# Characteristic Equations – Definition, General Form, and Application

Using characteristic equations is important when working with homogeneous differential equations. Knowing how to find characteristic equations and knowing how to utilize them are prerequisites if you want to know linear homogeneous differential equations by heart.

The characteristic equations are essential when solving linear homogeneous differential equations. We use the roots of the characteristic equation to establish the general solution of the homogeneous differential equation.

Our discussion’s goal is to make sure that you know how to find the characteristic equations from linear homogeneous differential equations. We’ll also show you the significance of the characteristic equations when solving differential equations and have provided some examples for you to work on as well.

## What Is the Characteristic Equation?

The characteristic equation of a linear and homogeneous differential equation is an algebraic equation we use to solve these types of equations. Here’s an example of a pair of a homogeneous differential equation and its corresponding characteristic equation:

\begin{aligned} y^{\prime\prime} -2y^{\prime} + y &= 0\\\downarrow\phantom{x}\\r^2 – 2r + r &= 0\end{aligned}

Now, let’s generalize this for all second order linear homogeneous differential equations with a general form, as shown below.

\begin{aligned}ay^{\prime \prime} + by^{\prime} + cy &= 0\end{aligned}

The easiest solution we can try out for this equation is $y = e^{rx}$, so we’ll also have $y^{\prime} = re^{rx}$ and $y^{\prime\prime} = r^2e^{ex}$.  Substitute these expressions into the differential equation.

\begin{aligned}a r^2e^{rx}+ bre^{rx} + ce^{rx}  &= 0\\e^{rx} (ar^2+ br + c) &= 0\end{aligned}

Since $e^{rx}$ will never be equal to zero, the next factor, $ar^2+ br + c$, must be equal to zero for $y = e^{rx}$ to be a zero.

\begin{aligned} ar^2+ br + c &= 0\end{aligned}

We can solve for $r$ using algebraic techniques we’ve learned in the past. These roots will determine the form of the homogeneous solution’s nature. The process for establishing rules for characteristic equations for higher order and more complex homogeneous equations will be similar.

## How To Find the Characteristic Equation and Apply It in Differential Equations?

From the previous section, we find the characteristic equation by assigning the differentials as a variable. For example, when working with second order homogeneous equations, $ay^{\prime\prime} + by^{\prime} + cy =0$.

We’ve shown that we can rewrite the following: $y^{\prime \prime}$ with $r^2$, $y^{\prime}$ with $r$, and $y$ with $1$. Here are a few more examples of linear homogeneous differential equations and their corresponding characteristic equations:

 Linear Homogeneous Differential Equations Characteristic Equations \begin{aligned}y^{\prime \prime}–4y^{\prime} -5y&= 0\end{aligned} \begin{aligned}r^2–4r-5 &= 0\end{aligned} \begin{aligned}y^{\prime \prime}+6y^{\prime} + 9y&= 0\end{aligned} \begin{aligned}r^2+ 6r+9 &= 0\end{aligned} \begin{aligned}4y^{\prime \prime} – 12y^{\prime} + 9y &= 0\end{aligned} \begin{aligned}4r^2–12r+9&= 0\end{aligned}

Now, let’s work on the first given differential equation, $y^{\prime \prime} – 4y^{\prime} + 4y = 0$. We’ve established that $y =e^{rx}$, $y^{\prime} =re^{rx}$, and $y^{\prime\prime} =r^2e^{rx}$. This equation reduces to a quadratic equation, as shown below:

\begin{aligned} y^{\prime \prime} – 4y^{\prime} – 5y &= 0\\r^2e^{rx} – 4re^{rx} -5e^{rx}&= 0\\r^2-4r -5&= 0\\(r -5)(r+1)&= 0\\r= -1, r&= 5\end{aligned}

This means that both $y = e^{-x}$ and $y = e^{5x}$ are two independent solutions, so our homogeneous differential equation will have a general solution of:

\begin{aligned}y&= C_1e^{-x} + C_2e^{5x}\end{aligned}

Now, let’s generalize this process for finding the characteristic equation and solution for the $n$-th order linear homogeneous equation,

\begin{aligned}a_ny^{(n)} + a_{n -1} y^{(n -1)} + …+ a_1y^{\prime} + a_0y &= 0,\end{aligned}

where $a_n, a_{n -1}, …, a_1, a_0$ are constants and $a_0$ is not equal to zero. We can write down the characteristic equation as shown below.

\begin{aligned}a_nr^n + a_{n -1}r ^{n -1} + …+ a_1r + a_0 &= 0\end{aligned}

We use the characteristic equations and the roots to find the solutions for linear homogeneous equations. In our discussion of homogeneous equations, we’ve laid out how to find the general solutions of the homogeneous differential equations based on their characteristic equations.

