# Higher Order O.D.E.’s Real, Distinct Roots of The Characteristic Equation Examples 1

Consider the following $n^{\mathrm{th}}$ order linear homogenous differential equation:

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Recall from the Higher Order Homogenous Differential Equations Real, Distinct Roots of The Characteristic Equation page that if the roots $r_1$, $r_2$, …, $r_n$ to the characteristic equation $a_0r^n + a_1r^{n-1} + … + a_{n-1}r + a_n = 0$ are distinct real numbers, then $y_1(t) = e^{r_1t}$, $y_2(t) = e^{r_2t}$, …, $y_n = e^{r_nt}$ form a fundamental set of solutions to our differential equation and general solution to our differential equation for constants $C_1$, $C_2$, …, $C_n$ is:

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We will now look at some examples of solving linear homogenous differential equations of this type.

## Example 1

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

The characteristic equation for this differential equation is $r^3 + r^2 – 2r = 0$. We immediately see that one root is $r_1 = 0$ is a root to the characteristic equation, and so we can factor it out as $r(r^2 + r – 2) = 0$. The quadratic term can be factored as well to get:

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Therefore $r_1 = 0$, $r_2 = 1$ and $r_3 = -2$. Therefore the general solution to this differential equation is given by:

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## Example 2

**Find the general solution to the differential equation $2\frac{d^4y}{dt^4} – 2\frac{d^3y}{dt^3} -14 \frac{d^2y}{dt^2} + 2\frac{dy}{dt} + 12y = 0$.**

The characteristic equation for this differential equation is:

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By trial and error we can see that $r_1 = 1$ is a solution to the characteristic equation. We can then do polynomial long division to factor our characteristic equation as:

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Using trial and error once again, we can see that $r_2 = -1$ makes the factor $r^3 – 7r – 6 = 0$, and so by using polynomial long division again, we have that:

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We can now easily factor the quadratic term as $(r – 3)(r + 2)$ and so:

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Therefore $r_3 = 3$ and $r_4 = -2$. Therefore the general solution to this differential equation is:

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