 Now, let's use the fact that we know now what a Laplace transform is and an inverse Laplace transform. We know how to deal with partial fractions. Let's solve this problem. y prime plus 3, y equals 13, sine of 2t, it's an initial value problem we have that the y of 0 equals 6. Can I use the Laplace transform and the inverse Laplace transform, which is something I do to an equation, to see if I can solve this linear equation, first order linear differential equation non-homogeneous. Let's take the Laplace transform of both sides. I remember my linearity property. So if I take the Laplace transform of this left hand side, it means I take the Laplace transform of each of these. And then of that, I remember that I can take a constant outside of that Laplace transform. So what am I left with here? We've got the Laplace transform of y prime of t, in other words, dy dt or whatever. Let's leave it y prime plus 3 times the Laplace transform of this y. That's going to be 13 times the Laplace transform of the sine of 2t. Where does that leave? Remember right in the beginning we wrote that the Laplace transform of the f of t equals f of s. That's what we wrote. Let's use that same notation. So let's now write this as the capital Y of s. Now these are functions in t. They might as well be functions in x, whatever. We have it y of s. That is the Laplace transform of y prime. Is it ys though? This is Sydney ys there, but can you remember what the Laplace transform is of the first derivative? Can you remember what the Laplace transform was of the first derivative? That was s times ys minus the f of 0. Wasn't it? That was the Laplace transform of the y prime of t. So we've got to remember that one. So in here I'm going to write s times ys minus y of 0 plus 3 times the Laplace transform of ys, ys. And that equals 13 times, what is the Laplace transform of the sign of 2t? Well that is 2 over s squared plus 4. 2 over s squared plus 4. Let's clear this up. Let's clear this up. What is y of 0? Y of 0 is 6. So I've got that. Let's clear this up. I have y of s. Y of s, I can take that out as a common factor and I'm left with s plus 3 on this side minus y of 0 is 6. That equals 26 over s squared plus 4. In other words, y of s equals, I've got this 26 over s squared plus 4. And I've got the plus 6, plus 6 there and I'm dividing both sides by s plus 3. I'm dividing both sides by s plus 3. Let's just rewrite that. Let's get these all on a common, this on a common s squared plus 4. So here I'm going to be left with 26 plus 6 times that. So that is 6s squared plus 24. And I've got to divide the whole log still by s plus 3. I'm going over this fairly quickly, this basic school algebra. So I'm not going to get stuck with that. So that equals, what does that equal? 6s squared plus 4t divided by s plus 3 and s squared plus 4. And there you have it. Now just look at what happened here. y of s equals that. But what did the Laplace transform do for me? The Laplace transform changed a differential equation into normal algebra. Normal algebra. I have y of s, s, so it might as well have been y and an x. I'm back to school algebra. So the beauty of Laplace transform is it takes me from a differential equation into a standard algebra. I have the f of x there and I have my x values there. There's s, which is just x. I'm going back to y equals x. Nothing different from this. So here I have basic algebra. I have the y of s equals this. Now in the previous video I showed you this partial fraction, this fraction written as partial fractions. So let's get on with that. What if I now have this? I have y of s, y of s on this side and on that side of the equation I have that. Let me take the inverse Laplace transform of both sides. The inverse Laplace transform of both sides. I remember my linearity property. So let's take the inverse Laplace transform of each of these and bring the constants out. So that's going to be 8 times the inverse Laplace transform of 1 over s minus negative 3. I'm writing it like that. Negative times negative is positive. Minus 2 times the inverse Laplace transform of... It's over. What about here? It's 3 plus 4. It's permanent. After the phone interruption I was left with taking the inverse transform of Laplace transform of both sides. This being the constant out, submitting the linearity property so I can take it of each one individually. And where was I? Plus 3 times the inverse Laplace transform of 2 over s squared plus 4. I've got to move this. If I take the inverse Laplace transform, remember this is the inverse Laplace transform of the Laplace transform of t. So what happens if I take the inverse Laplace transform of the Laplace transform and I'm just left with the f of t? I'm just left with the f of t. So on this side I've got 8. Here I've got 8 equals negative 3. So that's e to the power of negative 3t minus 2 times. Here I have my k equals 2. k equals 2 there. So it's s over s squared plus k squared. What is that? That's the cosine of kt. So that's the cosine of 2t plus I'm left with 3 times. Here I have my k equals 2 as well. 2, so it's k over s squared plus k squared. That leaves me as a sine of 2t. And there beautifully I've used the Laplace transform to change my differential equation into a algebraic equation. Normal, normal, normal algebra. I used my initial value problem. It was an initial value problem so at least I could get rid of this. I then had this fraction which I then had to use partial fractions as in the first video to rewrite it as 3 different fractions. And then I just reversed the whole thing because I took the Laplace transform. I then take the inverse Laplace transform and I'm beautifully left with an answer. That is fantastic. That is absolutely phenomenal.