/*! 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 8 Show, by substituting into the d... [FREE SOLUTION] | 91Ó°ÊÓ

91Ó°ÊÓ

Show, by substituting into the difference equation, that \(Y_{t}=A\left(b^{t}\right)+D \quad\) where \(\quad D=\frac{c}{1-b}\) is a solution of $$ Y_{t}=b Y_{t-1}+c \quad(b \neq 1) $$

Short Answer

Expert verified
Substituting and solving for \(D\) shows that the given \(Y_t = A(b^t) + D\) is a solution.

Step by step solution

01

Understand the Given Solution

The proposed solution is given as \[Y_t = A(b^t) + D\]where \(D = \frac{c}{1-b}\).
02

Substitute the Proposed Solution into the Difference Equation

The given difference equation is \[Y_t = bY_{t-1} + c.\]Substitute \(Y_t\) and \(Y_{t-1}\) from the proposed solution. This gives:\[Y_t = A(b^t) + D\]and\[Y_{t-1} = A(b^{t-1}) + D.\]
03

Substitute \(Y_{t-1}\) into the Difference Equation

Substitute \(Y_{t-1}\) into the difference equation:\[A(b^t) + D = b[A(b^{t-1}) + D] + c.\]
04

Simplify the Equation

Distribute \(b\) on the right side:\[A(b^t) + D = A(b^t) + bD + c.\]
05

Isolate Terms

For the equation to hold true for all \(t\), isolate \(D\) and \(c\):\[D = bD + c.\]
06

Solve for \(D\)

Rearrange the equation to solve for \(D\):\[D - bD = c\]\[D(1 - b) = c\]\[D = \frac{c}{1-b}.\]
07

Verify the Solution

Verify that \(D = \frac{c}{1-b}\) satisfies the original equation. This completes the validation that the proposed solution \(Y_t = A(b^t) + D\) is correct.

Unlock Step-by-Step Solutions & Ace Your Exams!

  • Full Textbook Solutions

    Get detailed explanations and key concepts

  • Unlimited Al creation

    Al flashcards, explanations, exams and more...

  • Ads-free access

    To over 500 millions flashcards

  • Money-back guarantee

    We refund you if you fail your exam.

Over 30 million students worldwide already upgrade their learning with 91Ó°ÊÓ!

Key Concepts

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

discrete time series
A discrete time series is a sequence of values or observations, indexed by time, which occur at specific, separate intervals. Imagine observing temperature at noon every day. Each recorded temperature represents a point in a discrete time series.

Time here is divided into distinct, non-overlapping periods such as hours, days, or years. In mathematical terms, we often denote a discrete time series as \(Y_t\), where \(t\) is an integer denoting the time period.

In such series, the values can evolve according to specific rules or equations, helping us predict or understand future values based on past data. This predictability makes discrete time series crucial in various fields, like economics, finance, weather forecasting, and even population studies.
homogeneous solutions
Homogeneous solutions are those that satisfy the corresponding homogeneous equation of a given difference equation.

Take the difference equation \(Y_t = bY_{t-1} + c\). The corresponding homogeneous equation is \(Y_t = bY_{t-1}\), which means solving for \(Y_t\) without the constant term \(c\).

The solution form for the homogeneous equation \(Y_t = bY_{t-1}\) typically looks like \(Y_t = A(b^t)\), where \(A\) is a constant determined by initial conditions. This homogeneous solution represents the part of the solution that is influenced solely by previous values in the series, without any external input (the \(c\) term).
Understanding homogeneous solutions helps in simplifying and solving more complex difference equations by dealing with one part of the equation at a time.
particular solutions
Particular solutions target the non-homogeneous part of a difference equation. These solutions directly address the equation with the constant term included.

For the given equation \(Y_t = bY_{t-1} + c\), the particular solution needs to include the effect of the \(c\), which is not present in the homogeneous part.

Typically, the particular solution adds a constant value \(D\), defined as \(D = \frac{c}{1-b}\). This \(D\) effectively satisfies the equation when included with the homogeneous solution, \(Y_t = A(b^t) + D\).

The constant \(D\) represents a steady-state value that the time series tends towards as \(t\) increases, assuming \(b < 1\).
Particular solutions are vital for understanding the complete behavior of the series, especially in applications where external factors (represented by \(c\)) influence changes over time.
validation of solutions
Validation of solutions involves confirming that proposed solutions satisfy the original difference equation.

