Download Lecture 10.4a

Calculus 221, section 10.4a Applications of First Order Linear DEs
notes prepared by Tim Pilachowski
The foundation of section 10.4 is a recognition that we can often gather information about how something is
changing (rate of change = first derivative) and use that information to derive an equation to describe an
amount.
In Lecture 10.4b we’ll concentrate on medical-biological scenarios. Today’s examples will all be economicsrelated. In all examples below, we will assume that deposits, withdrawals and interest calculations are all done
continuously.
Example A: You deposit $3000 into a retirement account which receives 5% interest annually. a) Set up and
solve a differential equation that is satisfied by P(t), the amount (i.e. principal) in the retirement account after t
years. b) Estimate the account balance after 30 years (to the nearest penny).
answers: P = 3000e 0.05t ; $13445.07
Example B: Each year, you deposit $3000 into a retirement account which receives 5% interest annually. a) Set
up and solve a differential equation that is satisfied by P(t), the amount in the retirement account after t years. b)
Estimate the account balance after 30 years (to the nearest penny).
answers: P = −60000 + 63000e 0.05t ; $222346.41
Example C: You deposit $3000 into a retirement account which receives 5% interest annually. Each year
thereafter annual income rises and you deposit 3000 + 500t dollars into the account. a) Set up and solve a
differential equation that is satisfied by P(t), the amount in the retirement account after t years. b) Estimate the
account balance after 30 years (to the nearest penny).
answers: P = −260000 − 10000t + 263000e 0.05t ; $618684.23
Example D: When you turn 50 you have $600,000 in a retirement account which receives 5% annually. You
plan to withdraw $50,000 each year. a) Set up and solve a differential equation that is satisfied by P(t), the
amount in the retirement account after t years. b) Estimate the number of years your retirement fund will last.
a) The differential equation is similar to that of Example B, except that where deposits are a “ + ”, withdrawals
are a “ – ”
P ′ = 0.05P − 50000
P ′ − 0.05 P = −50000
a(t ) = −0.05 ⇒
A(t ) = ∫ − 0.05 dt = −0.05t ⇒ i.f. = e − 0.05t
(e − 0.05t )P′ − (0.05e − 0.05t )P = −50000e − 0.05t
d − 0.05t
[
e
P ] = −50000e − 0.05t
dt
d − 0.05t
P ] dt = ∫ − 50000e − 0.05t dt
∫ dt [e
e − 0.05t P =
− 50000 − 0.05t
5000000 − 0.05t
=−
+ C = 1000000e − 0.05t + C
e
e
− 0.05
5
P = 1000000 + Ce 0.05t
We have an initial value condition, P(0) = 600000, to use in determining C.
600000 = 1000000 + Ce 0.05∗ 0
− 400000 = C
P = 1000000 − 400000e 0.05t
b) To estimate the number of years your retirement fund will last, set P = 0.
0 = 1000000 − 400000e 0.05t
− 1000000 = −400000e 0.05t
2.5 = e 0.05t
ln (2.5) = 0.05t
ln(2.5)
= 20 ln(2.5) = t
0.05
This is the exact answer. The fund will last approximately 18.3 years.
Note that the mathematics of withdrawing from a retirement account is the same as for calculating loan
amortization—regular amounts lower the balance as time passes.