Laplace Transform Solved Problems

\[\begin\mathcal\left\ & = s Y\left( s \right) - y\left( 0 \right)\\ \mathcal\left\ & = Y\left( s \right) - sy\left( 0 \right) - y'\left( 0 \right)\end\] Notice that the two function evaluations that appear in these formulas, \(y\left( 0 \right)\) and \(y'\left( 0 \right)\), are often what we’ve been using for initial condition in our IVP’s.

So, this means that if we are to use these formulas to solve an IVP we will need initial conditions at \(t = 0\).

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.

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.

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.

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.

And so I get the Laplace Transform of y, times s squared, plus 5s, plus 6, is equal to-- let's add these terms to both sides of this equation-- is equal to 2s plus 3 plus 10-- oh, that's silly-- plus 13.


Comments Laplace Transform Solved Problems

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