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Homework Help: ODE Modelling question

  1. Apr 18, 2010 #1
    1. The problem statement, all variables and given/known data
    A college graduate borrows $100,000 at an interest rate of 9% to buy a house. The college graduate plans to make payments at a monthly rate of 800(1+t/120), t is the number of months since the loan was made. Assuming this payment schedule can be maintained, when will the loan be fully paid?

    3. The attempt at a solution
    Let S(t) be the amount to be paid at some t
    S(0) = 100,000
    r = interest rate = .09
    k = constant payment rate = -(800 + 20t/3)

    So the differential equation modeling this problem is:

    (i) [tex] \frac{dS}{dt}\right = rS +k[/tex]

    The equation is linear and its integrating factor is u = Exp[-rt] so multiplying eq. (i) through by u, rewriting LHS by as the derivative of the product S(t)*Exp[-rt], and integrating across t:

    (ii) [tex]S(t)*e^-^r^t=\int-e^-^r^t(800+20t/3)dt + C[/tex]

    Integrating the RHS integral by parts letting U = 800+20t/3, dV = -Exp[-rt]:

    (iii) S(t)*Exp[-rt] = (1/r)*(800+20t/3) - (20/3r) \int Exp[-rt]dt + C

    Evaluating the last integral, and multiplying eq. (iii) through by Exp[rt]:

    (iv) S(t) = (1/r)(800+20t/3) + (20/3r^2) + CExp[rt]

    Exploiting the initial condition, C = 100,000 - (1/r)(800+20t/3) - (20/3r^2), so:

    S(t) = (1/r)(800+20t/3) + (20/3r^2) + (100,000 - (1/r)(800+20t/3) - (20/3r^2))Exp[rt]

    So the loan should be paid at some time t` when S(t`) = 0. Since the expression for S(t) has arguments of t as a scalar multiple and raised as an exponent, I solved this problem numerically. Using mathematica, it spits a value of t =~ -131.12 months. The answer provided by my book says t =~ 135.36 months.

    I'm pretty stumped on this problem... Any help would be greatly appreciated.

    PS Sorry latex doesn't seem to be working for me.
    Last edited: Apr 18, 2010
  2. jcsd
  3. Apr 19, 2010 #2


    Staff: Mentor

    I think that what's going on here is that you modeled the situation by a continuous function in your differential equation, while the actual situation involves payments made at discrete time intervals.

    To give you an idea of the difference between the two kinds of processes, consider the following sum and integral, both of which involve the same function, f(x) = 1/x.
    [tex]\sum_{n = 2}^{10} \frac{1}{n}[/tex]


    The sum comes out to about 1.929, while the integral comes out to about 2.303. The sum represents a lower bound on the area beneath the graph of y = 1/x.

    I suspect that what the author of your text wrote a finite sum that represented all of the payments, and then solved the resulting equation for the time t. I believe that's the reason for the discrepancy between what you got and the answer shown in the book.
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