/*! This file is auto-generated */ .wp-block-button__link{color:#fff;background-color:#32373c;border-radius:9999px;box-shadow:none;text-decoration:none;padding:calc(.667em + 2px) calc(1.333em + 2px);font-size:1.125em}.wp-block-file__button{background:#32373c;color:#fff;text-decoration:none} Problem 54 An annuity is a fund into which ... [FREE SOLUTION] | 91Ó°ÊÓ

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An annuity is a fund into which one makes equal payments at regular intervals. If the fund earns interest at rate \(r\) compounded continuously, and deposits are made continuously at the rate of \(d\) dollars per year (a continuous annuity), then the value \(y(t)\) of the fund after \(t\) years satisfies the differential equation \(y^{\prime}=d+r y .\) (Do you see why?) Solve the differential equation in the preceding instructions for the continuous annuity \(y(t),\) where \(d\) and \(r\) are unknown constants, subject to the initial condition \(y(0)=0\) (zero initial value).

Short Answer

Expert verified
The solution is \( y(t) = \frac{d}{r}(e^{rt} - 1) \).

Step by step solution

01

Identify the Type of Differential Equation

The given differential equation is \( y'(t) = d + ry(t) \). This is a first-order linear differential equation of the form \( y' + P(t)y = Q(t) \). Here, \( P(t) = -r \) and \( Q(t) = d \).
02

Find the Integrating Factor

To solve the differential equation, we first find an integrating factor. The integrating factor \( \mu(t) \) is given by \( e^{\int P(t)\,dt} = e^{-rt} \).
03

Multiply the Differential Equation by the Integrating Factor

Multiply both sides of the differential equation by the integrating factor \( e^{-rt} \): \[ e^{-rt}y'(t) + e^{-rt}ry(t) = de^{-rt} \] This can be rewritten using product rule as \[ \frac{d}{dt}[e^{-rt}y(t)] = de^{-rt}. \]
04

Integrate Both Sides

Integrate both sides with respect to \( t \):\[ \int \frac{d}{dt}[e^{-rt}y(t)] \, dt = \int de^{-rt} \, dt. \] The left-hand side yields \( e^{-rt}y(t) \). The right-hand side evaluates to \[ -\frac{d}{r} e^{-rt} + C. \] So, we have: \[ e^{-rt}y(t) = -\frac{d}{r} e^{-rt} + C. \]
05

Solve for y(t)

Multiply through by \( e^{rt} \) to obtain \( y(t) \): \[ y(t) = -\frac{d}{r} + Ce^{rt}. \] This is the general solution of the differential equation.
06

Apply Initial Condition

Use the initial condition \( y(0) = 0 \) to find \( C \):\[ 0 = -\frac{d}{r} + Ce^{0} \] \[ 0 = -\frac{d}{r} + C \] \( C = \frac{d}{r} \).
07

Substitute Back the Constant C

Substituting \( C = \frac{d}{r} \) back into our expression for \( y(t) \), we get:\[ y(t) = -\frac{d}{r} + \frac{d}{r}e^{rt}. \] Thus, the solution simplifies to:\[ y(t) = \frac{d}{r}(e^{rt} - 1). \]

