The Laplace transform offers a new way to solve constant coefficient linear (or "linear, time-independent," LTI, in EE parlance) ODEs which is efficient when the behavior of the forcing term is different in different segments of the time domain. It is especially efficient when the forcing term is discontinuous or even an "impulse" (a term we'll return to later). If the method of undetermined coefficients is applicable, on the other hand, it is usually more efficient than the Laplace transform.
A standard convention in the use of Laplace transform in ODEs is to start the clock at t=0, so initial conditions are given at t=0 and all functions are of interest only for nonnegative t.
Example: the "unit step function" or "Heaviside function" u_a(t) is 1 for t bigger than or equal to a, and 0 for t less than a. This can be used to build other functions; for a less than b, for example, the function u_a - u_b turns on at time a and off at time b.
These functions are examples of "piecewise continuous functions." A function f on [0,infinity) is piecewise continuous if there is a sequence 0 = a0 less than a1 less than .... such that (1) ai goes to infinity; (2) f is continuous in (ai,a(i+1)) for each i; and (3) lim_{t approaches ai from above} f(t) and lim_{t approaches ai from below} f(t) both exist for all i. (Of course the upper and lower limits at a both exist and equal f(a) if f is continuous at a.)
Definition. The Laplace transform of f(t) is
F(s) = integral_0^infinity f(t)e^{-st}dt.
I will also write L(f(t);s) for this function. L is a "transform"; it begins with a function of time, t, and ends with a function of a new variable, s. For each value of s you have to evaluate an integral to obtain the value of F(s).
First example: L(e^{at};s) = 1/(s-a). (1)
In particular, L(1;s) = 1/s (0)
The integral involved only converges for s greater than a; but the resulting formula makes sense for any s (as long as we allow an infinite value here and there, such as we get at a). In fact it makes sense for s complex, and we will think of F(s) as a complex-valued function of a complex variable s.
After the first few applications of the definition, everything will follow from properties of the Laplace transform. Here's a start.
LT0. L(f) is not changed if we alter the value f(a) for any given a.
LT1. L is linear: L(f+g)=L(f) + L(g), L(cf) = cL(f) (c constant).
Application: L(e^t+2e^{3t};s) = 1/(s-1) + 2/(s-3).
LT2. If f is piecewise linear and L(f)=0, then f(a)=0 whenever f is continuous at a.
Application: If f and g are piecewise continuous functions with the same Laplace transform, then f(a)=g(a) whenever f and g are both continuous at a.
LT3. L may be applied to complex-valued functions, and it respects real and imaginary parts.
Application: Take a = i omega in (1):
L(e^{i omega t);s) = 1/(s - i omega) = (s + i omega)/(s^2 + omega^2),
while on the other hand
L(e^{i omega t);s) = L(cos(omega t);s) + i L(sin(omega t);s),
so
L(cos(omega t);s) = s/(s^2 + omega^2) (2)
L(sin(omega t);s) = omega/(s^2 + omega^2). (3)
LT4. -(d/ds)L(f(t);s) = L(tf(t);s).
For the proof, move the diffentiation under the integral.
Application: Take f(t) = 1: L(t;s) = -(d/ds)(1/s) = 1/s^2. Take f(t) = t: L(t^2;s) = -(d/ds)(s^{-2}) = 2/s^3. Continuing:
L(t^n;s) = n!/s^{n+1} , n = 0,1,2,.... (4)
We'd like to extend this to nonintegral values of n. For this we return to the definition and use the change of variables u = st, assuming s greater than 0:
L(t^{a-1};s) = integral_0^infinity t^{a-1} e^{-st} dt
= integral_0^infinity (u/s)^{a-1} e^u du/s
= (1/s^a) Gamma(a) (5)
where
Gamma(a) = integral_0^infinity u^{a-1} e^{-u} du
is the famous Gamma function. Comparing this with (2), we find
Gamma(n) = (n-1)! , n = 1,2,....