Book edition 5th

Author(s) Otto Bretscher

Pages 442 pages

ISBN 9780321796974

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Short Answer

Ngozi opens a bank account with an initial balance of 1,000 Nigerian naira. Let b(t) be the balance in the account at time t; we are told that b(0) = 1,000. The bank is paying interest at a continuous rate of 5% per year. Ngozi makes deposits into the account at a continuous rate of s(t) (measured in naira per year). We are told that s(0) = 1,000 and that s(t) is increasing at a continuous rate of 7% per year. (Ngozi can save more as her income goes up over time.)
(a) set up a linear system of the form $| \begin{matrix} \frac{d b}{d t} = ? b + ? s \\ \frac{d s}{d t} = ? b + ? s \end{matrix} |$
(b) Find $b (t)$ and $s (t)$ .

(a) The solution is $| \begin{matrix} \frac{d b}{d t} = 0.05 b + 1 s \\ \frac{d s}{d t} = 0 b + 0.07 s \end{matrix} |$

(b) The solution $b (t) = - 49000 e^{0.05 t} + 50000 c_{2} e^{0.07 t}, s (t) = 1000 e^{0.07 t}$

See the step by step solution

Step by Step Solution

TABLE OF CONTENTS :

TABLE OF CONTENTS

Step1: Explanation of the solution

Let $b (t)$ is the balance in the account at time $t$ and $b (0) = 1000$ .

The bank is paying interest at a continuous rate of $5 %$ per year.

Ngozi makes deposits in to the account at a continuous rate of $s (t)$ .

And $s (t)$ is increasing at a continuous rate of $7 %$ per year also that $s (0) = 1000$ .

The setup of a linear system as follows.

$| \begin{matrix} \frac{d b}{d t} = ? b + ? s \\ \frac{d s}{d t} = ? b + ? s \end{matrix} |$

Since, the bank is already paying interest at a continuous rate of $5 %$ per year and Ngozi makes deposits into the account at a continuous rate of $s (t)$ per year.

The first equation can written as follows.

$| \begin{matrix} \frac{d b}{d t} = 0.05 b + 1 s \\ \frac{d s}{d t} = ? b + ? s \end{matrix} |$

Since, $s (t)$ is increasing at a continuous rate of $7 %$ per year.

Therefore, the equation can be written as follows.

$| \begin{matrix} \frac{d b}{d t} = 0.05 b + 1 s \\ \frac{d s}{d t} = 0 b + 0.07 s \end{matrix} |$

Hence, the setup of the linear system is $| \begin{matrix} \frac{d b}{d t} = 0.05 b + 1 s \\ \frac{d s}{d t} = 0 b + 0.07 s \end{matrix} |$

Step 2: Explanation of the solution

Consider the linear system of equation as follows.

$| \begin{matrix} \frac{d b}{d t} = 0.05 b + 1 s \\ \frac{d s}{d t} = 0 b + 0.07 s \end{matrix} |$

Rewrite the system in the form of $\frac{d \vec{x}}{d t} = A \vec{x}$ as follows.

$[\begin{matrix} \frac{d b}{d t} \\ \frac{d s}{d t} \end{matrix}] = [\begin{matrix} 0.05 & 1 \\ 0 & 0.07 \end{matrix}] [\begin{matrix} b \\ s \end{matrix}]$

Since the coefficient matrix $A = [\begin{matrix} 0.05 & 1 \\ 0 & 0.07 \end{matrix}]$ is a diagonal matrix, the eigenvalues are the diagonal entries.

The eigenvalues are as follows.

$\begin{array}{l} λ_{1} = 0.05 \\ λ_{2} = 0.07 \end{array}$

Now, find the corresponding eigenvectors for eigenvalue $λ_{1} = 0.05$ as follows.

$\begin{matrix} (A - λ I) = 0 \\ [\begin{matrix} 0.05 - (0.05) & 1 \\ 0 & 0.07 - (0.05) \end{matrix}] [\begin{matrix} u_{1} \\ u_{2} \end{matrix}] = [\begin{matrix} 0 \\ 00 \end{matrix}] \\ [\begin{matrix} 0 & 1 \\ 0 & 0.02 \end{matrix}] [\begin{matrix} u_{1} \\ u_{2} \end{matrix}] = [\begin{matrix} 0 \\ 00 \end{matrix}] \\ [\begin{matrix} 0 u_{1} + 1 u_{2} \\ 0 u_{1} + 0.02 u_{2} \end{matrix}] = [\begin{matrix} 0 \\ 0 \end{matrix}] \end{matrix}$

The corresponding equations are as follows.

$\begin{matrix} 0 u_{1} + 1 u_{2} = 0 \\ 0 u_{1} + 0.2 u_{2} = 0 \end{matrix}$

Solving the equations as follows.

$\begin{matrix} 0 u_{1} + 1 u_{2} = 0 \\ u_{2} = - 0 u_{1} \\ u_{2} = 0 \end{matrix}$

Therefore, the eigenvectors as follows.

$\begin{matrix} \vec{u} = [\begin{matrix} u_{1} \\ u_{2} \end{matrix}] \\ = [\begin{matrix} u_{1} \\ 0 \end{matrix}] \\ \vec{u} = u_{1} [\begin{matrix} 1 \\ 0 \end{matrix}] \end{matrix}$

Similarly, find the corresponding eigenvectors for eigenvalue $λ_{1} = 0.07$ as follows.

