_0} \: (**) \end{align}
• If we solve both $(*)$ and $(**)$ for $C$ and $D$, we get that:
(4)
\begin{align} \quad C = \frac{y'_0 - y_0r_2}{r_1 - r_2} e^{-r_1t_0} \end{align}
(5)
\begin{align} \quad D = \frac{y_0r_1 - y'_0}{r_1 - r_2} e^{-r_2t_0} \end{align}
• Since $r_1$ and $r_2$ are distinct real roots, we have that $r_1 \neq r_2$ and so constants $C$ and $D$ exist for the initial value problem given above when the characteristic equation $ar^2 + br +c = 0$ has distinct real roots. $\blacksquare$