Real, Distinct Roots of The Characteristic Equation

# Higher Order Homogenous Differential Equations Real, Distinct Roots of The Characteristic Equation

Consider the following $n^{\mathrm{th}}$ order linear homogenous differential equation with the constant coefficients $a_0, a_1, ..., a_n \in \mathbb{R}$:

(1)
\begin{align} \quad a_0 \frac{d^{n}y}{dt^{n}} + a_1 \frac{d^{n-1}y}{dt^{n-1}} + ... + a_{n-1} \frac{dy}{dt} + a_n y = 0 \end{align}

Recall from the Higher Order Homogenous Differential Equations - Constant Coefficients page that the characteristic equation to this differential equation is:

(2)
\begin{align} \quad a_0r^n + a_1r^{n-1} + ... + a_{n-1}r + a_n = 0 \end{align}

Let $r_1$, $r_2$, …, $r_n$ be the roots to the characteristic equation and suppose that these roots are real and distinct, that is $r_i \neq r_j$ for $i \neq j$, $i, j = 1, 2, ..., n$. Then $y_1(t) = e^{r_1t}$, $y_2(t) = e^{r_2t}$, …, $y_n(t) = e^{r_nt}$ are all solutions to our differential equation. Thus since each root id real and distinct, then the next question to ask is whether or not $y_1$, $y_2$, …, $y_n$ form a fundamental set of solutions. The answer boils down to whether or not $y_1$, $y_2$, …, $y_n$ are a linearly independent set of functions. The following proposition tells us that indeed they are provided that the roots $r_1$, $r_2$, …, $r_n$ are indeed distinct.

 Proposition 1: If $r_1, r_2, ..., r_n \in \mathbb{R}$ are $n$ distinct real numbers, then the set of functions $y_1(t) = e^{r_1t}$, $y_2(t) = e^{r_2t}$, …, $y_n(t) = e^{r_nt}$ are linearly independent on all of $\mathbb{R}$.
• Proof: To show that $y_1$, $y_2$, …, $y_n$ are a linearly independent set of functions on all of $\mathbb{R}$, we must show that the following equation implies that the constants $k_1 = k_2 = ... = k_n = 0$:
(3)
\begin{align} \quad k_1e^{r_1t} + k_2e^{r_2t} + ... + k_ne^{r_nt} = 0 \end{align}
• First multiply both sides of the equation above by $e^{-r_1t}$:
(4)
\begin{align} \quad e^{-r_1t} \left ( k_1e^{r_1t} + k_2e^{r_2t} + ... + k_ne^{r_nt} \right ) = 0 \\ \quad k_1\frac{e^{r_1t}}{e^{r_1t}} + k_2\frac{e^{r_2t}}{e^{r_1t}} + ... + k_n\frac{e^{r_nt}}{e^{r_1t}} = 0 \\ \quad k_1 + k_2e^{(r_2 - r_1)t}+ ... + k_ne^{(r_n - r_1)t} = 0 \\ \end{align}
• We will now differentiate the function above with respect to $t$ to get that:
(5)
\begin{align} \quad k_2(r_2 - r_1)e^{(r_2 - r_1)t} + k_3(r_3 - r_1)e^{(r_3 - r_1)t} + ... + k_n(r_n - r_1)e^{(r_n - r_1)t} = 0 \end{align}
• We will now multiply both sides of the equation above by $e^{-(r_2 - r_1)t}$:
(6)
\begin{align} \quad k_2(r_2 - r_1)\frac{e^{(r_2 - r_1)t}}{e^{(r_2 - r_1)t}} + k_3(r_3 - r_1)\frac{e^{(r_3 - r_1)t}}{e^{(r_2 - r_1)t}} + ... + k_n(r_n - r_1)\frac{e^{(r_n - r_1)t}}{e^{(r_2 - r_1)t}} = 0 \\ \quad k_2(r_2 - r_1) + k_3(r_3 - r_1) e^{(r_3 - r_2)t} + ... + k_n(r_n - r_1) e^{(r_n - r_2)t} = 0 \end{align}
• We will now differentiate both sides of the equation above with respect to $t$ again to get:
(7)
\begin{align} \quad k_3(r_3 - r_2)(r_3 - r_1)e^{(r_3 - r_2)t} + k_4(r_4 - r_2)(r_4 - r_1)e^{(r_4 - r_2)t} + ... + k_n(r_n - r_2)(r_n - r_1)e^{(r_n - r_2)t} = 0 \end{align}
• If we continue this process over and over again, we eventually get that:
(8)
\begin{align} \quad k_n(r_n - r_{n-1})(r_n - r_{n-2})...(r_n - r_1)e^{(r_n - r_{n-1})t} = 0 \end{align}
• Note that $e^{(r_n - r_{n-1})t} \neq 0$ as the exponential function, and $(r_i - r_j) \neq 0$ for each $i \neq j$, $i, j = 1, 2, ..., n$ since $r_1$, $r_2$, …, $r_n$ are distinct numbers. Thus we must have that $k_n = 0$ and so:
(9)
\begin{align} \quad k_1e^{r_1t} + k_2e^{r_2t} + ... + k_{n-1}e^{r_{n-1}t} = 0 \end{align}
• Repeating the steps from above, we have that $k_n = k_{n-1} = ... = k_2 = k_1 = 0$, and so $y_1$, $y_2$, …, $y_n$ are a linearly independent set of functions. $\blacksquare$

From Proposition 1 above, we see that the general solution to an $n^{\mathrm{th}}$ order linear homogenous differential equation with constant coefficients and whose roots to the characteristic equation are real and distinct is of the form:

(10)
\begin{align} \quad y = C_1e^{r_1t} + C_2e^{r_2t} + ... + C_ne^{r_nt} \end{align}

## Example 1

Find the general solution to the differential equation $\frac{d^3y}{dt^3} - 6 \frac{d^2y}{dt^2} + 11\frac{dy}{dt} -6y = 0$.

We first note that the characteristic equation for the differential equation is:

(11)
\begin{align} \quad r^3 - 6r^2 + 11r -6 = 0 \end{align}

We can immediately see that $r = 1$ is a solution to this differential equation, and in applying polynomial long-division, it is not too hard to see that $r = 2$ and $r = 3$ are also solutions to the characteristic equation as you should verify. Thus the general solution to our differential equation is:

(12)
\begin{align} \quad y = C_1e^t + C_2e^{2t} + C_3e^{3t} \end{align}