# Compound interest Differential Equation

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! • Are you asked to build a differential equation ? Sep 20 '14 at 2:59 • @ClaudeLeibovici It should be some recurrence equation hopefully !!!... Sep 20 '14 at 5:04 • Yes, it does asks to build a DE. – Wade Sep 20 '14 at 17:28 ## 2 Answers$\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} $${\rm B}\pars{z} ={s_{0}\ z/\pars{1 - z} + \Delta s\ z^{2}/\pars{1 - z}^{2} \over 1 - \pars{r + 1}z}$$ $$b_{n}= \frac{\left[\left(r + 1\right)^{n} - n r-1\right]\Delta s + \left[\left(r + 1\right)^{n} - 1\right] r\,s_{0}} {r^{2}}$$ $$\color{#66f}{\large b_{n}} =\color{#66f}{\large 12000\braces{1220\bracks{\pars{1201 \over 1200}^{n} - 1} - n}}$$ $$b_{284} \approx 499,325.84\,,\qquad b_{\color{#c00000}{\Large 285}} \approx 502,781.94\,,\qquad b_{286} \approx 506,250.93$$ $$\color{#c00000}{\Large 285} = \color{#66f}{\Large 5} \times 48 + \color{#66f}{\Large 11} \times 4 + \color{#66f}{\Large 1}$$ $$\color{#66f}{\large% 5\ \mbox{years}, 11\ \mbox{months and 1 week}. }$$ • 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? – Wade Sep 20 '14 at 17:31 • @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. Sep 20 '14 at 19:47
• Aren't there 52 weeks in a year ? Sep 22 '15 at 13:03

Solution as an image: