00:01
Okay, so we can write the differential equation, dq, dt, is equal to k times, well, alpha minus q, times beta minus q, where alpha is greater than beta.
00:38
Svea of alpha is greater than beta.
00:45
And our initial condition is, well, q of zero.
00:49
So q of zero is equal to zero.
00:53
Okay.
00:54
So separating the variables and integrating, we've got the integral of the q over alpha minus q times beta minus q is equal to.
01:18
To the integral of k d t.
01:25
Okay, so how do we solve this integral? we can use well partial fractions.
01:30
So we've got, um, whoops, we've got one over, well, alpha minus q times beta minus q is equal to, well, some constant a over alpha minus q, alpha minus q plus b over beta minus q.
02:04
Okay.
02:07
So what we've got here, we've got, well, one is equal to, so we get one is equal to a times beta minus q plus b times alpha minus q.
02:28
Okay, so we got one plus one plus four.
02:32
Well, q times, so we get one plus, well, q times zero, is equal to a beta plus b alpha minus q times a plus b.
02:57
Okay.
02:58
So combining like terms, we get a beta plus b alpha equals 1.
03:06
So we've got a beta plus b alpha is equal to 1.
03:19
And we have a plus b, we have a plus b, is equal to 0...