Repeated Roots of The Characteristic Equation

Repeated Roots of The Characteristic Equation

Recall that if $a\frac{d^2y}{dt^2} + b \frac{dy}{dt} + c = 0$ is a second order linear homogenous differential equation, then the characteristic equation for this differential equation is $ar^2 + br + c = 0$. We saw that if the roots of this equation, call them $r_1$ and $r_2$ are both real and distinct, then for $C$ and $D$ as constants, $y = Ce^{r_1t} + De^{r_2t}$ gives us solutions to our differential equation. We also saw that if the roots of this equation are complex numbers with $r_1 = \lambda + \mu i$ and $r_2 = \lambda - \mu i$ then for $C$ and $D$ as constants, $y = Ce^{\lambda t} \cos (\mu t) + e^{\lambda t} \sin (\mu t)$ is the general solution to this differential equation.

We will now look at the case in which the roots of the characteristic polynomial are real and not distinct, that is $r_1 = r_2$. Suppose that $r_1 = r_2$. Then we must have that the descriminant of the characteristic equation $ar^2 + br + c = 0$ is zero, that is $b^2 - 4ac = 0$. Thus by applying the quadratic formula, we see that the solutions are:

(1)
\begin{align} \quad r = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} = -\frac{b}{2a} \end{align}

Thus we can see that $y_1 = e^{-bt/2a}$ is a solution to our differential equation. We need two different solutions to possibly form a fundamental set of solutions though. In order to find another solution, for some function $v(t)$ assume that $y = v(t)y_1(t) = v(t)e^{-bt/2a}$ is a solution to our differential equation. We note that the derivatives of this solution are:

(2)
\begin{align} \quad y' = \frac{-b}{2a} e^{-bt/2a} v(t) + e^{-bt/2a} v'(t) \end{align}
(3)
\begin{align} \quad y'' = \frac{b^2}{4a^2} e^{-bt/2a} v(t) - \frac{b}{2a} e^{-bt/2a} v'(t) - \frac{b}{2a} e^{-bt/2a} v'(t) + e^{-bt}{2a} v''(t) \end{align}

Substituting the values of $y$, $y'$, and $y''$ into our differential equation and we have that:

(4)
\begin{align} \quad a \frac{d^2 y}{dt^2} + b \frac{dy}{dt} + c = 0 \\ \quad a \left [ \frac{b^2}{4a^2} e^{-bt/2a} v(t) - \frac{b}{2a} e^{-bt/2a} v'(t) - \frac{b}{2a} e^{-bt/2a} v'(t) + e^{-bt}{2a} v''(t) \right ] + b \left [ \frac{-b}{2a} e^{-bt/2a} v(t) + e^{-bt/2a} v'(t) \right ] + c \left [ v(t)e^{-bt/2a} \right ] = 0 \\ \quad \frac{b^2}{4a} v(t) - \frac{b}{2} v'(t) - \frac{b}{2} v'(t) + av''(t) - \frac{b^2}{2a} v(t) + bv'(t) + cv(t) = 0 \\ \quad av''(t) + (-b + b)v'(t) + \left ( \frac{b^2}{4a} - \frac{b^2}{2a} + c \right ) v(t) = 0 \\ \quad av''(t) + \left ( \frac{b^2}{4a} - \frac{b^2}{2a} + c \right ) v(t) = 0 \\ \quad av''(t) + \left ( \frac{b^2 - 2b^2}{4a} + c \right ) v(t) = 0 \\ \quad av''(t) + \left ( c - \frac{b^2}{4a} \right ) = 0 \end{align}

Note that since $b^2 - 4ac = 0$ we have that $b^2 = 4ac$ and $\frac{b^2}{4a} = c$. Therefore $c - \frac{b^2}{4a} = 0$ and for some constants $C$ and $D$, the equation above becomes simply:

(5)
\begin{align} \quad av''(t) = 0 \Leftrightarrow v''(t) = 0 \Leftrightarrow v'(t) = D \Leftrightarrow v(t) = C + Dt \end{align}

Since $y = v(t) e^{-bt}{2a}$ we have that:

(6)
\begin{align} \quad y = (C + Dt) e^{-bt}{2a} \\ \quad y = Ce^{-bt/2a} + Dte^{-bt/2a} \end{align}

Therefore we have obtained another solution to our differential equation, namely $y_2(t) = te^{-bt/2a}$. If we look at the Wronskian of $y_1(t) = e^{-bt/2a}$ and $y_2(t) = te^{-bt/2a}$ we see that:

(7)
\begin{align} \quad W(y_1, y_2) = \begin{vmatrix} e^{-bt/2a} & te^{-bt/2a}\\ -\frac{b}{2a} e^{-bt/2a} & e^{-bt/2a} - \frac{bt}{2a} e^{-bt/2a} \end{vmatrix} = e^{-bt/a} - \frac{b}{2a} e^{-bt/a} + \frac{b}{2a}te^{-bt/a} = e^{-bt/a} \end{align}

Note that the Wronskian $W(y_1, y_2)$ is never equal to zero, and so $y_1(t) = e^{-bt/2a}$ and $y_2(t) = te^{-bt/2a}$ form a fundamental set of solutions and so the general solution for the differential equation $a \frac{d^2 y}{dt^2} + b \frac{dy}{dt} + cy = 0$ for some constants $C$ and $D$ is given by:

(8)
\begin{align} \quad y = Ce^{-bt/2a} + Dte^{-bt/2a} \end{align}