Homogeneous Equations with Constant Coefficients
Here, for the first time, I'll introduce the notion of a characteristic equation (which I will define later). This is an important concept, and this type of differential equation is important. You'll find yourself using the technique learned in this section frequently in the course. Also, before we start, I'd like to introduce a bit of notation. Instead of constantly referring to the given differential equation as, well, "the given differential equation", let's simplify things. Each given differential equation can be written as a function of y. For instance, the differential equation
$ xy''+y'=0 $
can be written as
$ L(y)=xy''+y' $
and any solution α to the differential equation will have the property that
$ L(\alpha)=0 $
Also, before we move on, I just wish to point out that the word "homogeneous" in this case refers to the fact that the independent variable does not appear by itself in the differential equation. So for instance, y'+y=t is not a homogeneous differential equation (because of the t by itself), but y'+y=0 is.
With that out of the way, let's discuss homogeneous equations with constant coefficients.
Suppose your differential equation is of the form
$ ay''+by'+cy=0 $
where a, b and c are arbitrary real constants. Using our new notation, we can say that:
$ L(y)=ay''+by'+cy=0 $
It turns out that exponential functions of the form exp(rt) satisfy the equation. If we assume that y=exp(rt), then we have
$ \begin{align} y&=e^{rt}\\ y'&=re^{rt}\\ y''&=r^2e^{rt} \end{align} $
Therefore, assuming that y has the form exp(rt), and setting L(y)=0, we obtain
$ \begin{align} L(y)=ar^2e^{rt}+bre^{rt}+ce^{rt}&=0\\ (ar^2+br+c)e^{rt}&=0\\ ar^2+br+c&=0 \end{align} $
As you can see, the exponential terms can be factored out and then divided out of the equation (the exponential function is never zero, so we don't have to worry about dividing it out). Now, here we have an equation that is quadratic in r. This equation is called the characteristic equation. To find the solution to L, we need to find the roots of this equation. The roots, obviously, depend upon the values of a, b and c. In this section, we are only considering differential equations whose characteristic equations have distinct, real roots. Let's call those roots r1 and r2. The solution to the differential equation, then, is
$ y=c_1e^{r_1t}+c_2e^{r_2t} $
where c1 and c2 are arbitrary coefficients.
Example
Suppose your differential equation is
$ y''+y'-6y=0 $
Let the solution be of the form y=e^{rt}. Then L(y) is
$ \begin{align} r^2e^{rt}+re^{rt}-6e^{rt}&=0\\ r^2+r-6&=0\\ (r+3)(r-2)&=0\\ r_1=-3;r_2&=2 \end{align} $
Therefore, our solution is
$ y=c_1e^{-3t}+c_2e^{2t} $
