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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.