00:01
That y is equal to the natural log of x plus the integral from x to e of the square root of t squared plus y of t squared d t.
00:09
So this question is asking us two different things.
00:12
First is asking us to find the equivalent first order differential equation of y and to find initial condition of y.
00:20
So for the first part differential equation, i want to take the derivative of both sides.
00:24
However, i'm seeing something that's a little bit off.
00:28
I know for this integral when i take the derivative of it, i want to use fundamental theorem calculus.
00:34
However, my problem here is that this x is on the lower bound.
00:38
I really want it to be on the upper bound.
00:41
However, what we learned from properties of integrals is that i can flip my bounds here if i take the negative of that integral.
00:52
So i take the negative, and now i can flip my bounds...