**Unformatted text preview: **ing an initial principal of P0 .
Solution:
24 2 Differential Equations
We must solve (2.8) with initial condition P(0) = P0 . The method of separation of variables can be
applied. Separating the variables gives
dP
= rdt .
P
Integrating both sides of the latter equation yields
ln P = rt + C .
Solving for P by exponentiation (see Example 2.5) we get
P = Ce rt .
Substituting P0 = P(0) and putting t = 0 we have P0 = Ce0 = C so the formula for the principal
becomes
P(t ) = P0 ert . (2.9) !
There are a number of interesting financial lessons that can be learned from (2.9).
Example 2.10:
a) What is the effective annual interest rate if money is compounded continuously at an annual
rate r ?
b) How many years are needed to double any starting principal if money earns interest at an
annual rate r compounded continuously?
Solution:
a) By definition, the effective annual interest rate is the relative change in the principal in one
year. In other words, it is the quantity
P(1) − P(0) P0 er − P0
=
= er − 1 .
P(0)
P0
When r is small (as it usually is when we are dealing with interest rates) this quantity is just
slightly larger than the nominal rate r . The increase is due to the compounding. The table
below shows this for some representative interest rates. For example, at 7% annual interest
you would actually earn 7.25% over the course of a year. 25 2 Differential Equations
r .03 .05 .07 .10 .15 Effective Annual Rate .0304 .0513 .0725 .1052 .1618 Doubling Time(years) 23.1 13.9 9.9 6.9 4.6 Table 2.1
b) We need to find how long it takes for the principal to reach double its initial amount. We must
find the time t for which 2 P0 = P0 ert . Canceling the term P0 , we are left with the equation
2 = ert . To solve this equation we take the natural logarithm of both sides obtaining
ln 2 = rt ln e = rt , since ln e = 1 . Denoting the solution by t2 , we have
t2 = ln 2
.
r (2.10) Some representative values of the doubling time are given in Table 2.1. We can derive a
useful rule of th...

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