The Characteristic Equation Method

Oct 13, 2025 by pinkandtrash.com 49 views

Introduction

In this article, we will focus on solving the initial value problem of a second-order linear homogeneous differential equation. The given differential equation is $y^{\prime \prime} + 2y^{\prime} + 2y = 0$ , with the initial conditions $y\left(\frac{\pi}{2}\right) = 0$ and $y^{\prime}\left(\frac{\pi}{2}\right) = 2$ . We will use the characteristic equation method to find the general solution of the differential equation and then apply the initial conditions to determine the specific solution.

The Characteristic Equation Method

The characteristic equation method is a powerful tool for solving second-order linear homogeneous differential equations. The characteristic equation is obtained by substituting $y = e^{rt}$ into the differential equation, where $r$ is a constant. This substitution leads to the characteristic equation $r^2 + 2r + 2 = 0$ . We can solve this quadratic equation using the quadratic formula: $r = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}$ , where $a = 1$ , $b = 2$ , and $c = 2$ .

Solving the Characteristic Equation

Using the quadratic formula, we get:

$r = \frac{-2 \pm \sqrt{2^2 - 4(1)(2)}}{2(1)}$ $r = \frac{-2 \pm \sqrt{4 - 8}}{2}$ $r = \frac{-2 \pm \sqrt{-4}}{2}$ $r = \frac{-2 \pm 2i}{2}$

Simplifying, we get two complex roots:

$r_1 = -1 + i$ $r_2 = -1 - i$

The General Solution

Since the characteristic equation has two distinct complex roots, the general solution of the differential equation is given by:

$y(t) = c_1 e^{r_1 t} + c_2 e^{r_2 t}$ $y(t) = c_1 e^{(-1 + i)t} + c_2 e^{(-1 - i)t}$

Applying the Initial Conditions

We are given the initial conditions $y\left(\frac{\pi}{2}\right) = 0$ and $y^{\prime}\left(\frac{\pi}{2}\right) = 2$ . We can use these conditions to determine the values of $c_1$ and $c_2$ .

First, we substitute $t = \frac{\pi}{2}$ into the general solution:

$y\left(\frac{\pi}{2}\right) = c_1 e^{(-1 + i)\frac{\pi}{2}} + c_2 e^{(-1 - i)\frac{\pi}{2}}$ $0 = c_1 e^{-\frac{\pi}{2} + \frac{i\pi}{2}} + c_2 e^{-\frac{\pi}{2} - \frac{i\pi}{2}}$ $0 = c_1 e^{-\frac{\pi}{2}} e^{i\frac{\pi}{2}} + c_2 e^{-\frac{\pi}{2}} e^{-i\frac{\pi}{2}}$ $0 = c_1 e^{-\frac{\pi}{2}} i + c_2 e^{-\frac{\pi}{2}} (-i)$

Simplifying, we get:

$0 = c_1 e^{-\frac{\pi}{2}} i - c_2 e^{-\frac{\pi}{2}} i$ $0 = (c_1 - c_2) e^{-\frac{\pi}{2}} i$

Since $e^{-\frac{\pi}{2}} \neq 0$ , we can divide both sides by $e^{-\frac{\pi}{2}} i$ :

$0 = c_1 - c_2$

Finding the Value of $c_2$

We can use the second initial condition $y^{\prime}\left(\frac{\pi}{2}\right) = 2$ to find the value of $c_2$ . First, we differentiate the general solution with respect to $t$ :

$y^{\prime}(t) = c_1 (-1 + i) e^{(-1 + i)t} + c_2 (-1 - i) e^{(-1 - i)t}$

Substituting $t = \frac{\pi}{2}$ , we get:

$y^{\prime}\left(\frac{\pi}{2}\right) = c_1 (-1 + i) e^{(-1 + i)\frac{\pi}{2}} + c_2 (-1 - i) e^{(-1 - i)\frac{\pi}{2}}$ $2 = c_1 (-1 + i) e^{-\frac{\pi}{2} + \frac{i\pi}{2}} + c_2 (-1 - i) e^{-\frac{\pi}{2} - \frac{i\pi}{2}}$ $2 = c_1 (-1 + i) e^{-\frac{\pi}{2}} e^{i\frac{\pi}{2}} + c_2 (-1 - i) e^{-\frac{\pi}{2}} e^{-i\frac{\pi}{2}}$ $2 = c_1 (-1 + i) e^{-\frac{\pi}{2}} i + c_2 (-1 - i) e^{-\frac{\pi}{2}} (-i)$

Simplifying, we get:

$2 = c_1 (-1 + i) e^{-\frac{\pi}{2}} i + c_2 (-1 - i) e^{-\frac{\pi}{2}} i$ $2 = (-c_1 + c_2) e^{-\frac{\pi}{2}} i$

