(a) Explanation Given information: In exercise 26 , the initial amount is $ 20 , 000 , the annual interest rate is r = 0.04 , and withdrawals at the rate of $ 25 per day Formula used: The solution for initial value problem y ′ = r y + K , y 0 = P 0 is y = P t = P 0 + K r e r t − K r . Calculation: The variable y is the amount of money in the account after t years. The rate of payments per day is $ 25 . Therefore, the rate of payment per year is $ 25 × 365 = $ 9 , 125 . So the differential equation is, y ′ = 0.04 y − 9125 . The initial amount is $ 20000 . Therefore, the initial condition is y 0 = 20 , 000 . Thus, the differential equation and initial condition is, y ′ = 0.04 y − 9125 , y 0 = 20 , 000 Compare the equation y ′ = 0.04 y − 9125 , y 0 = 20 , 000 with the equation y ′ = r y + K , y 0 = P 0 to obtain the value of r , K , P 0 (b) To determine To find: The time is required before the account is depleted.

15th Edition

ISBN: 9780137590438

Author: Goldstein

Publisher: PEARSON

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Chapter 5.4, Problem 34E

(a)

To determine

To find: The value of Pt in exercise 26.

(b)

To determine

To find: The time is required before the account is depleted.

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Chapter 5.4, Problem 33E

Chapter 5.4, Problem 35E

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The value of P t in exercise 26.

Solution Summary: The author explains how to find the value of P(t) in exercise 26.