# Laplace transforms and ODE

1. Jun 11, 2009

### Niles

Hi all.

Lets solve the following ODE by using Laplace transforms

$$y'' - y' = \exp(-x),$$

where y depends on x. Laplace-transforming we obtain (where Y denotes the Laplace-transformed of y)

$$s^2Y(s)-sy(0)-y'(0) - sY(0) - y(0) = \ldots$$

My question is, why we are allowed to do this? Because when I Laplace transform y'', then we get some expression dependent on the variable s, but when we Laplace transform y', then we get an expression dependent on some other variable t, and likewise for the RHS. Who says that s=t? And then there's also the variable from transforming the RHS.

Last edited: Jun 11, 2009
2. Jun 11, 2009

### jeffreydk

I think you got caught up in some simple confusion. The Laplace Transform transforms a function of some variable (it could be x or t or whatever) to a function of s by the rule
$$F(s)=\int_{0}^{\infty}f(t)e^{-st}dt$$
So in your case $f(x)\longrightarrow F(s)$ and your equation will go

$$\left[s^2Y(s)-sy(0)-y'(0)\right]-\left[sY(s)-y(0)\right]=\frac{1}{s+1}$$

The RHS since the Laplace Transform of an exponential goes
$$e^{\alpha x}\longrightarrow \frac{1}{s-\alpha}$$

3. Jun 11, 2009

### Mute

when you do something like a Laplace transform, you don't do it to each term separately - what you're doing is:

LHS = RHS

$$\Righarrow \mathcal{L}[\mbox{LHS}] = \mathcal{L}[\mbox{RHS}]$$.

This is why the variable is the same for each term: the laplace transform is linear, so although you're transforming the whole LHS, you can split it up into transforms of each of the terms on the LHS, so they all have the variable s. Similarly for the RHS - you had to apply the transform to both sides of the equation, so it has the same variable.

It's pretty much the same idea as

if a + b = c, then (constant)(a + b) = (constant)a + (constant)b = (constant)c

You would never say (constant1)a + (constant2)b = (constant3)c, as in general that just wouldn't be true! From the first equation, the constants must be the same, just as for your Laplace transform case the transform variable in each case must be the same.

4. Jun 11, 2009

### Niles

I see, thanks. I guess I was momentarily confused.

Thanks for clarifying.

5. Jun 11, 2009

### BobbyBear

Um, I think Niles's question has been answered from the point of view he meant it, so I don't mean to create any confusion, it's just that the question:

also makes me think of it in the following terms:

to be able to take the Laplace transform of a funcion, we have to make certain assumptions about the function (that it is of exponentional order or some such:P), so, by taking the Laplace transform of the whole equation, we are making assumptions about the function y and its derivatives, right? So for a general linear differential equation, aren't we limiting the solutions we can obtain using the Laplace Transform? What if the actual solution of the differential equaion is such that it does not have a Laplace Transform? What happens when in such a case we try using this method?

Plus of course we need the non-homogenous term of the equation to be Laplace Transformable to even be able to attempt to apply this method. . .

Last edited: Jun 11, 2009
6. Jun 14, 2009

### HallsofIvy

I don't understand why you say "when I Laplace transform y'', then we get some expression dependent on the variable s, but when we Laplace transform y', then we get an expression dependent on some other variable t". The Laplace transform of of f(x) is, by definition,
$$L(f)(x)= \int_0^\infty e^{-sx}f(x)dx$$
I see no reason why you would get "t" as the variable for one transform and "s" for the other. Essentially, what you are doing is multiplying both sides of the differential equation by $e^{-sx}$ and then integrating with respect to x. Doing that, all Laplace transforms resulting will have variable "s". Of course, you could use $e^{-tx}$ and have variable "t" but still all Laplace transforms will have the same variable.

I must confess that I consider the whole "Laplace Tranform" method to be more trouble than it is worth! It only works for linear equations with constant coefficients for which there are easier methods. In any case, it can be shown that all solutions to such equations are "of exponential order"- exactly the kind that have Laplace Transforms.

Last edited by a moderator: Jun 14, 2009
7. Jun 14, 2009

### Niles

It was a moment of confusion, and I wasn't thinking. A simple argument would just be that the Laplace transform is linear (which got mentioned). I have to think more carefully about these matters before asking questions from now on.

8. Jun 15, 2009

### BobbyBear

mhm, indeedy *chucks tables of Laplace Transforms into recycling bin* :P

9. Jun 15, 2009

### BobbyBear

Oh, but do you know how much you learn from overcoming moments of confusion!
I say keep posting, I learnt a few things from this thread too :)

"Je pense, donc je suis"
(I think, therefore I am)
— René Descartes

10. Jul 11, 2009

### Petar Mali

When we use
$$F(s)=\int_{-\infty}^{\infty}f(t)e^{-st}dt$$?

And in which cases is better use Laplace and in which Fourier transformation?