Higher Order O.D.E.'s Real, Distinct Roots Examples 1

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:

(1)

\begin{align} \quad \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 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:

(2)

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

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:

(3)

\begin{align} \quad r(r+2)(r-1) = 0 \end{align}

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

(4)

\begin{align} \quad y = C_1e^{0t} + C_2e^{t} + C_3e^{-2t} \\ \quad y = C_1 + C_2e^t + C_3e^{-2t} \end{align}

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:

(5)

\begin{align} \quad 2r^4 - 2r^3 - 14r^2 + 2r + 12 = 0 \\ \quad r^4 - r^3 - 7r^2 + r + 6 = 0 \end{align}

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:

(6)

\begin{align} \quad (r - 1)(r^3 - 7r - 6) = 0 \end{align}

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:

(7)

\begin{align} \quad (r - 1)(r + 1)(r^2 - r - 6) = 0 \end{align}

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

(8)

\begin{align} \quad (r - 1)(r + 1)(r - 3)(r + 2) = 0 \end{align}

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

(9)

\begin{align} \quad y = C_1e^{t} + C_2e^{-t} + C_3e^{3t} + C_4e^{-2t} \end{align}