4.3, or the definitely (for real) beginning of our journey through chapter 4.

Section 4.3 is entitled “Linear, Homogeneous Equations with Constants Coefficients.”

These are equations of the form y” + py’ + qy = 0. In this case, p and q are constants. First, we’re going to look at the first-order equation y’ + py = 0. This is an exponential equation, in which its general solution is y(t) = Ce^-pt. In this case, C is just a constant. We’re going to look for a solution of the type y(t) = e^λt, where λ is a constant. When you put this type of solution into the original differential equation,

So Word apparently doesn't like λt...

That final equation is called the characteristic equation for our original second-order differential equation. The polynomial is called the characteristic polynomial, and the roots of this equation are called characteristic roots. Then if λ is a characteristic root, then y = e^λt is a solution to the differential equation.

If you write both the differential equation and the characteristic equation together, then

They look very similar.

The roots for the characteristic polynomial are given by the quadratic formula:

When you look at the discriminant (p² – 4q), there are three cases to consider:

1. Two real roots (p² – 4q > 0)

2. Two complex roots (p² – 4q < 0)

3. One repeated real root if p² – 4q = 0

For the three cases (and the summary), I have a lot of quotes for you. I was doing so well without direct quotes! Oh well…such is life, I suppose.

For two real roots: “If the characteristic equation λ² + pλ + q = 0 has two distinct real roots λ₁ and λ₂, then the general solution to y” + py’ + qy = 0 is

where C₁ + C₂ are arbitrary constants” (151).

For the complex roots, whenever there’s a variable with strikethrough (like this) and it’s also surrounded by brackets, I changed it from an overstrike (which I can’t show in the regular words because overstrikes are hard) to a strikethrough. I thought I’d let you know before you started to wonder why I have brackets around strikethrough stuff.

For two complex roots: “Suppose the characteristic equation λ² + pλ + q = 0 has two complex conjugate roots, λ = a + ib and [λ] = a – ib.

1. The functions

form a complex-valued fundamental set of solutions, so the general solution is

where C₁ and C₂ are arbitrary constants.

2. The functions

form a real-valued fundamental set of solutions, so the general solution is

where A₁ and A₂ are constants” (153). 

For our third case, with the repeated roots: “If the characteristic equation λ² + pλ + q = 0 has only one double root λ₁, then the general solution to y” + py’ + qy = 0 is

where C₁ and C₂ are arbitrary constants” (155).

And finally, the summary of all of this (put in bold and italics because summaries):

“If p² – 4q > 0, the characteristic equation has two distinct, real roots λ₁ and λ₂. A fundamental set of solutions is

If p² – 4q = 0, the characteristic equation has one repeated real root λ. A fundamental set of solutions is

If p² – 4q < 0, the characteristic equation has two complex conjugate roots a ± ib. A fundamental set of solutions is…” (156).

And that’s it for 4.3! I seriously can’t believe I got this done tonight. I’m super excited about this. I don’t have to do this section tomorrow! Hooray!

Enjoy all of the quotes. Also enjoy the beginnings of our journey through Chapter 4.

I shall see you in 4.4! :D