 So in this video, I want to solve the differential equation y prime equals y And so said it stated in another way. I want to find a function which is equal to its own derivative Ah, that sounds a little bit familiar, right? Although we haven't yet in this series developed the tools necessary to solve a wide variety of differential equations We are still able to solve this differential equation using Some mathematical facts that we already know. Can we think of a function who is equal to its own derivative? And hopefully we can think of one We've seen it we've seen it before right take y equals e to the x, right? Because e to the x is this really important function in calculus because e to the x is equal to its own derivative And as such it's a solution to this differential equation the left-hand side and the right-hand side are both e to the x it works and more generally We get that our general solution our general solution here Would just be y equals c e to the x Because if you take this function right here and you take its derivative you get c e to the x by our usual calculus facts with derivatives And so it might not be apparent yet, but it turns out that we'll see later in this in this chapter This this this unit about differential equations that the fact that e to the x is its own derivative that it solves is very basic differential equation This is actually the reason why the number e is a so-called Natural exponential right the natural log of what's so natural about e what's natural about e is that e to the x Solves this fundamental differential equation a function which is equal to its own derivative We'll see very shortly why that's so important. Alright, let's take a look at an example of a difference equation That might not be obvious at the get-go to us So in this example take dy over dx equals 3x squared minus 2x or another word y prime equals 3x squared minus 2x now in this situation We need to look for a function y which solves this differential equation But in this differential equation what we have here is we know that the derivative of y is 3x squared minus 2x That is the derivative y I can actually express without any reference to y itself In this situation If you take this dy over dx Actually, I'll just put it up here Instead of using this y prime notation will use the Leibniz notation dy over dx I kind of like this notation here because it looks like a fraction and as such it could be manipulated like a fraction if You times both sides of the equations by a dx you get dy equals 3x squared minus 2x dx like so The right-hand side I have to I have a dx and I have a function with respect to x I could integrate the right-hand side with respect to x and I could integrate the left-hand side with well Just with respect to dy there with respect to y the left-hand side would equal y and the right-hand side Let's see we get 3x squared its antiderivative is x cubed we get negative 2x its antiderivative x squared And there is a plus C. I don't know what that constant would be but notice right here We've now found the general solution to this differential equation right here This right here is a differential equation, right? We're looking for a function whose derivative is 3x squared minus 2x as we try to find the Antiderivative we actually solved a differential equation So one thing I want you to realize here is that solving Our calculate antiderivatives is actually a type of differential equation More generally if you have the equation the differential equation y prime equals f of x then the solution It's going to be y equals the antiderivative of f of x that is y equals the integral of f of x dx And don't forget the plus C because we want the general solution If you if you have no plus C there you're actually taking a particular solution You're assuming C is equal to zero and so what I want to point out to you is that as we in this series have been Calculating antiderivatives we've actually been solving Differential equations and so I want to make an analogy here and calculus one you learned about this idea called Explicit differentiation, but honestly explicit differentiation. We didn't learn about Well, let's pause right there. We didn't learn about explicit differentiation until we started learning about implicit Implicit differentiation Right the idea here is this if you had a specific function y equals f of x we could calculate the derivative dy over dx Through a combination of rules like the chain rule product rule, etc Now on the other hand implicit differentiation. You don't have the equation Y equals f of x that's actually forbidden to us, but we can still find y prime Y prime is yada yada yada using basically the chain rule. We're able to do this and So while implicit differentiation was a little bit different We could find the derivative even if we didn't have the function y equals yada yada yada like we do in the explicit case here Now differential equations are essentially What we mean by implicit Anti-differentiation All right, that's what our differential equations are really turning out to be in this situation The examples of antiderivatives we've been doing thus far in this series would be what we call explicit anti Differentiation finding an antiderivative when the function y equals Y equals f of x is given to you. I should say y prime equals f of x. We know the derivative explicitly Let's find the function. Well, if we don't know the derivative explicitly we could still know it implicitly by a differential equation and Thus what we've done in the past is actually a special case of what we're trying to do now Let's look at let's look at a Initial value problem associated to this Let's find the particular solution of the differential equation dy over dx minus 2x equals 5 when Given that y equals 2 when x equals negative 1. That's our initial value. We'll come back to that in a second Initial initial value We will we'll come back to the initial value when we're near the end We have to first find the general solution and it turns out that we're basically there already y prime minus 2x equals 5 While this is a differential equation. It's not an explicit Differential equation that is we just have to find the antiderivative It is implicit but with a little bit of algebraic manipulation We could just add 2x to both sides, right and we end up with y prime Equals 2x plus 5 and since we can solve for y prime as a function of x then taking the antiderivative Take the antiderivative 2x plus 5 here dx We can end up with y equals x squared plus 5x plus c, right? And so we have a general solution The general solution but we need to find a particular solution that's where this initial value comes into play right here So we know that when y equals 2 x that is y will equal to when x equals negative 1 So we get negative 1 squared plus 5 times negative 1 plus c we can solve for c here We get 2 equals 1 minus 5 plus c That is c minus 4 Let's add 4 to both sides. We get c equals 6 2 plus 4 and hence our particular solution should equal y equals x squared plus 5 x plus 6 Pretty simple. We've been doing this type of stuff many times previously, right this idea of solving Solving an explicit anti differentiation problem is what we have been doing. It's just a special case of this Implicit anti differentiation aka differential equations that we've tried to do now