Lesson 8.3: Second-order Linear Equations: Homogeneous
Introduction
Welcome to Lesson 8.3! Today, we are diving deep into second-order linear equations, particularly focusing on homogeneous equations. 🎉 The main goal is to understand how these equations are structured and how to solve them.
Learning Objectives
By the end of this lesson, you (students) should be able to:
- Understand the auxiliary (characteristic) equation.
- Solve second-order equations for real distinct, repeated, and complex roots.
- Write the general solution as a combination of complementary functions.
- Understand the link between these equations and simple harmonic motion as well as damped motion.
- Form and solve the auxiliary equation correctly.
What is a Second-order Linear Homogeneous Equation?
A second-order linear homogeneous equation is an equation of the form:
$$ a y'' + b y' + c y = 0 $$
where:
- $ y $ is the unknown function,
- $ a, b, $ and $ c $ are constants,
- $ y' $ is the first derivative of $ y $, and
- $ y'' $ is the second derivative of $ y $.
The term "homogeneous" means that there are no terms that are not dependent on $ y $ (i.e., no constants or functions of $ x $ outside of $ y $).
The Auxiliary (Characteristic) Equation
To solve a second-order homogeneous linear equation, we first derive the auxiliary equation, which is obtained by assuming a solution of the form:
$$ y = e^{rt} $$
where $ r $ is a constant and $ t $ is the variable. When we substitute $ y = e^{rt} $ into the original differential equation, we get:
$$ a (r^2 e^{rt}) + b (r e^{rt}) + c (e^{rt}) = 0 $$
Dividing through by $ e^{rt} $ (which is non-zero), we obtain the characteristic polynomial:
$$ a r^2 + b r + c = 0 $$
This is the auxiliary equation we need!
Solving the Auxiliary Equation
We can use the quadratic formula to find the roots of the auxiliary equation:
$$ r = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} $$
The nature of the roots will help us determine the general solution to the associated differential equation. Let's consider the three possible cases for the roots:
Case 1: Distinct Real Roots
If $ b^2 - 4ac > 0 $, the roots $ r_1 $ and $ r_2 $ are distinct and real. In this case, the general solution is:
$$ y(t) = C_1 e^{r_1 t} + C_2 e^{r_2 t} $$
where $ C_1 $ and $ C_2 $ are constants determined by the initial conditions (if available).
Example 1:
Consider the equation:
$$ y'' - 5y' + 6y = 0 $$
Here, $ a = 1, b = -5, c = 6 $. The auxiliary equation is:
$$ r^2 - 5r + 6 = 0 $$
The roots are:
$$ r = \frac{5 \pm \sqrt{(-5)^2 - 4(1)(6)}}{2(1)} $$
$$ r = \frac{5 \pm \sqrt{1}}{2} = 3, 2 $$
Thus, the general solution is:
$$ y(t) = C_1 e^{3t} + C_2 e^{2t} $$
Case 2: Repeated Real Roots
If $ b^2 - 4ac = 0 $, there is one repeated real root $ r $. The general solution in this case is:
$$ y(t) = (C_1 + C_2 t)e^{rt} $$
Example 2:
Consider the equation:
$$ y'' - 4y' + 4y = 0 $$
Here, the auxiliary equation is:
$$ r^2 - 4r + 4 = 0 $$
Thus, $ r = 2 $ is repeated. The solution is:
$$ y(t) = (C_1 + C_2 t)e^{2t} $$
Case 3: Complex Roots
If $ b^2 - 4ac < 0 $, we have complex conjugate roots. Let’s say the roots are $ r = \alpha \pm \beta i $. The general solution in this case involves sine and cosine:
$$ y(t) = e^{\alpha t}(C_1 \cos(\beta t) + C_2 \sin(\beta t)) $$
Example 3:
Consider:
$$ y'' + 4y = 0 $$
The auxiliary equation gives:
$$ r^2 + 4 = 0 $$
Thus, $ r = \pm 2i $. The solution is:
$$ y(t) = C_1 \cos(2t) + C_2 \sin(2t) $$
Conclusion
In today’s lesson, we explored the concept of second-order linear homogeneous equations, understood how to derive their auxiliary equation, and solved for various types of roots. Mastering these concepts allows you (students) to analyze real-world systems modeled by such equations, linking them to phenomena like simple harmonic motion.
Study Notes
- A second-order linear homogeneous equation is of the form: $a y'' + b y' + c y = 0$.
- The auxiliary equation is: $a r^2 + b r + c = 0$.
- Use the quadratic formula: $r = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}$ to find roots.
- Distinct real roots: $y(t) = C_1 e^{r_1 t} + C_2 e^{r_2 t}$.
- Repeated real roots: $y(t) = (C_1 + C_2 t)e^{rt}$.
- Complex roots: $y(t) = e^{\alpha t}(C_1 \cos(\beta t) + C_2 \sin(\beta t))$.