I've been reading on these questions, but still cannot solve this, please tell me where I did wrong.
The question asks: A borrows "100,000 dollars" at "annual interest 9%", plans to make payment at a "monthly rate 800(1+t/120)". When will the debt be paid?
The way I approach it was: rate of change of debt=annual interest rate*current debt-yearly payment
So: dp(t)/dt = 0.09*p(t)-(yearly payment), which is 9600+960t-440
So: dp(t)/dt = 0.09*p(t)-(960t+9160) dt
Integrating Factor: u(t) = exp∫-0.09 = e^(-0.09t)
And p(t)= e^(0.09t)∫e^(-0.09t)(-(960t+9160))dt, integration by parts:
p(t)=((960t+9160)/0.09+960/(0.09)^2+C), plug in p(0)=100000
C is roughly -120296.28
p(t)=(9160/0.09)+960/(0.09)^2-(120296.28)
When I put these numbers into directional field software, it makes no sense. The solution have positive slopes, meaning the debt will grow. I could think of a few steps where I might got it wrong:
The payment was originally in months, and I add them up into yearly intervals, I'm not sure if it's correct.
My intergrating factor should be ok, but maybe the integrals are done incorrectly, please help.
I would set this up as an net present value problem.
The NPV of the future cash-flows equals the loan amount.
$100,000 = \sum_\limits{t=1}^{n} \frac {800(1+\frac {t}{120})}{(1+\frac {9\%}{12})^t}\\ 100,000 = 800\sum_\limits{t=1}^{n} \frac {1}{1.0075^t}+\frac {800}{120}\sum_\limits{t=1}^{n} \frac {t}{1.0075^t}\\ 100,000 =800\frac {1-1.0075^{-n}}{0.0075} + \frac {800}{120}(\frac {1.0075(1-1.0075^{-n})}{(0.0075)^2}-\frac {n(1.0075^{-n})}{0.0075})$
I don't think you will find an exact solution for $n$. But you can find an approximate solution.
You want a diff eq.?
$y'=0.09y - 9600-80t\\ y'-0.09y = -9600-80t\\ e^{-0.09t}y' -0.09e^{-0.09t} = -9600e^{-0.09t} - 80te^{-0.09t}\\ e^{-0.09t}y = 10666te^{-0.09t} + 889te^{-0.09t} + 9877 e^{-0.09t} + C\\ y = 116,543.2 + 889t + Ce^{0.09t}\\ y(0)=100,000\\ C = -16,543.2\\ 116,543.2 + 889t -16,543.2e^{0.09t} = 0\\ t\approx 23.5 \text{ years}$
vs. $135$ months using the other technique. The big difference is that one method uses monthly compounding and the other uses continuous compounding.