# 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 $a_0, a_1, …, a_n \in \mathbb{R}$:

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

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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 $k_1 = k_2 = … = k_n = 0$:

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- First multiply both sides of the equation above by $e^{-r_1t}$:

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- We will now differentiate the function above with respect to $t$ to get that:

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- We will now multiply both sides of the equation above by $e^{-(r_2 – r_1)t}$:

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- We will now differentiate both sides of the equation above with respect to $t$ again to get:

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- If we continue this process over and over again, we eventually get that:

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- 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:

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- 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:

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## 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:

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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:

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