 So let's try to solve this differential equation. So earlier we found the Laplace transform of y, where y of 0 was equal to 0. Now let's use this information to actually solve the differential equation. So we have our Laplace transform. So we'll pull in our library of Laplace transforms, and look for one that's 1 over s minus 1 times s plus 1, and we find there isn't one. Now it's possible our library isn't big enough. And in fact, this is a very small library. We can find much larger libraries without too much effort. And of course, because the Laplace transform of a function is a well-defined operation, we could find the Laplace transform of other functions and expand our library. This would be a major undertaking. Fortunately for this problem, we don't have to. A useful thing to remember is that the problem of solving a differential equation is very closely related to the problem of finding an antiderivative. And this means that most of the things we did to find antiderivatives will also apply to differential equations. And one of those techniques was a technique of integration by partial fractions. So using a partial fraction decomposition, we assume that this fraction can be expressed as a sum of fractions whose denominators are the individual factors. We'll multiply this out and then solve for a and b. Make sure the math police aren't watching because we're not allowed to do what we're about to do next, even though everybody does it. If all your friends integrated transcendental functions, would you do it too? If s equals negative 1, our equation becomes which tells us that b equals minus 1 half. Likewise, if s equals 1, our equation becomes which tells us that a is equal to 1 half. Quick, it's a math police! Get rid of the evidence! Nothing to see here, we just found a equals 1 half, b equals minus 1 half using rigorous mathematical techniques that are not involved in dividing by zero. And that gives us the partial fraction decomposition of a rational expression. We'll rewrite this a little bit, and then we'll look to see if our fractions show up in our library. And 1 over s minus 1 and 1 over s plus 1 are in our library. They give us the Laplace transforms. And since l is a linear operator, this gives us the Laplace transform of and this suggests that y itself is equal to 1 half e to power x minus 1 half e to minus x. And it's useful to remember we've already incorporated our initial condition, so this is the particular solution to our differential equation. Let's solve a much more complicated equation. How about this? So we'll apply our Laplace transform. Since the Laplace transform is a linear operator, the Laplace transform of a sum is the sum of the Laplace transforms. Since the Laplace transform is a linear operator, the Laplace transform of a constant times a function is that constant times the Laplace transform of the function. Our reduction formula allows us to rewrite the Laplace transform of a second derivative. Likewise, we can reduce the Laplace transform of a first derivative. And everything else is still there. We can do a little algebra, then rearranging and using our initial conditions. We can find the Laplace transform of e to power 2t directly by evaluating the integral. And everything else is the same, so now we can solve for the Laplace transform of y. Now one of our s plus 1 is the Laplace transform of a function in our library, so we could leave that alone. For the rest, we'll have to rewrite our rational expression using partial fractions. So we'll assume that our fraction can be rewritten as a sum of fractions with denominators s minus 2, s plus 1, and s plus 1 squared. Rearranging. So if s equals negative 1, our equation becomes solving. If s is 2, our equation becomes, and we can solve this for a. And to find b, we can pick any other value of s. So if s is equal to 0, we can substitute this and our known values for a and c to get the equation. And so the Laplace transform of y is equal to this sum of fractions. And we can simplify, look up these Laplace transforms in our library, use the linearity of the Laplace transform to simplify, and so that tells us the Laplace transform of y is the Laplace transform of e to 2x minus 3x e to minus x, and so that suggests that y is equal to this function. And again, we've already incorporated our initial conditions, and so this should be regarded as the particular solution to this differential equation.