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Homework Help: First Order Differential Equation

  1. May 19, 2005 #1
    "A recent college graduate borrows $100,000 at an interest rate of 9% to purchase a condominium. Anticipating steady salary increases, the buyer expects to make payments at a monthly rate of 800(1 + t/120), where t is the number of months since the loan was made. Assuming that this payment schedule can be maintained, when will the loan be fully paid?"

    Please, help me find where I made a mistake. Here's what I've got:

    [tex]S_0 = \$ 100,000 [/tex]

    [tex]r = 0.09[/tex]

    [tex]k(t) = \$ 800 \left( 1 + \frac{t}{120} \right) / \mbox{month}[/tex]

    [tex]\frac{dS}{dt}=rS-k(t), \qquad S(0)=S_0[/tex]


    [tex]\mu = \exp \left( -r \int dt \right) = e ^{-rt}[/tex]

    [tex]S(t)=e^{rt}\int -800 \left( 1 + \frac{t}{120} \right) e ^{-rt} \: dt[/tex]

    [tex]S(t)=e^{rt}\left( \frac{20e^{-rt}}{3r^2} + \frac{800e^{-rt}}{r} + \frac{20e^{-rt}t}{3r} + \mathrm{C} \right)[/tex]

    [tex]S(t) = \frac{20t}{3r} + \frac{800}{r} + \frac{20}{3r^2} + \mathrm{C} e^{rt}[/tex]

    [tex]t=0 \Rightarrow \frac{800}{r} + \frac{20}{3r^2} + \mathrm{C} = S_0 \Rightarrow \mathrm{C} = \frac{-20-2400r+3r^2 S_0}{3r^2} [/tex]

    [tex]S(t)=\frac{20}{3r^2} - \frac{20e^{rt}}{3r^2} + \frac{800}{r} - \frac{800e^{rt}}{r} + S_0 e^{rt} + \frac{20t}{3r}[/tex]

    [tex]S(t)=0 \Rightarrow t \approx - 131 \mbox{ months} [/tex]

    which is WRONG!!!!

    Any help is highly appreciated.
  2. jcsd
  3. May 19, 2005 #2


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    I can't say I completely follow what you are doing, but it seems to me you have two different time scales going on and you may have lost track of the initial value a few equations into your analysis. If you were making no payments, the amount owed would be increasing every month. In the continuous interest limit you would have

    [tex]S(t)=S_0 e^{rt/12}[/tex]

    where r is the annual interest rate and t is in months. The payments were already expressed in terms of t in months and of course they reduce the value of S(t). So I think you need

    [tex]\frac{dS}{dt}=rS/12 -k(t), \qquad S(0)=S_0[/tex]
  4. May 19, 2005 #3
    Your integration is all correct, so it is something small, OlderDan seems to be making sense to me.
  5. May 20, 2005 #4
    You're right. It's something pretty small. I should have divided r by 12, which ultimately gives the correct answer: t = 135.36 months.

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