
a)
dp/dt = exp(p) means dp exp(p) = dt. Integration
on both sides gives  exp(p) = t c, where c is a constant
so that p(t) = log(ct).
b)
Integrating dp/(exp(p)+1) = dt gives plog(1+exp(p)) = t+c so
that p = t+c + log(1+exp(p)), or exp(p) = exp(t+c) (1+exp(p))
Solving for p gives exp(p(t))=exp(t+c)/(1exp(t+c))=1/(exp(tc)1) which can also
be rewritten as p(t) = log(1+exp(ct)) for some constant c.
c) To integrate dp/(p(1p)) = dt we write 1/(p(1p) = 1/p + 1/(1p)
and get log(p) + log(1p) = t+c or p = exp(t+c) (1p) so that
p = exp(t+c)/(1exp(t+c)) which can also be written as
p(t) = exp(t)/(exp(t)C) for some other constant C.
 In biology , p(t) might be the number of sick people as a
function of time at the start of an infectious epidemic. Here, a is proportional to the
number of people that a given sick individual will interact with (when there are a lot of
sick people, this equation will no longer be a good approximation to p(t) since sick
people will by more likely interact with other sick people.)
In Sociology p(t) might be the number of people at Harvard as a function of time
who have heard a particular joke if the joke has't been around too long. In this case,
a represents the average number of friends that a given person has.
Again, if the joke has been around for a while, there will be a greater chance that a
reasonable percentage of friends of any given individual
will have heard the joke.)
In economics , p(t) might represent the number of people as a
function of time (if not too much time has gone by) who are involved in a chain letter
type of scheme to make money (a pyramid scheme, such as that which bankrupted
Albania a few years ago.) Here, a is going to be proportional to the average number of
acquaintances of a given individual.
In environmental science
p(t) is the number of isotopes of a radioacitive container stored
at the Hanford facility ( checkout this for
more information). The number of decays in some time interval is proportional to p(t).
Here a is negative. The time for 1/2 of the isotopes to be decayed is log(2)/a.

a)u(t, x) = e^{2t} e^{4(x3t)} = e^{4x+10t}.
b)u(t, x) = e^{2t} e ^{ (x3t)2}.
 The solution that provides the sketches is u(t, x) = 1/(1 + (x  3t)^{2}).
 The function r(x) is the death rate of people at age x. Thus, it should be roughly
constant until x = 15 and then increase to a peak at x = 19, decrease again until x = 30
and then slowly increase for x larger than 30, maybe more steeply for x larger than 55. The speed is one
since each person gets one year older after one year.
 u_{t} = r u + (2 m t + x^{2})/4 m t^{2}) u
u_{xx} = (2 m t + x^{2})/4 m t^{2}) u
 a) u_{t} = l u, u_{xx} = (lr)/m u
b) u_{t} = l u, u_{xx} = (lr)/m u
c) u_{t} = r u, u_{xx} = 0
d) u_{t} = l u, u_{xx} = (lr)/m u
e) u_{t} = l u, u_{xx} = (lr)/m u
 a = u(1, 0)^{2}/u(2, 0)), r = log[ u(1, 0)/(u(2, 0)) ], m = 1/(log[ u(1, 0)/u(1, 2)]).
 In this case, the standard choice takes q=c u  m u_{x}. Here is how one can
argue for this term: without random motions, the net number of particles which cross x
between time tdt and t (for very small dt) is going to be roughly the number of
particles with positions between x c dt and x at time t, which is roughly u(t,x) c dt.
With random motion, this number must be adjusted because some particles with x
coordinate slightly greater than xc dt at time tdt will have velocity less than c and will
not make it to x in time dt, while others with x coordinate slightly less than xc dt will
have velocity greater than c, and so make it past x in the time dt. If the velocities are
assigned randomly, then this correction factor to the term u(t,x  c dt) should be roughly
proportional to u_{x}(tdt,xc dt) dt. In particular, when dt is very small, then the
the xderivative of u at (tdt, xc dt) is almost the same as that of u at (x,t). Thus, the
correction to the original c u dt term gives (c u + m u_{x}) dt as the small dt
approximation for the net number of particles which cross x between tdt and t.
Granted that such is the case, this leads to q = c u + m u_{x} by the arguments used
previously in the notes.
