# 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)

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)

(3)

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

(4)

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)

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

(6)

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)

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)