*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.\[\left( \right)Y\left( s \right) s - 12 = \frac\] Solve for \(Y(s)\).

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.

And we could get rid of this right here, because we've used it as much as we need to. And notice, using the Laplace Transform, we didn't have to guess at a general solution or anything like that. 5 times-- this is 2 right here-- so 5 times 2, plus 6 times the Laplace Transform of y. Now, let's group our Laplace Transform of y terms and our constant terms, and we should be hopefully getting some place. Notice that the coefficients on the Laplace Transform of y terms, that those are that characteristic equation that we dealt with so much, and that is hopefully, to some degree, second nature to you.

Even when we did a characteristic equation, we guessed what the original general solution was. So we get s squared, times the Laplace Transform of y-- I'm going to write smaller, I've learned my lesson-- minus s times y of 0. y of 0 is 2, so s times y of 0 is 2 times s, so 2s, distribute that s, minus y prime of 0. So minus 3, plus-- so we have 5 times s times the Laplace Transform of y, so plus 5s times the Laplace Transform of y, minus 5 times y of 0. So let's see, my Laplace Transform of y terms, I have this one, I have this one, and I have that one. Well let me factor out the Laplace Transform of y part. So let's see, I have 1s, so minus 2s, minus 3, minus 10, is equal to 0. So that's a little bit of a clue, and if you want some very tenuous connections, well that makes a lot of sense. And actually, let me just give you the big picture here, because this is a good point.

I've been doing a ton of videos on the mechanics of taking the Laplace Transform, but you've been sitting through them always wondering, what am I learning this for?

And now I'll show you, at least in the context of differential equations. And those are excellent questions and you should strive for that.We are trying to find the solution, \(y(t)\), to an IVP.What we’ve managed to find at this point is not the solution, but its Laplace transform.So, in order to find the solution all that we need to do is to take the inverse transform.Before doing that let’s notice that in its present form we will have to do partial fractions twice.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?These are going to be invaluable skills for the next couple of sections so don’t forget what we learned there.Before proceeding into differential equations we will need one more formula.We will need to know how to take the Laplace transform of a derivative. \(f^\) are all continuous functions and \(f^\) is a piecewise continuous function.First recall that \(f^\) denotes the \(n^\) derivative of the function \(f\). Then, \[\mathcal\left\ = F\left( s \right) - f\left( 0 \right) - f'\left( 0 \right) - \cdots - s\left( 0 \right) - \left( 0 \right)\] Since we are going to be dealing with second order differential equations it will be convenient to have the Laplace transform of the first two derivatives.

## Comments Laplace Transform Solved Problems

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## Marcel B. Finan Arkansas Tech University All Rights Reserved

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## Laplace Transform to Solve a Differential Equation, Ex 1, Part 1/2.

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