00:01
Here we want to solve this differential equation for the given values of k, y, and the particular function f of t equals e to the t.
00:09
But first we're going to find the general form of the solution in terms of k and f, so then we can just substitute in and solve it.
00:16
And this will also help when we solve the next problem with a different function f.
00:22
So we're going to want to use an integrating factor.
00:30
And to do that, we'll first need to rearrange a little bit.
00:34
Just bringing over the y term.
00:41
And then remember for integrating factor, we take e to the integral of the expression in front of y and multiply both sides.
00:51
And our integrating factor will end up being e to the minus kt.
01:10
Okay.
01:11
And then remember, the trick is that this left -hand side is just the derivative of the product of the integrating factor.
01:21
And y.
01:30
So if we want an expression for y, we would need to multiply by e to the kt, and then take the integral of e to the minus kt times f of t...