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.

  • $\begingroup$ The method really helps nothing in solving differential equations, since one cannot write down explicitly the inverse Laplace transform. Anyways, one should remember Heaviside who apparently first devised a method that should be considered equivalent. Mathematicians came afterwards. $\endgroup$
    – John B
    Aug 11, 2017 at 10:50
  • $\begingroup$ Sure. That makes sense, but it is a method that nevertheless works in differential equations, and I would like to know why exactly it works. Sure, the characteristic equation and the method of undetermined coefficients work too, where we have to be careful with the particular solution, making sure it's linearly independent from the homogeneous solution. But it seems the Laplace transform works in solving them too, and I can't see directly why that would be the case. $\endgroup$
    – 14tim4
    Aug 12, 2017 at 23:20
  • $\begingroup$ No, it almost never works, and for that reason it is impossible to answer to your question. You are assuming that you can always compute explicitly the inverse of the Laplace transform, but you almost never can. Even more important: just take the simplest nonlinear equation that you can think of... $\endgroup$
    – John B
    Aug 13, 2017 at 0:09
  • $\begingroup$ @JohnB Ahh I see your point! Would be it be good to clarify my question for the cases that do work? $\endgroup$
    – 14tim4
    Aug 13, 2017 at 0:13
  • 2
    $\begingroup$ Here's a link to a fantastic video on where the formula for the Laplace transform comes from: ocw.mit.edu/courses/mathematics/… $\endgroup$
    – BobaFret
    Aug 13, 2017 at 1:18

1 Answer 1


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.

  • $\begingroup$ The Fourier transform is its own inverse. The Laplace transform is a regularized Fourier transform in disguise. $\endgroup$
    – reuns
    Aug 13, 2017 at 12:09
  • $\begingroup$ Can't we just say that a laplace transform is essentially a 'definite integral ' and a definite integral operator on both sides does no harm. $\endgroup$
    – llecxe
    Mar 23, 2021 at 16:11

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