To validate \(Y_t = A(b^t) + D\) for the given difference equation \(Y_t = bY_{t-1} + c\), we substitute \(Y_t\) and \(Y_{t-1}\) back into the equation.

This substitution and simplification of terms show that the left-hand side equals the right-hand side, proving the solution's correctness.

Specifically, the steps involve:
  • Writing down \(Y_t = A(b^t) + D\) and \(Y_{t-1} = A(b^{t-1}) + D\).
  • Substituting \(Y_{t-1}\) into the original equation, resulting in \(A(b^t) + D = b[A(b^{t-1}) + D] + c\).
  • Simplifying both sides to isolate \(D\) and confirm it equals \(D = \frac{c}{1-b}\).

By confirming these steps, we ensure the proposed solution matches the behavior described by the original difference equation, making it a valid solution.

One App. One Place for Learning.

All the tools & learning materials you need for study success - in one app.

Get started for free

Most popular questions from this chapter

The Harrod-Domar model of the growth of an economy is based on three assumptions. (1) Savings, \(S_{t}\), in any time period are proportional to income, \(Y_{t}\), in that period, so that $$ S_{t}=\alpha Y_{t} \quad(\alpha>0) $$ (2) Investment, \(I_{t}\), in any time period is proportional to the change in income from the previous period to the current period so that $$ I_{t}=\beta\left(Y_{t}-Y_{t-1}\right) \quad(\beta>0) $$ (3) Investment and savings are equal in any period so that $$ I_{t}=S_{t} $$ Use these assumptions to show that $$ Y_{t}=\left(\frac{\beta}{\beta-\alpha}\right) Y_{t-1} $$ and hence write down a formula for \(Y_{t}\) in terms of \(Y_{0}\). Comment on the stability of the system in the case when \(\alpha=0.1\) and \(\beta=1.4\), and write down expressions for \(S_{t}\) and \(I_{t}\) in terms of \(Y_{0}\).

(Excel) Consider the two-sector model $$ \begin{aligned} &C_{t}=100+0.6 Y_{t-1}^{0.8} \\ &Y_{t}=C_{t}+60 \end{aligned} $$ (a) Write down a difference equation for \(Y_{t}\). (b) Given that \(Y_{0}=10\), calculate the values of \(Y_{t}\) for \(t=1,2, \ldots 8\) and plot these values on a diagram. Is this system stable or unstable? (c) Does the qualitative behaviour of the system depend on the initial value of \(Y_{0}\) ?

Consider the supply and demand equations $$ \begin{aligned} &Q_{S t}=0.4 P_{t-1}-12 \\ &Q_{D t}=-0.8 P_{t}+60 \end{aligned} $$ Assuming that equilibrium conditions prevail, find an expression for \(P_{t}\) when \(P_{0}=70\). Is the system stable or unstable?

(Maple) The output, \(Q\), of an Internet firm depends in the short term on capital, \(K\), and time, \(t\). The production and savings functions are given by $$ Q=a t K \quad \text { and } \quad S=Q-b t $$ respectively, where \(a\) and \(b\) are positive constants. Assuming that capital accumulation is equal to savings, show that $$ \frac{\mathrm{d} K}{\mathrm{~d} t}=t(a K-b) $$ If the initial capital is \(c\), solve this equation to obtain an expression for \(K\) in terms of \(t\) and the constants, \(a, b\) and \(c\). Write down the corresponding expression for \(Q\). Comment on the qualitative behaviour of this solution in the case when (a) \(c=b / a\) (b) \(c>b / a\) (c) \(c

(Maple) In the absence of any withdrawals, the value of an investment fund, \(y(t)\), varies according to $$ \frac{\mathrm{d} y}{\mathrm{~d} t}=y(1-y) $$ where \(y\) is measured in millions of dollars and \(t\) is measured in years. Money is taken out of the fund at a constant rate of \(\$ 250000\) per year so that $$ \frac{\mathrm{d} y}{\mathrm{~d} t}=y(1-y)-0.25 $$ (a) Find the solution of this differential equation when the initial value of the fund is \(\$ 1\) million. Plot a graph of this solution over the range \(0 \leq t \leq 20\). (b) Find the solution of this differential equation when the initial value of the fund is \(\$ 250000\). Plot a graph of this solution over the range \(0 \leq t \leq 2\). (c) Compare the solutions obtained in parts (a) and (b).

See all solutions

Recommended explanations on Math Textbooks

View all explanations

What do you think about this solution?

We value your feedback to improve our textbook solutions.

Study anywhere. Anytime. Across all devices.