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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 expressions that relate a function to its derivatives. They are useful for modeling how a quantity changes over time, among other things. In the context of a continuous annuity, we want to understand how the value of an investment increases with regular deposits and compounded interest.
This specific problem involves a first-order linear differential equation, which takes the form \( y' + P(t) y = Q(t) \). Here, \( y'(t) = d + ry(t) \), meaning the rate of change of the annuity fund is influenced by both steady deposits (\(d\)) and existing fund value being amplified by the interest rate (\(r\)).
Solving such equations typically involves the use of an integrating factor, which helps transform the differential equation into a more manageable form. This transformation relies heavily on the ability to manipulate exponential functions, as seen with the integrating factor \( \mu(t) = e^{\int -r \, dt} = e^{-rt} \).
This shows the power of differential equations to model real-world scenarios by establishing relationships between influencing factors and rates of change.
Compounded Interest
Compounded interest is a core financial concept, especially in continuous annuities. Unlike simple interest, compounded interest is calculated on the initial principal and also on the accumulated interest from previous periods.
In the context of an annuity, we deal with continuous compounding, which means interest is added to the principal continuously, rather than at discrete intervals like daily, monthly, or annually. This leads to an exponential increase in the value of the annuity over time. The relationship is depicted as part of the differential equation \( y'(t) = d + ry(t) \), where \( ry(t) \) represents the compounded interest aspect.
The continuous compounding formula is \( y(t) = P e^{rt} \), where \( P \) is the principal amount, \( r \) is the interest rate, and \( t \) is time. In continuous annuities, the solution obtained \( y(t) = \frac{d}{r}(e^{rt} - 1) \) mirrors this logic, blending it with regular deposits. This formula showcases how investments accumulate faster through continuous compounding.
Initial Value Problem
An initial value problem in differential equations refers to finding a specific solution that satisfies both the differential equation and an initial condition. The initial condition provides the value of the function at a particular point, which guides the uniquely determined solution.
For continuous annuities, the initial value condition is typically set at \( y(0) = 0 \), meaning the annuity starts without any initial funds. This initial condition is essential for solving our differential equation completely, as it helps determine the constant in the solution.
In step 6 of our solution, plug in the initial condition \( y(0) = 0 \) into the expression \( 0 = -\frac{d}{r} + C \) to derive the constant \( C = \frac{d}{r} \). This constant reflects the scenario where no funds are present initially, ensuring the annuity value accumulates only through deposits and compounded interest.
Therefore, integrating the initial condition into the equation facilitates obtaining a tailored solution, ensuring accuracy in predictive financial modeling with differential equations.

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Most popular questions from this chapter

BIOMEDICAL: Heart Function In the reservoir model, the heart is viewed as a balloon that swells as it fills with blood (during a period called the systole), and then at time \(t_{0}\) it shuts a valve and contracts to force the blood out (the diastole). Let \(p(t)\) represent the pressure in the heart at time \(t\) a. During the diastole, which lasts from \(t_{0}\) to time \(T\), \(p(t)\) satisfies the differential equation $$\frac{d p}{d t}=-\frac{K}{R} p$$ Find the general solution \(p(t)\) of this differential equation. (K and \(R\) are positive constants determined, respectively, by the strength of the heart and the resistance of the arteries. The differential equation states that as the heart contracts, the pressure decreases \((d p / d t\) is negative) in proportion to itself. \()\) b. Find the particular solution that satisfies the condition \(p\left(t_{0}\right)=p_{0} . \quad\left(p_{0}\right.\) is a constant representing the pressure at the transition time \(t_{0}\).) c. During the systole, which lasts from time 0 to time \(t_{0}\), the pressure \(p(t)\) satisfies the differential equation $$ \frac{d p}{d t}=K I_{0}-\frac{K}{R} p $$ Find the general solution of this differential equation. (I \(_{0}\) is a positive constant representing the constant rate of blood flow into the heart while it is expanding.) [Hint: Use the same \(u\) -substitution technique that was used in Example \(7 .\) d. Find the particular solution that satisfies the condi- $$ \text { tion } p\left(t_{0}\right)=p_{0} $$ e. In parts (b) and (d) you found the formulas for the pressure \(p(t)\) during the diastole \(\left(t_{0} \leq t \leq T\right)\) and the systole \(\left(0 \leq t \leq t_{0}\right) .\) Since the heart behaves in a cyclic fashion, these functions must satisfy \(p(T)=p(0) .\) Equate the solutions at these times (use the correct formula for each time) to derive the important relationship $$ R=\frac{p_{0}}{I_{0}} \frac{1-e^{-K T / R}}{1-e^{-K t_{0} / R}} $$.

A cell receives nutrients through its surface, and its surface area is proportional to the two-thirds power of its weight. Therefore, if \(w(t)\) is the cell's weight at time \(t,\) then \(w(t)\) satisfies \(w^{\prime}=a w^{2 / 3}\) where \(a\) is a positive constant. Solve this differential equation with the initial condition \(w(0)=1\) (initial weight 1 unit).

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An annuity is a fund into which one makes equal payments at regular intervals. If the fund earns interest at rate \(r\) compounded continuously, and deposits are made continuously at the rate of \(d\) dollars per year (a continuous annuity), then the value \(y(t)\) of the fund after \(t\) years satisfies the differential equation \(y^{\prime}=d+r y .\) (Do you see why?) Solve the differential equation in the preceding instructions for the continuous annuity \(y(t)\) with deposit rate \(d=\$ 1000\) and continuous interest rate \(r=0.05,\) subject to the initial condition \(y(0)=0\) (zero initial value).

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