(adsbygoogle = window.adsbygoogle || []).push({}); 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.

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# Homework Help: ODE Modelling question

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