Ignoring Jonathan's idiocy for the moment (and permanently if he
settles down about his nonsense), let's check to see if Tim got the
right answer to this set of coupled differential equations. We solve
these kind by taking the derivative of one of the equations so that
we'll end up with the same variables in both equations.
Let's take the derivative of dR/dt; since dR/dt is equal to -kB, we
can write the derivative of dR/dt as -kdB/dt. However, another way of
writing the derivative of dR/dt is d^R/dt^2. So we have
d^2R/dt^2 = -kdB/dt;
after dividing through by -k, we find that dB/dt = -(1/k)d^2R/dt^2
The second equation is another expression for dB/dt, though; we can
see from that equation that dB/dt is equal to -lR. So we'll set our
two expressions for dB/dt equal to each other, which will give us a
single differential equation that we can then solve using standard
techniques:
dB/dt = -lR = -(1/k)d^2R/dt^2
Multiplying through by -k gives
d^2R/dt^2 = klR
We now have one of the most fundamental relations in all of
differential equations. If you don't know the solution to this
equation by heart, it might be that remedial work could be in store
for you until you get up to speed.
The solution (that is, R) will be an exponential function, because
these functions have the special property that when you take their
derivative, you get the same thing back again. So an equation
involving a function whose second derivative is the same thing as the
function itself (to within a constant of proportionality) is going to
involve exponential functions.
Or, the way they would teach it in diffy-eq, I guess, if you wanted to
be more rigorous about it, would be, try a trial solution of the form
R = e^rt, because you suspect exponentials are involved.
Take its derivative once and you get
dR/dt = re^rt, by the properties of the exponential function
Take its derivative again and you get
d^2R/dt^2 = r^2e^rt;
but because our trial solution was e^rt, this equation is the
same one as the one we were trying to solve if we set r^2 = kl; that
is,
d^2R/dt^2 = r^2 R is the same thing as
d^2R/dt^2 = klR
if kl is the same thing as r^2
So let's write that last one out as an equation so we can write our
solution in terms of the variables given us:
kl = r^2, so
r =(+/-) sqrt(kl)
We have two solutions, apparently, one for each sign.
So which one is the solution? The one where r = +sqrt(kl), or
the one where r = -sqrt(kl)?
The answer is that either will do, but serious mathematicians
generally want us to write the answer in the most general form
possible; that is, we want *the* solution rather than one of the
solutions that will work. To do this, we use the linearity property
of the differentiation operator and sum the two solutions.
We left out a multiplication constant on the trial exponential
solutions before we did the derivation for the value of r, because it
was really irrelevant at that point (because ignoring it was
equivalent to dividing the equation through by a constant, which
wouldn't affect anything; i.e., for our purposes at that time, the
trial solution R = e^rt and R = Ae^rt were really equivalent).
But when we sum these two solutions, we're going to have to deal with
the fact that those two constants we ignored did not have to be the
same, in general. We'll call one of them A and one of them B, that
is,
one of our solutions is
R1 = Ae^sqrt(kl)
and the other is
R2= Be^-sqrt(kl),
where we substituted the positive root value of r into the first type
of solution for R (which we've now denoted as R1, to distinguish it
from the general solution R, which, as I said, will be the sum of
these two possible solutions) , and the negative root value of r into
the second type of solution for R, which we call R2.
So let's sum the two to get our general solution:
R = R1 + R2 = Ae^sqrt(kl) + Be^-sqrt(kl)
That's it, and now we just need to find out whether that's the same
thing as the hyperbolic sine and hyperbolic cosine functions that Tim
was talking about. Which means, "What the hell *were* those things?
Sums of exponentials, right?" That would imply that Tim was right.
According to Google,
sinh = 1/2 (e^t - e^-t)
and
cosh = 1/2 (e^t + e^-t)
Which is the answer we derived, with A = 1/2 and B = -1/2 in one case
and A = 1/2 and B= 1/2 in the other.
So Tim was right, for suitable values of k and l, but I give him a
9/10 for not putting his answer in the most general form possible.