00:01
So we have an initial value problem here, consisting of a differential equation, y ' is equal to f of x and y, in this case f of x and y is 2x times y squared, with y of 0 equals 1 fourth.
00:14
I'd like to classify it, verify a solution, and talk about its existence, or its uniqueness, rather.
00:23
I'm going to classify it, what is its degree? its degree is 1, because the greatest derivative here we see is the first derivative of y.
00:30
Is it autonomous? it is not autonomous, because you'll notice that f of x, y is not equal to just a function g of y, or more precisely, the derivative of f of x and y with respect to x is not uniquely equal to 0.
00:50
Just say it depends on x, and that makes it not autonomous.
00:55
And it is nonlinear, because a linear equation looks something like y ' equals f of x times y plus g of x, which is just a function of x times y, plus a function of x times, just plus a function of x, but here we're multiplying y by itself, so it's not linear in y, which makes it a nonlinear differential equation.
01:18
So we've got that.
01:21
Now we want to show that y of x here, so that's the initial value problem.
01:26
Notice that y of x is equal to 1 over x composed with 4 minus x squared...