# Laplace Transform Solved Problems

And actually, you end up having a characteristic equation. Well, we could use this once again, so let's do that.And the initial conditions are y of 0 is equal to 2, and y prime of 0 is equal to 3. So this over here-- I'll do it in magenta-- this is equal to s times what? Well that's s times the Laplace Transform of y, minus y of 0, right?So, in order to find the solution all that we need to do is to take the inverse transform.

And actually, you end up having a characteristic equation. Well, we could use this once again, so let's do that.

And then I'm going to say, boy, what functions the Laplace Transform is at something? If that confuses you, just wait and hopefully it'll make some sense. Let's divide both sides by this s squared plus 5s plus 6. Now, if we just had this in our table of our Laplace Transforms, we would immediately know what y was, but I don't see something, or I don't remember anything we did in our table that looks like this expression of s.

From here until that point it's just some fairly hairy algebra. I think it's some kind of marketing phone call. So I get the Laplace Transform of y is equal to 2s plus 13, over s squared plus 5s plus 6. I'm essentially out of time, so the next video we're going to figure out what functions Laplace Transform is this.

$\mathcal\left\ - 10\mathcal\left\ 9\mathcal\left\ = \mathcal\left\$ Using the appropriate formulas from our table of Laplace transforms gives us the following.

$Y\left( s \right) - sy\left( 0 \right) - y'\left( 0 \right) - 10\left( \right) 9Y\left( s \right) = \frac$ Plug in the initial conditions and collect all the terms that have a $$Y(s)$$ in them.

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.

While Laplace transforms are particularly useful for nonhomogeneous differential equations which have Heaviside functions in the forcing function we’ll start off with a couple of fairly simple problems to illustrate how the process works.

The first step in using Laplace transforms to solve an IVP is to take the transform of every term in the differential equation.

It’s now time to get back to differential equations.

We’ve spent the last three sections learning how to take Laplace transforms and how to take inverse Laplace transforms.

## Comments Laplace Transform Solved Problems

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