Page 313 - A Course in Linear Algebra with Applications

8.3: Applications to Systems of Linear Differential Equations 297

kx kx
where U = I 1. Hence u = ce and v = de , with c and
W
d arbitrary constants. The general solution is Y = SU, which
yields
kt kt
j a= ce + de~
\b = -ce kt 4- de~ kt
Now the initial conditions are a(0) = ao and b(0) = bo, so

c + d = ao
—c + d=bo

Solving we obtain c = (ao — bo)/2, d = (a 0 + &o)/2. Therefore
the numbers of tanks surviving at time t in Divisions A and
B are respectively

a = {2^y^ ^o + bo\ -kt
+
e
a fe
o - o\ ekt + (ao + b 0\ e_ fc(

It is more convenient to write a(t) and b(t) in terms of the
x
hyperbolic functions cosh(:r) = ^(e x + e~ ) and sinh(x) =
x x
\{e — e~ ). Then the solution becomes
a = aocosh(kt) — 6osinh(H)
b = bocosh(kt) — aosinh(A;i)

Now Division B will have lost all its tanks when 6 = 0,
i.e., after time
1
t=itanh- (-).
v
k ao
Observe also that

2 t2 2 1,2
a — b — an — bf,