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Differential equation involving money

  1. Feb 6, 2008 #1
    A recent college student borrowed $100,000 at an interest rate of 9%. With salary increases, he expects to make payments at a monthly rate of 800*(1+t/120), where t is number of months since the loan was made.

    (a.) Assuming that this payment schedule can be maintained, when will the loan be fully paid?

    I figured that I would need to differentiate this equation (800*(1+t/120)), and i got:
    (96,000)/(14,400) = 6.66

    I have a feeling that I am going about this all wrong. any suggestions?
  2. jcsd
  3. Feb 7, 2008 #2


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    Your title refers to a differential equation but I see no differential equation here. Also I don't see where you have included the 9% interest. Assuming that is annual interest, that would be the same as 9/12= 3/4% per month. Each month the amount he must pay increases by 3/4% and decreases by 800(1+ t/120). I have no idea why you would want to differentiate that or what you set the derivative equal to. If we let A(t) be the amount still owed after t months, then the change, from month t to t+1, is [itex]\delta[/itex]A= (1.0075) A(t)- 800(1+ t/120), giving you a "difference" equation to solve for A(t) and then solve A(t)= 0.

    If you really intend using a differential equation, then we can "average" the amount he pays over the month. Also the interest of 9% per year, if we treat it as "compound continuously" becomes an amount of Ae0.09t. The differential equation is dA/dt= Ae0.09t- 800(1+ t/120). Solve that for A(t) (just integrate) and then solve A(t)= 0 for t.
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