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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.
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 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.
However, if we combine the two terms up we will only be doing partial fractions once.
Not only that, but the denominator for the combined term will be identical to the denominator of the first term.
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.
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.