Okay, we have another concept of the solution, so a relation Phi(x, y) = 0. We call it an implicit solution of the differential equation four, okay? On the interval I, if there is at least one function little phi(x), which satisfies both Phi(x) and phi(x), is identical equal to 0. And capital F(x), phi prime, and nth derivative of phi, is identically equal to 0 on the interval I. This second phrase means phi(x) is the explicit solution of differential equation four, right, Then we call Phi(x,y) = 0 as an implicit solution of differential equation four, okay? Let me show you a couple of examples, okay, for example, first, I claim that x squared plus y squared is equal to 4, right. This is the equation of a circle having its center at the origin and radius 2, okay. This is the implicit solution of simple first order differential equation xdx + ydy=0 on the interval (-2, 2). Let's rely to the differential equation the in another way for first. We are starting from xdx + ydy = 0, it simply means y' = -x over y, okay. Now let's look at the implicit function given going there, x squared + y squared = 4, right? This is implicit form of a function, so now we have a candidate of a solution given implicitly by this equation, right? That means y squared = 4- x squared, that means you can solve it for y, okay, as plus or minus the square root of 4- x squared, right. Now, from this implicit form of the function, we have two explicit form of the function. One is plus square root of 4- x squared, and the other one is y = minus square root of 4- x squared. You may call this one to be phi(x). To check that, okay, this is an implicit solution of that differential equation here, right. We must see that this is an explicit solution of this differential equation. Okay, so why not compute the y prime from here, then y prime is equal to, okay. Plus or minus, right, okay, one-half of 4- x squared to the negative one-half. And product, by the chain rule, -2 over x, right, are you following me, right? So this is equal to negative and plus, plus times negative, that is negative, negative times negative, that makes plus. This 2 over 2 cancels out, so you get x times 4- x squared and to the negative one-half, that's y prime, okay? Is it the same as -x over y, on the other hand, what is -x over y? That is -x times 1 over y, this is equal to, okay, 1 over y is equal to plus or minus 1 over square root of 4- x squared, right, okay. So that is negative plus over x times 4- x squared and to the negative one-half, all right? So from this computations, you can confirm it, these two are identical, right? In other words, okay, this implicit form of the function is a solution to implicit form of the solution of that differential equation, okay? We have two such choices, as we have seen it down here, either plus square root of 4- x squared, or negative square root 4- x squared. They are the explicit solution of the given differential equation, okay, Where does this interval come from, okay? As you know, okay, the inside of this square root sign should be non-negative, okay? In other words, 4- x square must be greater than or equal to 0, okay? Or equivalently, x must be between- 2 and 2, right, okay, they aren't excluding the two endpoints, -2 and 2, okay, because, Outside of those two endpoints, right, okay, outside of those two endpoints, -2 or 2, right. In other words, if x is less than -2 or x is greater than 2, okay, you don't have the function at all. And at the endpoints, okay, the differentiation is not well-defined. Or more precisely, only the one side of derivative is defined, so I just exclude the two end points down there, okay? So, okay, this x squared + y squared = 4 is an implicit solution of that differential equation on the open interval (-2, 2). As one more example, I claim that 4x squared- y squared = c, where c is an arbitrary constant. Is a one parameter family of implicit solutions of the nonlinear, can you see it? Nonlinear first order differential equation, yy'- 4x = 0, on suitable interval. Here we have implicit functions given by the 4x squared- y squared = c. And that means y squared = 4x squared- c, right, and that means y can be either plus or minus square root of 4x squared- c, okay. And as you can see, which is valid only when 4x squared- c is non-negative, or rather strictly positive, okay? Excluding those two endpoints, as I did in this example number five, okay? So on any interval satisfying this inequality, okay, the explicit form of the function y = plus or minus square root of 4x squared- c is an implicit form of the function, which is a differentiable, okay, And my claim is, this is an implicit form of the solution of that differential equation. Or rather, this explicit form of the function is a solution to this differential equation, okay. You can compute y' from this expression, and plug in y and y' into that equation, confirm that. Confirming that this is a solution, but I will do it in a little bit different way, now, okay? From this expression directly, without knowing what is y in terms of x, right?. You can compute the y' by the so-called implicit differentiation, okay?. So let's take a differentiation on both sides of this expression, right? I'm taking d over dx, right, then what will you get then? Derivative of 4x squared, that is 8x, derivative of y squared is enforced by the chain rule, 2y. And because y is a function of x implicitly by this relation, so you need y', okay? And derivative of any constant, that is equal to 0, right? Divide it through by 2, right, then finally, you will get, right, 4x- 2yy' = 0. I'm sorry, I'm dividing by 2 so I do not have this 2 in front, we have 4x- yy' = 0. That is exactly the given differential equation. That means we just confirmed that, through the implicit differentiation of this expression. We just showed that this implicit form of the function satisfied the given differential equation, on the interval where 4x squared- c is strictly positive.