Since $e^{-\frac{\pi}{2}} \neq 0$ , we can divide both sides by $e^{-\frac{\pi}{2}} i$ :

$2 = -c_1 + c_2$

Solving for $c_1$ and $c_2$

We have two equations:

$0 = c_1 - c_2$ $2 = -c_1 + c_2$

We can solve this system of equations by adding the two equations:

$2 = 2c_2$

Dividing both sides by 2, we get:

$1 = c_2$

Substituting this value into the first equation, we get:

$0 = c_1 - 1$

Solving for $c_1$ , we get:

$c_1 = 1$

The Specific Solution

Now that we have found the values of $c_1$ and $c_2$ , we can substitute these values into the general solution:

$y(t) = c_1 e^{r_1 t} + c_2 e^{r_2 t}$ $y(t) = 1 e^{(-1 + i)t} + 1 e^{(-1 - i)t}$

Simplifying, we get:

$y(t) = e^{(-1 + i)t} + e^{(-1 - i)t}$

Behavior of the Solution

To analyze the behavior of the solution, we can use the Euler's formula:

$e^{ix} = \cos x + i \sin x$

Substituting $x = t$ , we get:

$e^{it} = \cos t + i \sin t$

Using this formula, we can rewrite the solution as:

$y(t) = e^{(-1 + i)t} + e^{(-1 - i)t}$ $y(t) = e^{-t} (\cos t + i \sin t) + e^{-t} (\cos t - i \sin t)$

Simplifying, we get:

$y(t) = 2 e^{-t} \cos t$

This is the specific solution of the initial value problem.

Conclusion

In this article, we have solved the initial value problem of a second-order linear homogeneous differential equation. We used the characteristic equation method to find the general solution and then applied the initial conditions to determine the specific solution. The specific solution is given by $y(t) = 2 e^{-t} \cos t$ . We have also analyzed the behavior of the solution using Euler's formula.

Q: What is the characteristic equation method?

A: The characteristic equation method is a powerful tool for solving second-order linear homogeneous differential equations. It involves substituting $y = e^{rt}$ into the differential equation, where $r$ is a constant, and then solving the resulting characteristic equation to find the roots.

Q: How do I find the roots of the characteristic equation?

A: To find the roots of the characteristic equation, you can use the quadratic formula: $r = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}$ , where $a$ , $b$ , and $c$ are the coefficients of the characteristic equation.

Q: What if the characteristic equation has complex roots?

A: If the characteristic equation has complex roots, the general solution of the differential equation will be given by $y(t) = c_1 e^{r_1 t} + c_2 e^{r_2 t}$ , where $r_1$ and $r_2$ are the complex roots.

Q: How do I apply the initial conditions to find the specific solution?

A: To apply the initial conditions, you can substitute the values of $y(0)$ and $y^{\prime}(0)$ into the general solution and then solve for the constants $c_1$ and $c_2$ .

Q: What if the initial conditions are given at a point other than $t = 0$ ?

A: If the initial conditions are given at a point other than $t = 0$ , you can use the general solution and the initial conditions to find the specific solution.

Q: How do I analyze the behavior of the solution?

A: To analyze the behavior of the solution, you can use Euler's formula: $e^{ix} = \cos x + i \sin x$ . This will help you to rewrite the solution in a more convenient form and to understand its behavior.

Q: What is the significance of the specific solution?

A: The specific solution is the solution of the initial value problem that satisfies the given initial conditions. It is a particular solution of the differential equation that is valid for the given initial conditions.

Q: Can I use the characteristic equation method to solve other types of differential equations?

A: Yes, the characteristic equation method can be used to solve other types of differential equations, such as higher-order linear homogeneous differential equations and systems of linear differential equations.

Q: Are there any other methods for solving initial value problems?

A: Yes, there are other methods for solving initial value problems, such as the method of undetermined coefficients and the method of variation of parameters. However, the characteristic equation method is a powerful and widely used method for solving initial value problems.

Q: How do I choose the correct method for solving an initial value problem?

A: To choose the correct method for solving an initial value problem, you should consider the type of differential equation, the initial conditions, and the desired solution. The characteristic equation method is a good choice when the differential equation is a second-order linear homogeneous differential equation with constant coefficients.

Q: Can I use a computer algebra system (CAS) to solve initial value problems?

A: Yes, you can use a computer algebra system (CAS) to solve initial value problems. CASs can be used to solve differential equations and to find the specific solution that satisfies the given initial conditions.

Q: What are some common mistakes to avoid when solving initial value problems?

A: Some common mistakes to avoid when solving initial value problems include:

Not checking the initial conditions to ensure that they are consistent with the differential equation
Not using the correct method for solving the differential equation
Not checking the solution to ensure that it satisfies the initial conditions
Not analyzing the behavior of the solution to understand its properties.

By avoiding these common mistakes, you can ensure that you are solving initial value problems correctly and efficiently.