 So in the previous video, we showed that the function y equals c, e to the x squared over two, is in fact a solution to the differential equation y prime equals xy. And we can do this for any, any parameter c whatsoever. This right here represents what we call the general solution of our differential equation, general solution. That is, this right here gives us the complete generality of what we could expect, where you could choose any parameter c right here, and that would give us a solution here. And one thing that you're gonna see is that for first order differential equations, you're gonna have a single parameter you can choose, and every solution will be based upon that single parameter. With second order, third order differential equations, you can have more than one parameter. Again, we won't get into that in this lecture series. I would invite you to watch a lecture series for Math 2280 at SUU to learn some more about that, or just other references online. But first order differential equations will have a single parameter general solution. Now, if a specific value for the parameter c is chosen, like in the example we did before, f of x was e to the x squared over two, that's choosing the parameter to be one. And another example we did was g of x equals three, e to the x squared over two, we chose the parameter to be two. In that situation, we get what's called a particular solution, a particular solution to the differential equation. And so a big part of differential equations is about trying to first, we have the first, when we have a differential equation, we try to solve the general solution. We have to find that. This is, of course, the hard part, and in future videos we'll talk more about how to find the general solution. What I wanna talk about now though is that if you have a general solution, how does one find a particular solution? How do we know if it's one for c or three for c or some other value? Well, if you have a general solution in hand, what you can do is you can use some initial condition that as you know something about the function, like maybe you know f of zero is equal to some specific number, we'll call it why not. And so if we have a general solution, we can find the particular solution using some initial condition, some initial value, and this is often referred to as the initial value problem. So in a previous video, we looked at the differential equation y prime equals one half y squared minus one. And we saw that the general solution to this differential equation was one plus c e to the t over one minus c e to the t. So we checked that this was a, we saw that this was a solution to the differential equation right here. As this is a first order differential equation, there'll only be the single parameter c. And so we found the general solution. What we wanna do now is actually find a particular solution using the initial value y of zero equals two. And so once you have a general solution again, that's the heavy lifting. This is pretty powder puff at this moment. Let us take the initial condition, we're gonna plug in t equals zero. We're gonna stick in t equals zero in these situations right here, right? And we're gonna stick in for y a two. That's all that one has to do. So y equals two exactly when t equals zero. So we go one plus c e to the zero over one minus c e to the zero. And now it comes to simplify these expressions and solve for c here. Notice that e to the zero is equal to one and one times c is just gonna be c. So you get two equals one plus c over one minus c. I would clear the denominators at this moment times the right-hand side by one over one minus c. You have to do it to the left-hand side as well. The one minus c's would cancel on the left. This then gives us two times one minus c. I'm gonna distribute that and get two minus two c. This would equal the right-hand side, which is one plus c. Adding some like terms, we're gonna add two c to both sides. We're gonna subtract one from both sides. We end up with one on the left-hand side and we're gonna get three c on the right-hand side in which case then we see that c should equal one third like so. And so if we take that and plug it into the form we had before, we're gonna end up with, so this again, this was our general solution. We're now plugging in specifically, we're gonna get one plus one third e to the t over one minus one third e to the t. Now what we've written here in yellow is a particular solution. It solves this differential equation, not just a general. It also solves it using the initial value y of zero equals two. Now, in this particular form, I don't really like fractions inside of fractions. I get kind of messy. So if you times the whole fraction by three over three, distribute these threes throughout, you end up with the format I actually like here, y equals, you're gonna get three plus e to the t over three minus e to the t. And so this actually represents the particular solution to this differential equation we're looking for. It's a solution to the differential equation. We saw that previously and it also will satisfy the initial condition. If you plug in x equal t equals zero right here, you're gonna get three plus one, which is four over three minus one, which is two, four over two is two. And so this solves our initial value problem. So if we can find the general solution, finding the solving the initial value problem is super easy that is finding the particular solution. It's a cinch. The hard part of course is finding the general solution. And so in the next videos in this series, we'll start to give some hints on how one finds solutions to differential equations as opposed to them just appearing in front of you.