00:01
In this problem, we want to find a first order differential equation and the initial condition for y.
00:07
And so to do that, we're going to start by taking the differential of each side with respect to x.
00:14
So i'm going to take dy, dx, is equal to d, d, dx of negative 1, plus d, d, dx of this integral.
00:26
1 to x of t minus y of t d t d t and so we can start simplifying some things down here this d y d x stays the same the d d x of negative 1 will cancel out to 0 because there's no x term there negative 1 is just a constant and then for this integral the differential of an integral we can use the fundamental theorem of calculus and so i'm going to go ahead and write that out right here.
01:01
So the fundamental theorem of calculus states that if you have the differential of an integral from a, a constant a to x of f of t d t, that is equal to, sorry, that is equal to f of x.
01:21
And so here, if we look at this problem, one is our constant a...