 Roots’ Nature Solution’s General Form The characteristic equation’s roots are real and distinct. \begin{aligned}y(x) &= C_1e^{r_1 x} + C_2e^{r_2 x} +…+ C_ne^{r_n x} \end{aligned} There exists identical roots within the characteristic equation’s solutions.$m$ roots: $\{r_1, r_2, …,r_m\}$The multiplicity may vary, so let’s generalize the multiplicity as:multiplicity: $\{k_1, k_2, …,k_m\}$ \begin{aligned}y(x) &= C_1e^{r_1 x} + C_2xe^{r_1 x} + …+C_{k_1}x^{k_1 – 1}e^{r_1 x}+…\\&+C_{n -k_m +1}e^{r_m x} +C_{n -k_{m}+ 2}xe^{r_m x} + …+C_nx^{k_m -1}e^{r_m x}\end{aligned} The characteristic equation’s roots are complex and unique.\begin{aligned}r_{1,2} &= \alpha + \beta i\\ r_{3,4 }&= \gamma- \delta i\\&.\\&.\\&.\end{aligned} \begin{aligned}y= {e^{\alpha x}}\left( {{C_1}\cos \beta x + {C_2}\sin \beta x} \right) + {e^{\gamma x}}\left( {{C_3}\cos \delta x + {C_4}\sin \delta x} \right) + \cdots\end{aligned}

We’ve covered everything that we need to know about characteristic equations and homogeneous equations. It’s time that we try out different equations to further establish our knowledge on characteristic equations.

Example 1

Write down the characteristic equations of the linear homogeneous differential equations shown below.

a. $y^{\prime\prime} – 10y^{\prime}+25y = 0$
b. $\dfrac{d^2y}{dx^2} + 6\dfrac{dy}{dx} =- 9y$
c. $y^{\prime\prime\prime} +5y^{\prime\prime} –10y^{\prime} +25y = 0$

Solution

Since all three the equations are linear homogeneous differential equations, we can use the rules we’ve established in our discussion when writing down the characteristic equations.

\begin{aligned}y^{\prime\prime} –10y^{\prime}+25y&= 0\end{aligned}

Replace $y^{\prime\prime}$ with $r^2$, $y^{\prime}$ with $r$, and $y$ with $1$. Hence, we have the characteristic equation shown below.

\begin{aligned}(a)\phantom{x} r^2 – 10r + 25 &= 0\end{aligned}

For the second equation, rewrite it first so that it is in the standard form of homogeneous equations. We apply the same process to find the characteristic equation of the second differential equation.

\begin{aligned}\dfrac{d^2y}{dx^2} + 6\dfrac{dy}{dx} &=- 9y\\\dfrac{d^2y}{dx^2} + 6\dfrac{dy}{dx} +9y&= 0\\(b) \phantom{x}r^2 + 6r + 9&= 0\end{aligned}

For the third homogeneous differential equation, we use the same process but this time, we have a new term: $y^{\prime \prime\prime} =r^3$.

\begin{aligned}y^{\prime\prime\prime} + 5y^{\prime\prime} – 10y^{\prime} + 25y &= 0\\(c) \phantom{x}r^3 + 5r^2-10r+25&= 0\end{aligned}

Example 2

Find the solutions to the differential equation, $y^{\prime\prime}+7y^{\prime} + 10y = 0$.

Solution

The equation is a second order linear homogeneous differential equation, so we can use the characteristic equation to find the solution for the equation. We begin by writing down the characteristic of the equation.

\begin{aligned}y^{\prime\prime}+7y^{\prime}+10y &= 0 \\\downarrow \phantom{xxx}\\r^2 + 7r + 10&= 0\end{aligned}

Find the roots of the resulting quadratic equation by factoring the right-hand side of the expression.

\begin{aligned}(r + 2)(r + 5)&= 0\\r=-2, r&=-5\end{aligned}

This means that $y = e^{-2x}$ and $y = e^{-5x}$ are independent solutions, so we can use them to write down the general solution of the differential equation.

