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Differential Equations : Compound Interest

  1. Sep 24, 2014 #1
    1. The problem statement, all variables and given/known data
    A college student wants to start a new savings account with an initial balance of $0. He plans to save money at a

    continuous rate of $700 per month. Additionally, every month he plans to increase this rate by $7. (Such

    that for example in month 3 he is saving at rate $721 per month.) Also, he found a bank account that pays

    continuously compounded interest at a rate of 9% per year. Estimate how long it will take the college to save

    $500,000. Note that you will need to set up and solve a DE, and then you'll need to plot

    the solution to make the final estimate.


    2. Relevant equations



    3. The attempt at a solution

    Setting up the differential equation is the problem for me.

    Let S = the amount of money in his savings account
    let t = time (in months)

    dS/dt = (0.09/12)(S + 7t) I divided 0.09(bank interest) by 12 b/c of the college student adding in money every month

    I multiplied the bank's interest per month by (S+7t) because the interest is acted upon the amount of money in the savings account.

    When I solve for the general equation by using an integrating factor, I can't fulfill the initial conditions in the final general solution.

    What's wrong with my differential equation?

    Thanks!
     
    Last edited by a moderator: Sep 24, 2014
  2. jcsd
  3. Sep 24, 2014 #2

    SteamKing

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    One thing which is confusing: you use S to denote the total amount of funds in the account and also the amount which is deposited every month.

    I would suggest P(t) for the account balance at time t, while S, the amount deposited is a constant $700/mo. This problem is subtle, because the $700 deposit (plus the additional $7/mo/mo increase in the deposit is not paid into the account continuously, but only at one time every month, say on the first day of the month
     
  4. Sep 24, 2014 #3
    So would the DE be dP(t)/dt = (0.08/12)(700+7t) ??
     
  5. Sep 24, 2014 #4

    Ray Vickson

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    In the small time interval from ##t## to ##t + \Delta t## (##t, \Delta t## in units of months) the balance grows by
    [tex] \Delta S = r S \Delta t + (700 + 7 t) \Delta t,[/tex]
    where ##r = 0.09/12##. The first term above is the growth due to compound interest, while the second is the amount paid in from outside sources.
     
  6. Sep 25, 2014 #5

    BiGyElLoWhAt

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    @SteamKing I think that's why the question says estimate. I've seen problems like this that want exact answers, but you have to evaluate them piecewise.
     
  7. Sep 25, 2014 #6

    Ray Vickson

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    No. See my other post.
     
  8. Sep 25, 2014 #7
    Thanks!
     
  9. Sep 26, 2014 #8

    haruspex

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    That's not quite how I read it. The payment graph is a step function, constant during each month but incrementing at each whole month boundary. Maybe that's what you meant, but it's not what you've written.
    Also, it says the interest is paid continuously, in a compound manner, at a rate equivalent to 9% p.a. To me, that implies the interest paid in ##\delta t## is ##r \delta t## where r = ln(1.09).

    Correction: r = ln(1.09)/12. I forgot the units are months.
     
    Last edited: Sep 26, 2014
  10. Sep 26, 2014 #9

    Ray Vickson

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    Logically, what you say is true, but that is typically not how it is done in Finance. There, an annual interest rate of 12%, compounded monthly, is regarded as a rate of 1% per month, even thought that produces a true annual rate of ##100 \times (1.01^{12} -1) \doteq 12.68\%##. So, in this case the continuous rate (in just about any Finance textbook) would be ##r = 0.09##. That corresponds to a true annual rate of ##100 \times (e^{.09}-1) \doteq 9.417\%##. Of course, if one works in units of months rather than years the rate is 0.09/12.

    The question did say the $700 per month was paid in continuously. I suppose the issue is what to do about the $7/month increase in the savings rate. If one imposes it discretely---once per month---one ends up with a discrete model with sums instead of integrals and differences instead of derivatives. The suggestion to set up and solve a differential equation seems to indicate that something else was intended, and that was what I assumed: that the '7' was an instantaneous rate of increase.
     
  11. Sep 26, 2014 #10

    haruspex

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    I agree it's not clear.
    In my interpretation, you still need to integrate across a month, but then sum the months.
     
  12. Sep 26, 2014 #11

    Ray Vickson

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    There is one final issue: if we take the deposit rate as ##C + 7x## at time (month) ##x##, then between times ##t## and ##t+1## the total deposit is ##\int_t^{t+1} [C + 7x] \, dx = C + 7(t + \frac{1}{2}) = C+3.5 + 7t##. This does, indeed, increase at the rate of 7 $/mo, but the total deposit in month 1 is ##\$(C+3.5).## To match the problem description exactly we should not take ##C = 700## (as I did originally), but instead should use ##C = 696.5##. That would give a total deposit of $700 in month 1, $707 in month 2, etc. It does not make much of a difference to the solution: using ##700 + 7x## or ##696.5+7x## in the future-value formula yields a difference of about 1/4 month in the final answer; in other words, the first form gets to $500,000 about 1/4 month sooner than the second one.
     
  13. Sep 26, 2014 #12

    haruspex

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    No, in month m (counting from 0), t runs from m to m+1. The deposit in that time is 700+7(m+1). (I would have said 700+7m, but it explicitly says the rate is 721 for the third month, m=2.). The complication is that interest is earned conitinuouly through the month.
    One way to tackle it would be to calculate the end result that comes from the deposit in month m, then sum over m.
     
  14. Sep 26, 2014 #13

    Ray Vickson

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    I take month 1 to start at time t = 0 and end at time t = 1, because it seemed most natural to me to start the clock at time 0.

    Anyway, there is a bit of a mis-match between the statement of the problem and the "for example" part: one part says " He plans to save money at a
    continuous rate of $700 per month." while the other part says "Such that for example in month 3 he is saving at rate $721 per month." According to the latter the savings rate in month 2 is $714 and so in month 1 is $707. There would be no month in which the saving rate was $700. This problem arises whether or not one uses a continuous or discrete-time analysis. In the continuous time case one can match the 'for example' explanation by taking the saving rate at time ##t## (months) to be ##703.5 + 7t##.

    If we take the '700' to be a continuous rate but the '7' to be discrete (after dealing with the issue of whether the extra 7 comes at the start or the end of a month) we could get a DE, but it would be ugly:
    [tex] \frac{dS}{dt} = \begin{cases}rS + 700, & 0 < t <1\\
    rS + 707,& 1 < t < 2\\
    rS + 714, & 2 < t < 3 \\
    \vdots & \vdots
    \end{cases}
    [/tex]
    (or, maybe the first RHS should be ##rS + 707##, etc,)
    This is doable, but sufficiently complicated that one might as well go directly to a fully discrete-time model and avoid the DE altogether. Again, in view of the suggestion to use a DE, I would guess that this is not what the problem's poser had in mind.
     
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