3
$\begingroup$

A college student starts a savings account with an initial balance of $\$0$. He plans to save money at a continuous rate of $\$200$ per week. Also, at every week he plans to increase this rate by $\$10$. (ex. At the 4th month he would be saving at a rate of $\$240$ per week). Additionally, the college student finds a bank account that pays continuously compounded interest at a rate of $4\%$ per year. Estimate the time it'll take for the college student to save $\$500,000$. Hint: set up and solve a differential equation and plot the solution to make the final estimate.

My attempt:

The differential equation is hard to set up.

Let $S =$ amount saved.

Let $t =$ time.

$$\frac{dS}{dt} = \frac{0.04}{52}(200 + 10t)$$

I tried this differential equation but it doesn't satisfy the initial condition.

Can someone help me come up with the differential equation?

Thanks!

$\endgroup$
3
  • $\begingroup$ Are you asked to build a differential equation ? $\endgroup$ Sep 20 '14 at 2:59
  • $\begingroup$ @ClaudeLeibovici It should be some recurrence equation hopefully !!!... $\endgroup$ Sep 20 '14 at 5:04
  • $\begingroup$ Yes, it does asks to build a DE. $\endgroup$
    – Wade
    Sep 20 '14 at 17:28
3
$\begingroup$

$\newcommand{\angles}[1]{\left\langle\, #1 \,\right\rangle} \newcommand{\braces}[1]{\left\lbrace\, #1 \,\right\rbrace} \newcommand{\bracks}[1]{\left\lbrack\, #1 \,\right\rbrack} \newcommand{\ceil}[1]{\,\left\lceil\, #1 \,\right\rceil\,} \newcommand{\dd}{{\rm d}} \newcommand{\ds}[1]{\displaystyle{#1}} \newcommand{\expo}[1]{\,{\rm e}^{#1}\,} \newcommand{\fermi}{\,{\rm f}} \newcommand{\floor}[1]{\,\left\lfloor #1 \right\rfloor\,} \newcommand{\half}{{1 \over 2}} \newcommand{\ic}{{\rm i}} \newcommand{\iff}{\Longleftrightarrow} \newcommand{\imp}{\Longrightarrow} \newcommand{\pars}[1]{\left(\, #1 \,\right)} \newcommand{\partiald}[3][]{\frac{\partial^{#1} #2}{\partial #3^{#1}}} \newcommand{\pp}{{\cal P}} \newcommand{\root}[2][]{\,\sqrt[#1]{\vphantom{\large A}\,#2\,}\,} \newcommand{\sech}{\,{\rm sech}} \newcommand{\sgn}{\,{\rm sgn}} \newcommand{\totald}[3][]{\frac{{\rm d}^{#1} #2}{{\rm d} #3^{#1}}} \newcommand{\verts}[1]{\left\vert\, #1 \,\right\vert}$ $$ \mbox{Let}\quad \left\{\begin{array}{rclcl} b_{n} &:& \mbox{Balance after}\ n\ \mbox{weeks}.&& b_{0} = 0 \\[2mm] s_{0} & : & \mbox{Initial week saving} & = & 200 \\[2mm] \Delta s & : & \mbox{Amount added to the every week saving} & = & 10 \\[2mm] r & : & \mbox{Bank interest} \pars{~\mbox{per one per week}~} & = & {4/\pars{12\times 4} \over 100} = {1 \over 1200} \end{array}\right. $$ We assumed $4$ weeks per month.

$$ \begin{array}{rclc} b_{0} & = & 0 \\ b_{1} & = & s_{0} \\ b_{2} & = & b_{1}\pars{1 + r} + \pars{s_{0} + \Delta s} \\ b_{3} & = & b_{2}\pars{1 + r} + \pars{s_{0} + 2\Delta s} \\ b_{4} & = & b_{3}\pars{1 + r} + \pars{s_{0} + 3\Delta s} \\ \vdots & = & \vdots\quad\vdots\quad\vdots\quad\vdots\quad\vdots\quad\vdots\quad\vdots\vdots \end{array} $$

In general we have to solve: $$ b_{n} = b_{n - 1}\pars{1 + r} + \bracks{s_{0} + \pars{n - 1}\Delta s}\,,\quad n=2,3,4,\ldots\,;\qquad b_{1} = s_{0}\tag{1} $$

Lets $\quad\ds{{\rm B}\pars{z} \equiv \sum_{n = 1}^{\infty}b_{n}z^{n}}\quad$ with $\quad\ds{\verts{z} < {1 \over 1 + r}}$:

\begin{align} \sum_{n = 2}^{\infty}b_{n}z^{n} &= \pars{1 + r} \sum_{n = 2}^{\infty}b_{n - 1}z^{n} +s_{0}\sum_{n = 2}^{\infty}z^{n} + \Delta s\sum_{n = 2}\pars{n - 1}z^{n} \\[3mm]{\rm B}\pars{z} - b_{1}z &= \pars{1 + r}\ \underbrace{\sum_{n = 1}^{\infty}b_{n}z^{n + 1}}_{\ds{=\ z\,{\rm B}\pars{z}}} + s_{0}\,{z^{2} \over 1 - z} + \Delta s\,{z^{2} \over \pars{1 - z}^{2}} \\[5mm] \bracks{1 - \pars{r + 1}z}{\rm B}\pars{z}& =s_{0}\,{z \over 1 - z} + \Delta s\,{z^{2} \over \pars{1 - z}^{2}} \end{align}

$$ {\rm B}\pars{z} ={s_{0}\ z/\pars{1 - z} + \Delta s\ z^{2}/\pars{1 - z}^{2} \over 1 - \pars{r + 1}z} $$ $$ b_{n}= \frac{\left[\left(r + 1\right)^{n} - n r-1\right]\Delta s + \left[\left(r + 1\right)^{n} - 1\right] r\,s_{0}} {r^{2}} $$ $$ \color{#66f}{\large b_{n}} =\color{#66f}{\large 12000\braces{1220\bracks{\pars{1201 \over 1200}^{n} - 1} - n}} $$

$$ b_{284} \approx 499,325.84\,,\qquad b_{\color{#c00000}{\Large 285}} \approx 502,781.94\,,\qquad b_{286} \approx 506,250.93 $$

$$ \color{#c00000}{\Large 285} = \color{#66f}{\Large 5} \times 48 + \color{#66f}{\Large 11} \times 4 + \color{#66f}{\Large 1} $$

$$\color{#66f}{\large% 5\ \mbox{years}, 11\ \mbox{months and $1$ week}. } $$

enter image description here

$\endgroup$
3
  • $\begingroup$ Thanks for taking the time to answer! However, the different notations and the method you used is confusing to me. Do you know how to form a differential equation for this problem and possibly the general solution? $\endgroup$
    – Wade
    Sep 20 '14 at 17:31
  • $\begingroup$ @Wade You won't have a differential equation because your problem is a discrete one ( you're talking about $\color{#c00000}{\large 1}$ week, $\color{#c00000}{\large 2}$ weeks, $\color{#c00000}{\large 3}$ weeks, etc... ). If you want to have a differential equation you have to take the time interval between deposit as going to zero with some suitable limits for all the involucrated variables. Thanks. $\endgroup$ Sep 20 '14 at 19:47
  • $\begingroup$ Aren't there 52 weeks in a year ? $\endgroup$
    – Tom-Tom
    Sep 22 '15 at 13:03
-3
$\begingroup$

Solution as an image:

Solution

Final Answer:

Final Answer

$\endgroup$
1

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.