Math 308 Differential Equations Fall 2002
Solving The “Coffee Cup Problem”
The differential equation for H(t), the temperature at time t, is
dH
dt =k(70 −H).
Note that
H(t) = 70 (1)
(i.e. H(t) is the function with constant value 70) is a solution to the differential equation. Now assume that
H(t)6= 70. We treat dH
dt as if it really is a fraction, and rewrite the differential equation as
dH
H−70 =−k dt,
and then integrate both sides:
ZdH
H−70 =−Zk dt.
This gives us
ln |H−70|=−kt +C1.
Note the absolute value on the left; recall from Calculus that Rdx
x= ln |x|+c. Now exponentiate both sides
to obtain
|H−70|=e−kt+C1=eC1e−kt =C2e−kt,
where C2=eC1. (Note that C2>0.) Now, if H−70 >0, then |H−70|=H−70, and we have
H(t) = 70 + C2e−kt .(2)
If H−70 <0, then |H−70|=−(H−70), and so
H(t) = 70 −C2e−kt .(3)
Thus, the solutions to the differential equation are given by (1), (2), and (3). For this example, we see that
we can combine these three formulas into one expression that includes all the solutions:
H(t) = 70 + C3e−kt ,
where C3is an arbitrary constant.
If we are given an initial condition
H(0) = H0,
we can find C3in terms of H0by evaluating the solution at t= 0 and setting it equal to H0:
H(0) = 70 + C3e0= 70 + C3=H0,
so C3=H0−70, and the solution to the initial value problem is
H(t) = 70 + (H0−70)e−kt .
The following plot shows the solutions for H0= 90, H0= 70, and H0= 50, with k= 0.1, for the time
interval 0 ≤t≤25.
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