91Ó°ÊÓ

An endowment is an investment account in which the balance ideally remains constant and withdrawals are made on the interest earned by the account. Such an account may be modeled by the initial value problem \(B^{\prime}(t)=r B-m,\) for \(t \geq 0,\) with \(B(0)=B_{0} .\) The constant \(r>0\) reflects the annual interest rate, \(m>0\) is the annual rate of withdrawal, \(B_{0}\) is the initial balance in the account, and \(t\) is measured in years. a. Solve the initial value problem with \(r=0.05, m=\$ 1000 /\) year, and \(B_{0}=\$ 15,000 .\) Does the balance in the account increase or decrease? b. If \(r=0.05\) and \(B_{0}=\$ 50,000,\) what is the annual withdrawal rate \(m\) that ensures a constant balance in the account? What is the constant balance?

Step by step solution

01

Separable Equations and Integration

Since the given differential equation is linear and separable, we can rearrange the terms and integrate both sides w.r.t. \(t\): $$\frac{dB}{dt}=rB-m \Rightarrow \frac{dB}{B-m/r} = rdt$$ Now, integrate both sides with respect to \(t\): $$\int \frac{dB}{B-m/r} = \int rdt$$

02

Solve the Integrals

Use a substitution for the left side integral: \(u=B-m/r\), \(du=dB\). Then, the integrals become: $$\int \frac{du}{u} = \int rdt$$ Now, solve the integrals: $$\ln |u| = rt + C_1 \Rightarrow \ln |B-m/r| = rt + C_1$$ Exponentiate both sides: $$B-m/r = C e^{rt}$$

03

Solve the Initial Value Problem for Part (a)

For part (a), substitute the initial values into the solution: $$B(0)-m/r = 15000-1000/0.05 = Ce^0$$ Thus, \(C=14900\). So, the solution to the initial value problem is: $$B(t)=14900 e^{0.05t} + \frac{1000}{0.05}$$

04

Determine the Balance Change for Part (a)

To determine if the balance is increasing or decreasing, find the derivative of the balance with respect to time: $$\frac{dB}{dt} = 0.05(14900 e^{0.05t})$$ Since all terms in the derivative are positive, this means that the balance is always increasing with time.

05

Find the Annual Withdrawal Rate (\(m\)) for Part (b)

For part (b), we substitute the given initial balance and interest rate into the balance equation: $$B=50000$$ $$m=0.05B => m=0.05(50000) = 2500$$ The annual withdrawal rate that ensures a constant balance is \(m = 2500\) per year.

06

Find the Constant Balance for Part (b)

Since we have found the annual withdrawal rate that ensures a constant balance, we can use this value in the balance equation: $$B(t)=Ce^{0.05t} + \frac{2500}{0.05}$$ Using the initial balance for the given problem, we find C: $$B(0)=50000=Ce^0 + \frac{2500}{0.05}$$ $$C=0$$ Therefore, the constant balance is: $$B(t)=\frac{2500}{0.05} = 50000$$

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Differential Equations

Differential equations are mathematical equations that describe the relationship between a function and its derivatives. They are a powerful tool in expressing the change in various phenomena such as physics, chemistry, economics, and biology. An initial value problem in differential equations is a specific type of problem where the solution to a differential equation is sought under the condition that the function passes through a given point, known as the initial value. In the exercise, an initial value problem models the balance of an investment account over time, taking into consideration the interest earned and the withdrawals made.

Exponential Growth and Decay

In the context of the investment account problem, exponential growth and decay represent the ways in which the account balance might increase (exponential growth) or decrease (exponential decay). The growth or decay is determined by the constant interest rate applied to the account balance and the withdrawals made. When the rate of interest is greater than the withdrawals, the account experiences exponential growth; when the opposite is true, the account suffers exponential decay. This concept is crucial for understanding how the account balance evolves over time under constant economic conditions.

Separable Equations

Separable equations are a class of differential equations that can be separated into two parts: one that involves only the function and another that involves only the independent variable. Once separated, each side of the equation can be integrated separately. This method simplifies the process of finding the general solution to the differential equation. In our initial value problem for the endowment, separation of variables allows us to solve the equation by integrating both sides with respect to the respective variables, leading us closer to determining the account balance as a function of time.

Integration

Integration is a fundamental mathematical operation used to solve differential equations. It allows us to find the antiderivatives of functions, which are essential in solving initial value problems. Through integration, we obtain the general solution to a differential equation. In the provided exercise, integration is used once we have expressed the separated variables in integral form, which upon solving, gives us an expression for the account balance. This integration step is pivotal in connecting the rate of change of the account balance to the actual balance at any given time.