In the figure below we solve the equation indicated in the first line with the initial
condition that Q = 10 when t = 0.
- In the second line we separate the variables.
- We proceed to integrate both sides and, in the last line, we exponentiate to remove the
ln.
In the third line below we obtain the solutions the absolute value equation, assuming
that C is a positive constant.
- The corresponding solutions are shown in the fourth line, still assuming C to be a
positive constant.
- We generally do not split the solution into two cases, the rather simply allow C to be
an arbitrate constant, which can be positive or negative. The resulting solution is
equivalent of that given here.
- We finally apply the initial condition and determine that C = 4.
- Our solution will thus be Q(t) = 6 + 4 e-.003t.
We use similar techniques to solve the equation in the figure below, with the given
initial condition.
Applying the initial condition, we obtain C = -.9e for the R < 1 solution and C =
.9e for the R > 1 solution.
- The resulting solution is the same for both cases. For R < 1 we get 1 - -.9 e * e-1
= 1 + .9 e * e-1, while for R > 1 we get 1 + .9 3 * e-1.
- Thus we see that we are indeed justified in simply allowing C to be an arbitrary
constant.
We solve the equation below by separating the variables in the usual manner.
- We use the indicated substitution the second line.
- Integration yields the equation in the third line, which we solve for R in the fourth
line.
- In the bottom-line we apply the initial condition R(0) = 1 and obtain c = -6.
- The final solution is R = `sqrt( -6 / (sin(`theta^3) - 6)).
We consider a 1000 gallon container of water in which is dissolved 10 pounds of salt.
- We have three gallons per minute of water flowing into the container and three gallons
per minute flowing out.
- We assume that the water flowing in is pure and that the pure water has been completely
mixed with the water-salt solution in the container before flowing out.
We wish to determine the amount of salt as a function of time. To do so we will set up
a differential equation involving the rate dS / dt at which the salt leaves the system.
- As a preliminary calculation, we find the rate at which solve is leaving the system at t
= 0.
- At the initial instant there are 10 pounds of salt in 1000 gallons of water, or .01 lb /
gal.
- It follows that the 3 gallons that flow out per minute will carry 3 * .01 lb.
- We obtain the rate directly by multiplying the concentration by the 3 gal/min flow from
the container:
- rate at which salt is removed = concentration * outflow rate = .01 lb / gal * 3 gal /
min = .03 lb / min.
- This rate only applies to the very first instant; after this instant the concentration
of salt will be lower and the rate will therefore decrease.
More generally, if S is the amount of salt the concentration of salt will be S / V,
where V is the volume of the container.
- If the rate at which the mixed salt-water solution is flowing from the container is r,
then we have the differential equation dS / dt = (S / V) * r.
- For the present situation we have V = 1000 gal, S(0) = 10 and r = -3.
- We would thus solve the equation dS / dt = S / 1000 * -3 = .003 S, S(0) = 10.
The equation dS / dt = .003 S, S(0) = 10 is easily solved by the usual methods,
yielding S(t) = 10 e(-.003t).