00:02
All right, so here we're just verifying that this function, which is 2 times e to the power of negative t plus t minus 1, is a solution to the differential equation given by y dot is equal to t minus y, and the initial condition, y0 is equal to 1.
00:38
And of course, this is the case where y is going to be equal to, our function f here.
00:43
So there's two parts to verify.
00:46
The first thing is let's just do the initial condition.
00:49
It's nice and easy.
00:51
So 1 .0 is going to be f of 0.
00:54
Okay, so that's going to be equal to 2 times e to the power of 0 plus 0 minus 1.
01:00
Okay, well, e to the power of 0 is equal to 1.
01:04
So that's just going to be 2 times 1 is 2 minus 1 is equal to 1.
01:09
Okay, so right now we're getting that these two things match, that's a good sign.
01:15
And then next we want to verify that y dot is equal to t minus y.
01:19
So to find y dot, we need to take the derivative of our function f.
01:25
Okay.
01:26
And so we can just take the derivative of each of these added expressions one at a time...