It reduces the problem of solving differential equations into algebraic equations.For more information about the application of Laplace transform in engineering, see this Wikipedia article and this Wolfram article.So let's scroll down a little bit, just so we have some breathing room. And it actually turns out it's a sum of things we already know, and we just have to manipulate this a little bit algebraically.
In other words, we don’t worry about constants and we don’t worry about sums or differences of functions in taking Laplace transforms.
All that we need to do is take the transform of the individual functions, then put any constants back in and add or subtract the results back up. Okay, there’s not really a whole lot to do here other than go to the table, transform the individual functions up, put any constants back in and then add or subtract the results.
And I've gotten a bunch of letters on the Laplace Transform. It's hard to really have an intuition of the Laplace Transform in the differential equations context, other than it being a very useful tool that converts differential or integral problems into algebra problems.
But I'll give you a hint, and if you want a path to learn it in, you should learn about Fourier series and Fourier Transforms, which are very similar to Laplace Transforms. And that's good, because I didn't have space to do another curly L. So the Laplace Transform of y prime prime, if we apply that, that's equal to s times the Laplace Transform of-- well if we go from y prime to y, you're just taking the anti-derivative, so if you're taking the anti-derivative of y, of the second derivative, we just end up with the first derivative-- minus the first derivative at 0.
The only difference between them is the “\( \)” for the “normal” trig functions becomes a “\( - \)” in the hyperbolic function!
It’s very easy to get in a hurry and not pay attention and grab the wrong formula. \[F\left( s \right) = \mathcal\left\ = - G'\left( s \right),\hspaceg\left( t \right) = \cosh \left( \right)\] So, we then have, \[G\left( s \right) = \frac\hspace G'\left( s \right) = - \frac\] Using #30 we then have, \[F\left( s \right) = \frac\] \[H\left( s \right) = \mathcal\left\ = - F'\left( s \right),\hspacef\left( t \right) = t\sin \left( \right)\] Or we could use it with \(n = 2\).Now, to use the Laplace Transform here, we essentially just take the Laplace Transform of both sides of this equation. So we get the Laplace Transform of y the second derivative, plus-- well we could say the Laplace Transform of 5 times y prime, but that's the same thing as 5 times the Laplace Transform-- y prime. I took this part and replaced it with what I have in parentheses.So minus y prime of 0-- and now I'll switch colors-- plus 5 times-- once again the Laplace Transform of y prime. So 5 times s times Laplace Transform of y, minus y of 0, plus 6 times the Laplace Transform-- oh I ran out of space, I'll do it in another line-- plus 6 times the Laplace Transform of y. I know this looks really confusing but we'll simplify right now.We’ll do these examples in a little more detail than is typically used since this is the first time we’re using the tables.Make sure that you pay attention to the difference between a “normal” trig function and hyperbolic functions.Now we're just taking Laplace Transforms, and let's see where this gets us. So I get the Laplace Transform of y-- and that's good because it's a pain to keep writing it over and over-- times s squared plus 5s plus 6. Because the characteristic equation to get that, we substituted e to the rt, and the Laplace Transform involves very similar function. What I'm going to do is I'm going to solve this.And actually I just want to make clear, because I know it's very confusing, so I rewrote this part as this. I'm going to say the Laplace Transform of y is equal to something. We haven't solved for y yet, but we know that the Laplace Transform of y is equal to this. Due to the nature of the mathematics on this site it is best views in landscape mode.If your device is not in landscape mode many of the equations will run off the side of your device (should be able to scroll to see them) and some of the menu items will be cut off due to the narrow screen width.And that'll actually build up the intuition on what the frequency domain is all about. So let's say the differential equation is y prime prime, plus 5, times the first derivative, plus 6y, is equal to 0. So what are the Laplace Transforms of these things? Notice, we're already using our initial conditions. And then we end up with plus 5, times-- I'll write it every time-- so plus 5 times the Laplace Transform of y prime, plus 6 times the Laplace Transform of y. So just to be clear, all I did is I expanded this into this using this.Well anyway, let's actually use the Laplace Transform to solve a differential equation. And you know how to solve this one, but I just want to show you, with a fairly straightforward differential equation, that you could solve it with the Laplace Transform. So the Laplace Transform of 0 would be be the integral from 0 to infinity, of 0 times e to the minus stdt. Well this is where we break out one of the useful properties that we learned. I think that's going to need as much real estate as possible. So we learned that the Laplace Transform-- I'll do it here. The Laplace Transform of f prime, or we could even say y prime, is equal to s times the Laplace Transform of y, minus y of 0. So how can we rewrite the Laplace Transform of y prime?