00:01
We are told that the general solution to this differential equation is this function, and then we're given these initial values of y of 1 and y prime of 1, and we're asked to solve for c1 and c2.
00:14
So let's take advantage of this initial condition first.
00:17
If we plug in 1, then we get c1 e to the negative 5 plus, so 5 times 1 minus 1, so we get 4c2, and this is equal to 0.
00:35
And then to take advantage of this initial condition, let's take the derivative of y with respect to t, so we get negative 5c1 e to the negative 5t, and then this will just be plus 5c2.
00:55
So then we're told that y prime of 1 is equal to 2, but it's also equal to just negative 5c1 e to the negative 5, and then plus 5c2.
01:12
And we're told that this is equal to 2.
01:15
Okay, then we have these two equations here.
01:21
So there's more than one way to solve them.
01:24
That is to solve for c1 and c2.
01:25
One way is using this equation...