Laplace transformations for dummies

There are beautiful video lessons at MIT Opencourseware. I'm particularly in love with this presentation of the Laplace transform.

I have been teaching the Laplace Transform to a night degree class (mature) of civil engineers. They are good students but not great mathematicians. They couldn't follow the method of how we use the Laplace Transform to solve differential equations until I told them this story:

Suppose that you come across a poem written in English of whose meaning you don't understand. However suppose that you know a French-speaking gentleman who is a master of interpreting poems. So you translate the poem into French and send it to the French gentleman. The French gentleman writes a perfectly good interpretation of the poem in French and sends this back to you where you translate it back into English and you have the meaning of the poem.

Obviously these are simple difficulties that these students are having but I still think it's a nice story.

D'accord:

Poem in English = Differential Equation. Interpretation in English = Solution of Differential Equation. Translate to French = Take Laplace Transform. Poem in French (better interpreter) = Algebraic Equation (easier to solve). Interpretation in French = Laplace Transform of Solution of Differential Equation. Translate back into English = Inverse Laplace Transform

I'm going to come at this one from left-field. In quantum mechanics, we deal with infinite dimensional vector spaces (Hilbert spaces), so I tend to think of integral transforms in those terms. For instance,

$$\int_{-\infty}^{\infty} K(x,y) f(y) dy = F(x)$$

can be thought of as

$$ \mathbf{K} f = F $$

and $x$ and $y$ from the first equation are the indices of the infinite dimensional vectors and matrix (kernel) $f$, $F$, and $\mathbf{K}$. Using that interpretation, if $\mathbf{K}$ is unitary then the integral is just a changing the bases of the function (Hilbert) space. In other words, the integral can be viewed as the decomposition of original vector, $f$, in terms the new basis. For Fourier transforms the kernel is unitary, and while not true of Laplace transforms, the idea of it being a change of basis still holds. It should be noted that unlike in the finite case, in the infinite dimensional case care must be taken to ensure that the transform actually converges, but that is another problem entirely.