Differential equations and Fourier and Laplace transforms
I'll talk about the Laplace Transform. What I find important about the Laplace Transform in differential equations is that it makes operational calculus rigorous when dealing with DEs.
A well known engineer and mathemathician, Heaviside, considered $D = \frac{d}{dx}$ as an operator acting on $y$ to produce differential equations. He then thought about the following:
$$y' = f(t) $$
can be put as
$$Dy = f(t) $$
The if we thought about $\frac{1}{D}$ as the inverse operation of $D$ we would algebraically think about:
$$y = \frac{1}{D}f(t) $$
as the solution of our problem. But then you'd have the equivalence
$$ \int f(t) dt = \frac{1}{D}f(t) $$
So what about:
$y'+y = f(t) $
$(D+1)y = f(t) $
Our solution would now be
$$y = \frac{1}{D+1}f(t)$$
We know that if we use the integrating factor $e^t$ we have
$(e^t y)' = e^t f(t)$
Thus we finally get
$$y = e^{-t} \int {e^tf(t)dx}$$
As a consequence we might want to define
$$\frac{1}{D+1}f(t) = e^{-t} \int {e^tf(t)dt}$$
And if we think more generally, we would get
$$\frac{1}{D-m}f(t) = e^{mt} \int {e^{-mt}f(t)dt}$$
Again, what if we have a second order equation?
$$y''-y = f(t)$$
This means
$$(D^2-1)y = f(t)$$
And thus
$$ y = \frac{1}{D^2-1}f(t)$$
But what does this mean? Let's be blunt and write.
$$ \frac{1}{D^2-1} = \frac{1}{2} \left(\frac{1}{D-1}-\frac{1}{D+1}\right)$$
Thus this would mean the solution is
$$ y =\frac{1}{D^2-1}f(t) = \frac{1}{2} \left(\frac{1}{D-1}f(t)-\frac{1}{D+1}f(t)\right)$$
And from our last approach
$$ y =\frac{1}{D^2-1}f(t) = \frac{1}{2} \left(e^{t} \int {e^{-t}f(t)dt}-e^{-t} \int {e^{t}f(t)dt}\right)$$
Let's try this method with
$$y''-y = \sin t$$
Our solution would then be
$$y = \frac{1}{2}\left( {{e^t}\int {{e^{ - t}}\sin tdt} - {e^{ - t}}\int {{e^t}\sin tdt} } \right)$$
Just check this by yourself
$$\int {{e^t}\sin tdt} = \frac{1}{2}{e^t}\left( {\sin t - \cos t} \right) + {c_0}$$
$$\int {{e^{ - t}}\sin tdt} = - \frac{1}{2}{e^{ - t}}\left( {\sin t + \cos t} \right) + {c_1}$$
Then, after an algebraic manipulation you will get:
$$y = {c_1}{e^t} - {c_0}{e^{ - t}} - \frac{1}{2}\sin t$$
Which clearly satisfies the equation and $c_1$ and $c_2$ are to be determined.
This manipulations motivated a formal thoery of operational calculus that proved to be consistent by the analsis of the Laplace Transform. Note the similarities of the appereance of $$f(t) = e^t \Rightarrow F(s) = \frac{1}{s-1}$$ and the fact $e^t$ satisfies $(D-1)y = 0$. Similarily $$f(t) = e^{-t} \Rightarrow F(s) = \frac{1}{s+1}$$ and $e^{-t}$ satisfies $$(D+1)y = 0$$ And (!) you have that $$\sinh(t) = f(t) \Rightarrow F(s) = \frac{1}{s^2-1}$$ and this function satisfies $$(D^2-1)y=0$$
For more info on this check Spiegel's Applied Differential Equations (207-218), where you'll find theorems such as:
Let $\phi(D)$ be a polynomial in $D=\frac{d}{dx}$. Then
$$ \phi(D) \{ e^{at}f(t) \} = e^{at}\phi(D+a)\{f(t)\}$$
Another connection, and I guess this is the strong one is (Spiegel, p. 284)
$$\mathcal{L}^{-1}\left\{\frac{F(s)}{s-a}\right\} = e^{at} \int_0^t {e^{-au} f(u)du}$$
(Use the convolution theorem)
Rings a bell?
$$\frac{1}{D-m}f(t) = e^{mt} \int {e^{-mx}f(x)dx}$$
EDIT: The expression of $\phi(D)$ as linear factors $(D-p_1)$ can be done as long as the polynomial coefficients are constant. (This stems from the fact that $D$ isn't associative or commutative , i.e. $$(D f) g \neq D (fg)$$ and $$ D f \{\} = f' \neq f D\{\}$$ where you'd enter a function of $t$ inside $\{\}$.
I add: In the same manner the Fourier Transform makes an PDE an algebraic equation, as you state (I know almost nothing about the FT), the Laplace Transform makes an ODE an algebraic equation.
Check google:
- Answered here in another question
- and here has a great pdf explaining the difference.
- The Straight Dope answer
If you gave more details or be more specific it would help to get a more detail and quality answer. From you question, I think the links above will help you. The pdf is a good read.
I think the question asks for a practical answer. Electrical engineering furnishes some useful examples.
The use of Laplace and Fourier transforms allows for the solution of linear constant-coefficient integro-differential equations using little more than algebra and a table of transforms. For example,
$\frac{dy(t)}{dt} + 2y(t) + \int_{0}^{t}y(\lambda)e^{-2(t-\lambda)} d\lambda = 10u(t);$
$y(0) = 0$.
By consulting a table of transforms, taking advantage of the behavior of the equation under transformation to the "s" domain (in particular that convolutions in the time domain tend to correspond to simple multiplication in the "s" domain; this gives us enormous leverage), we usually get a rational expression whose partial fraction expansion is easy to transform back into the time domain.
[The problem I noted is solved in Examples 12.7,12.9 in Irwin, Basic Eng. Circuit Analysis at 506.]
The combination of integrals and derivatives that results from even simple AC circuit analysis often defies solution by other means, so in a sense the answer to the question is practical: any other method would be hideous. Even math texts (e.g., Diprima, Elem. Diff. Eq. at 279) tend to introduce Laplace transforms as "useful tools" for solving linear differential equations. The motivation is utilitarian. In a nutshell: a difficult calculus problem is transformed into a tractable algebra problem.
As for Laplace versus Fourier transforms, the short answer is that Laplace transforms are a generalization of Fourier transforms in the sense that there are Laplace transforms of functions which do not have Fourier transforms. By looking at the definition of the Laplace transform you can see that it addresses situations for t > 0 and that s = $\sigma + j\omega$. Comparing this with the definition of the Fourier transform--and comparing tables of transforms for each--should give you a sense that they are used in the same way but the particular problem would dictate the choice of method.