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. $\blacksquare$