Loan repayment with increasing monthly payments differential equation

[DivGradCurl]

"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:

$$S_0 = \$ 100,000$$

$$r = 0.09$$

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

$$\frac{dS}{dt}=rS-k(t), \qquad S(0)=S_0$$

$$\frac{dS}{dt}-rS=-k(t)$$

$$\mu = \exp \left( -r \int dt \right) = e ^{-rt}$$

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

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

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

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

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

$$S(t)=0 \Rightarrow t \approx - 131 \mbox{ months}$$

which is WRONG!

Any help is highly appreciated.

 

[OlderDan]

"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:

$$S_0 = \$ 100,000$$

$$r = 0.09$$

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

$$\frac{dS}{dt}=rS-k(t), \qquad S(0)=S_0$$

$$\frac{dS}{dt}-rS=-k(t)$$

$$\mu = \exp \left( -r \int dt \right) = e ^{-rt}$$

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

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

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

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

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

$$S(t)=0 \Rightarrow t \approx - 131 \mbox{ months}$$

which is WRONG!

Any help is highly appreciated.

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

$$S(t)=S_0 e^{rt/12}$$

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

$$\frac{dS}{dt}=rS/12 -k(t), \qquad S(0)=S_0$$

 

[whozum]

Your integration is all correct, so it is something small, OlderDan seems to be making sense to me.

 

[DivGradCurl]

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.

Thanks