Solving DEs with Distinct Real Roots (Step-by-Step)

Solving second order and higher order differential equations can be challenging, but for homogeneous, constant coefficient differential equations, we can use an easy-to-follow process that transforms our differential equation into a straightforward polynomial equation.

Jenn (B.S., M.Ed.) of Calcworkshop® teaching distinct real roots

Jenn, Founder Calcworkshop^®, 15+ Years Experience (Licensed & Certified Teacher)

How?

By the principal of superposition and the characteristic equation!

In fact, solving a second-order DE for distinct real roots has never been easier.

Let’s take a look.

Formulating the General Solution for Homogeneous Differential Equations

Okay, so a second-order differential equation is said to be linear if it can be written as

$\begin{array}{r} a_{2} y^{''} + a_{1} y^{'} + a_{0} y = f (x) \end{array}$

We call the function $f (x)$ on the right a forcing function since, in physical applications, it’s often related to a force acting on some system modeled by the differential equation. And if $f (x) = 0$ then DE is homogeneous, and nonhomogeneous if $f (x) \neq 0$

Now, if each coefficient $a_{2}, a_{1}, a_{0}$ is a constant and $f (x) = 0$ , then obtaining a solution to the differential equation is really quite simple because all we have to do is assume a possible solution.

First, let us guess that a possible solution of the equation $a_{2} y^{''} + a_{1} y^{'} + a_{0} y = 0$ is $y = e^{r x}$ . Next, we must ask ourselves, “For what values of $r$ will yield a solution?”

Well, we know that a value must be for which

$a_{2} \frac{d^{2}}{d x^{2}} e^{r x} + a_{1} \frac{d}{d x} e^{r x} + a_{0} e^{r x} = 0$

$a_{2} r^{2} e^{r x} + a_{1} r e^{r x} + a_{0} e^{r x} = 0$

And if we divide by $e^{r x}$ we obtain an algebraic equation, called the characteristic equation, in terms of $r$

$a_{2} r^{2} + a_{1} r + a_{0} = 0$

So, if we solve the characteristic equation above, we will find the value of $r$ for which $y = e^{r x}$ is a solution.

Wow!

Now, the characteristic equation, sometimes called the auxiliary equation, is easily obtained by replacing $y$ by $r$ and the order of the derivative by a numerically equal exponent.

In fact, this process can be extended to homogeneous linear higher order differential equations of the form $a_{n} y^{(n)} + a_{n - 1} y^{(n - 1)} + \dots + a_{2} y^{''} + a_{1} y^{'} + a_{0} y = 0$ , where the coefficients $a_{i}$ are real constants.

Three Possible Solutions to the Characteristic Equation

It is also important to note that when solving the characteristic equation, there are three possible solutions:

All roots are real and distinct
All roots are real, but some repeat
All roots are imaginary or complex

Steps for Solving Homogeneous Differential Equations with Constant Coefficients

And the steps for solving for the general solution to a second-order or higher-order homogeneous differential equation with constant coefficients are straightforward:

Step 1: Write the characteristic equation
Step 2: Solve the polynomial equation for factors by using factoring, quadratic formula, synthetic division, etc.
Step 3: Substitute each root into the general form:

$y = c_{1} e^{r_{1} t} + c_{2} e^{r_{2} t} + \dots + c_{n} e^{r_{n} t}$

Focus on Real and Distinct Roots

Please note that this lesson will focus on the roots of the characteristic equation that are real and distinct (different). In subsequent lessons, we will explore other possibilities, such as repeated roots and complex roots.

The main difference when dealing with complex roots lies in the characteristic equation, which requires a different general solution form involving trigonometric functions, sine and cosine, instead of exponential functions used for real and distinct roots.

Example: Solving a Second-Order Differential Equation

Let’s look at an example to see this process in action.

Find the general solution $y (t)$ of $y^{''} - 3 y^{'} + 2 y = 0$

First, we will transform our DE into the characteristic equation.

$a_{2} r^{2} + a_{1} r + a_{0} = 0$

$y^{''} - 3 y^{'} + 2 y = 0$

$r^{2} - 3 r + 2 = 0$

Next, we will solve the quadratic polynomial for our roots.

$\begin{array}{r} \begin{aligned} r^{2} - 3 r + 2 = 0 \\ (r - 2) (r - 1) = 0 \\ r_{1} = 2, r_{2} = 1 \end{aligned} \end{array}$

Now we plug our roots into the general solution formula of

$\begin{aligned} y = c_{1} e^{r_{1} t} + c_{2} e^{r_{2} t} + \dots + c_{n} e^{r_{n} t} \\ y = c_{1} e^{2 t} + c_{2} e^{1 t} \end{aligned}$

Easy, right! Okay, let’s look at a higher-order DE and apply the same process.

Example: Solving a Higher-Order Differential Equation

Find the general solution $y (t)$ of $y^{(4)} - y^{'''} - 4 y^{''} + 4 y^{'} = 0$

First, we will transform our DE into the characteristic equation.

$\begin{array}{r} \begin{aligned} y^{(4)} - y^{'''} - 4 y^{''} + 4 y^{'} = 0 \\ r^{4} - r^{3} - 4 r^{2} + 4 r = 0 \end{aligned} \end{array}$

Next, we will solve the quadratic polynomial for our roots.

$\begin{array}{r} \begin{aligned} r^{4} - r^{3} - 4 r^{2} + 4 r = 0 \\ r (r^{3} - r^{2} - 4 r + 4) = 0 \\ r (r - 1) (r^{2} - 4) = 0 \\ r (r - 1) (r - 2) (r + 2) = 0 \\ r_{1} = 0, r_{2} = 1, r_{3} = 2, r_{4} = - 1 \end{aligned} \end{array}$

Now we plug our roots into the general solution formula.

$\begin{array}{r} \begin{aligned} y = c_{1} e^{r_{1} t} + c_{2} e^{r_{2} t} + \dots + c_{n} e^{r_{n} t} \\ y = c_{1} e^{0 t} + c_{2} e^{1 t} + c_{3} e^{2 t} + c_{4} e^{- 2 t} \\ y = c_{1} + c_{2} e^{t} + c_{3} e^{2 t} + c_{4} e^{- 2 t} \end{aligned} \end{array}$

And that’s all there is to it! Nice!

Next Steps

Up next, you’ll:

Deepen your understanding of second-order linear differential equations, a crucial skill in mathematics and engineering.
Master the steps to find the general solution, enabling you to solve various types of differential equations.
Improve your problem-solving skills by learning how to avoid common mistakes and tackle complex problems confidently.
Expand your knowledge of real and complex roots, providing a comprehensive understanding of roots in differential equations.
Enhance your ability to work with initial conditions, an essential skill for solving real-world problems involving differential equations.

It’s going to be fun, so let’s jump right in!

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