$\begin{matrix} (A - λ I) = 0 \\ [\begin{matrix} 0.05 - (0.07) & 1 \\ 0 & 0.07 - (0.07) \end{matrix}] [\begin{matrix} v_{1} \\ v_{2} \end{matrix}] = [\begin{matrix} 0 \\ 0 \end{matrix}] \\ [\begin{matrix} 0.02 & 1 \\ 0 & 0 \end{matrix}] [\begin{matrix} v_{1} \\ v_{2} \end{matrix}] = [\begin{matrix} 0 \\ 0 \end{matrix}] \\ [\begin{matrix} 0.02 v_{1} + 1 v_{2} \\ 0 v_{1} + 0 v_{2} \end{matrix}] = [\begin{matrix} 0 \\ 0 \end{matrix}] \end{matrix}$

The corresponding equations are as follows.

$\begin{matrix} 0.02 v_{1} + 1 v_{2} = 0 \\ 0 v_{1} + 0 v_{2} = 0 \end{matrix}$

Solving the equations as follows.

$\begin{matrix} 0.02 v_{1} + 1 v_{2} = 0 \\ 0.02 v_{1} = - v_{2} \\ v_{1} = \frac{- 1 v_{2}}{0.02} \\ v_{1} = 50 v_{2} \end{matrix}$

Therefore, the eigenvectors as follows.

$\begin{matrix} \vec{v} = [\begin{matrix} v_{1} \\ v_{2} \end{matrix}] \\ = [\begin{matrix} 50 v_{2} \\ v_{2} \end{matrix}] \\ \vec{v} = v_{2} [\begin{matrix} 50 \\ 1 \end{matrix}] \end{matrix}$

Thus, the eigenvectors are as follows.

$[\begin{matrix} 1 \\ 0 \end{matrix}]$ and $[\begin{matrix} 50 \\ 1 \end{matrix}]$

The general solution to the system can be written as follows.

$\begin{matrix} [\begin{matrix} b (t) \\ s (t) \end{matrix}] = c_{1} e^{λ_{1} t} \vec{u} + c_{2} e^{λ_{2} t} \vec{v} \\ = c_{1} e^{0.05 t} [\begin{matrix} 1 \\ 0 \end{matrix}] + c_{2} e^{0.07 t} [\begin{matrix} 50 \\ 1 \end{matrix}] \\ = [\begin{matrix} c_{1} e^{0.05 t} \\ 0 \end{matrix}] + [\begin{matrix} 50 c_{2} e^{0.07 t} \\ c_{2} e^{0.07 t} \end{matrix}] \\ [\begin{matrix} b (t) \\ s (t) \end{matrix}] = [\begin{matrix} c_{1} e^{0.05 t} + 50 c_{2} e^{0.07 t} \\ c_{2} e^{0.07 t} \end{matrix}] \end{matrix}$

Now substitute the value 0 for in $[\begin{matrix} b (t) \\ s (t) \end{matrix}] = [\begin{matrix} c_{1} e^{0.05 t} + 50 c_{2} e^{0.07 t} \\ c_{2} e^{0.07 t} \end{matrix}]$ as follows.

$\begin{array}{l} [\begin{matrix} b (t) \\ s (t) \end{matrix}] = [\begin{matrix} c_{1} e^{0.05 t} + 50 c_{2} e^{0.07 t} \\ c_{2} e^{0.07 t} \end{matrix}] \\ [\begin{matrix} b (0) \\ s (0) \end{matrix}] = [\begin{matrix} c_{1} e^{0.05 (0)} + 50 c_{2} e^{0.07 (0)} \\ c_{2} e^{0.07 (0)} \end{matrix}] \\ [\begin{matrix} 1000 \\ 1000 \end{matrix}] = [\begin{matrix} c_{1} + 50 c_{2} \\ c_{2} \end{matrix}] \end{array}$

Now, equating the corresponding entries of the matrix on sides of the equation as follows.

$\begin{matrix} c_{1} + 50 c_{2} = 1000 \\ c_{2} = 1000 \end{matrix}$

Substitute the value $1000$ for $c_{2}$ in $c_{1} + 50 c_{2} = 1000$ as follows.

$\begin{matrix} c_{1} + 50 c_{2} = 1000 \\ c_{1} + 50 (1000) = 1000 \\ c_{1} + 50000 = 1000 \\ c_{1} = 1000 - 50000 \end{matrix}$

Simplify further as follows.

$c_{1} = - 49, 000$

Substitute the value $- 49, 000$ for $c_{1}$ and $1000$ for $c_{2}$ in $[\begin{matrix} b (t) \\ s (t) \end{matrix}] = [\begin{matrix} c_{1} e^{0.05 t} + 50 c_{2} e^{0.07 t} \\ c_{2} e^{0.07 t} \end{matrix}]$ as follows.

$\begin{array}{l} [\begin{matrix} b (t) \\ s (t) \end{matrix}] = [\begin{matrix} c_{1} e^{0.05 t} + 50 c_{2} e^{0.07 t} \\ c_{2} e^{0.07 t} \end{matrix}] \\ [\begin{matrix} b (t) \\ s (t) \end{matrix}] = [\begin{matrix} - 49000 e^{0.05 t} + 50 (1000) c_{2} e^{0.07 t} \\ 1000 e^{0.07 t} \end{matrix}] \\ [\begin{matrix} b (t) \\ s (t) \end{matrix}] = [\begin{matrix} - 49000 e^{0.05 t} + 50000 c_{2} e^{0.07 t} \\ 1000 e^{0.07 t} \end{matrix}] \end{array}$

Equating the corresponding entries on both sides of the equation as follows.

$\begin{array}{l} b (t) = - 49000 e^{0.05 t} + 50000 c_{2} e^{0.07 t} \\ s (t) = 1000 e^{0.07 t} \end{array}$

Thus, the equations are $b (t) = - 49000 e^{0.05 t} + 50000 c_{2} e^{0.07 t}, s (t) = 1000 e^{0.07 t}$ .

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