Laplace Transform - Why it works The definition of the Laplace Transform is $$F(s) = \int^{\infty}_0 f(t)e^{-st}  dt$$
It is very useful in terms of solving linear, constant coefficient ordinary differential equations, but why exactly does it work? Why does taking the Laplace transform of each term in in the differential equation, using the Laplace Transform's linearity to get each individual term in the differential equation in its own Laplace Transform work in solving differential equations?
Why does it work?
I am not so sure why I am hesitant to accept this. Having seen generating functions of recursive sequences as a way to describe the sequence of numbers  and aids in obtaining a (closed) formula for the recursive sequence - this is a similar analogy, we are essentially "transforming" sequences to a power series, but I am more accepting of this idea than the Laplace transform. 
 A: The two key properties that make Laplace transform so useful for solving certain ODEs are linearity and the relationship between the Laplace transform of a function and its derivative https://en.wikipedia.org/wiki/Laplace_transform $$\mathcal{L} \left\{f(t)\right\} = \int_{0^-}^\infty e^{-st} f(t)\, dt \\
                                 \left[\frac{f(t)e^{-st}}{-s} \right]_{0^-}^\infty -
                                       \int_{0^-}^\infty \frac{e^{-st}}{-s} f'(t) \, dt \\
                                   \left[-\frac{f(0^-)}{-s}\right] + \frac 1 s \mathcal{L} \left\{f'(t)\right\}$$ where use is made of integration by parts.
By this property, applied to each term of a linear, constant-coefficient ODE, you can communte for a differential equation to an algebraic one, hopefully easier to solve.
So, this is the "mechanics" of it: apply the transform to both sides of an ODE, use linearity, and convert from derivatives to powers.
As to "why" it works, one is tempted to say that it does so, as a result of the fact each operation above can be shown to be valid (although there might be a "circular argument" flavour hanging around here).
If you convicne yoursleve that linearity and the above relationship between transforms of a function and its derivative apply, by looking at the simple proofs, there should not be additional conceptual obstacles towards appreciating how it works. If I missed some point I would gladly expand upon it. 