\begin{aligned}y&= C_1e^{-2x}+C_2e^{-5x}\end{aligned}

Example 3

Find the solution to the differential equation, $y^{\prime\prime\prime} + 4^{\prime\prime} -7y^{\prime} -10y =0$.

a. Write down the general solution of the homogeneous differential equation.
b. Determine the particular solution of the equation with the following initial conditions:
\begin{aligned}\{y(0) = -2, y^{\prime}(0) =6, y^{\prime\prime}(0) = -12\}\end{aligned}

Solution

The equation is a linear homogeneous differential equation, so let’s write down its characteristic equation as shown below.

\begin{aligned} y^{\prime\prime\prime} + 4y^{\prime\prime}- 7y^{\prime} – 10y &= 0\\\downarrow \phantom{xxxxx}\\r^3 + 4r^2 – 7r – 10&= 0\end{aligned}

Factor the cubic expression then solve for the roots of the equation.

\begin{aligned}(r +1)(r + 5)(r -2) &= 0\\r =-1, r =-5, r &= 2 \end{aligned}

We have the following independent solutions for the differential equation: $y = e^{-x}$, $y = e^{-5x}$, and $y = e^{2x}$. Hence, we have the general solution for the equation.

\begin{aligned}y&= C_1e^{-x} + C_2e^{-5x} + C_3e^{2x}\end{aligned}

Now, use the general solution to find the expressions for $y^{\prime}$ and $y^{\prime \prime}$.

\begin{aligned}y&= C_1e^{-x} + C_2e^{-5x} + C_3e^{2x}\\y^{\prime}&= -C_1e^{-x} – 5C_2e^{-5x} + 2C_3e^{2x}\\y^{\prime \prime}&= C_1e^{-x} + 25C_2e^{-5x} + 4C_3e^{2x}\end{aligned}

Set up the system of three linear equations then find the values of $C_1$, $C_2$,a nd $C_3$.

\begin{aligned}-2 = y(0)&= C_1+C_2 +C_3\\6 = y^{\prime}(0)&= -C_1- 5C_2 +2C_3\\-12 =y^{\prime \prime}(0)&= C_1 +25C_2 +4C_3\end{aligned}

Apply algebraic techniques (or you can also use an equation solver for this part) to solve the system of linear equations. We have the following constants: $C_1 =-\dfrac{13}{6}$, $C_2= -\dfrac{1}{2}$, and $C_3 = \dfrac{2}{3}$. This means that our initial-value problem will have the particular solution shown below.

\begin{aligned}y&=-\dfrac{13}{6}e^{-x} -\dfrac{1}{2}e^{-5x} + \dfrac{2}{3}e^{2x}\end{aligned}

### Practice Questions

1. Write down the characteristic equations of the linear homogeneous differential equations shown below.
a. $6y^{\prime\prime} – 36y^{\prime} +54y = 0$
b. $\dfrac{d^2y}{dx^2} + 18\dfrac{dy}{dx} =- 81y$
c. $y^{\prime\prime\prime} + 8y^{\prime\prime} – 24y^{\prime}+36y = 0$
2. Find the solutions to the differential equation, $y^{\prime\prime} +2y^{\prime} -24y = 0$.
3. Find the solution to the differential equation, $2y^{(4)} +16y^{\prime\prime} +32y = 0$.
4. Find the solution to the differential equation, $2y^{(5)}+8y^{\prime\prime\prime} = 0$.
a. Write down the general solution of the homogeneous differential equation.
b. Determine the particular solution of the equation with the following initial conditions:
\begin{aligned}\{y(0) = -2, y^{\prime}(0) =6, y^{\prime\prime}(0) = -12\}\end{aligned}

1.
a. $6r^2 – 36r + 54 = 0$
b. $r^2+ 18r + 81=0$
c. $r^3 + 8r^2 – 24r+ 36 = 0$
2. $y = C_1e^{-4x} + C_2e^{6x}$
3.$y = C_1 \sin (2x) + C_2\cos(3x) +C_3x \sin (2x) + C_4x\cos (2x)$
4.
a. $y = C_1 \sin (2x) + C_2\cos (2x) +C_3 + C_4x + C_5x^2$
b. $y = \dfrac{1}{8} \sin (2x) + \dfrac{1}{16}\cos (2x) + \dfrac{31}{16} + \dfrac{1}{16}x + \dfrac{1}{8}x^2$