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Hello! I'm working on a finance-related problem, and I'm not sure that I'm solving the ODE well! Here's the prompt: A certain college graduate borrows $8,000 to buy a car. The lender charges interest at an annual rate of \(10\%\). Assuming that interst is compounded \(\sf continuously\), and that the borrower makes payments \(\sf continuously\) at a constant annual rate \(k\), determine the payment rate \(k\) that is required to pay off the loan in \(3\) years. Also, determine how much interest is paid during the \(3\)-year period. The answers: \(k=$3086.64\ /\text{year}\); \($1259.92\)

Differential Equations
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Here's what I did. Let \(D\) be the debt in dollars. Note that \(\dfrac{dD}{dt}=D'=.10D-k\implies D'-.10D=-k\). I'll use \(C\) variables as arbitrary constants. Then \(\Large {D=e^{-\int -.10dt}\int (-k)e^{\int -.10dt}dt\\~\\~\\ =-k\ e^{\int .10dt}\int e^{-\int .10dt}dt\\~\\~\\ =-k\ e^{.10t+C_1}\int e^{-.10t+C_2}dt\\~\\~\\ =-k\ e^{.10t+C_1}(-10 e^{-.10t+C_2}+C_3)\\~\\~\\ =-k\ e^{.10t}e^{C_1}(-10 e^{-.10t}e^{C_2}+C_3)\\~\\~\\ =-k\ e^{.10t}C_4(-10 e^{-.10t}C_5+C_3)\\~\\~\\ =-k\ e^{.10t}C_4(-10) e^{-.10t}C_5-k\ e^{.10t}C_4C_3\\~\\~\\ =-k\ e^0C_4(-10) C_5-k\ e^{.10t}C_4C_3\\~\\~\\ =10k\ C_6-k\ e^{.10t}C_7\\~\\~\\ =k\ C_8-k\ e^{.10t}C_7}\)
Where might I be making a mistake? This doesn't look right... I see that I can simplify: \(\Large\qquad\qquad\qquad = k(C_8-e^{0.10t}C_7)\)
I forget the reason behind it, but I'm pretty sure you just ignore the constant of integration when you find the integrating factor. I end up with \[D(t)=10k+Ce^{-1/10~t}\]

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Okay! I solved on Wolfram Alpha and I think I got something like that! I have to check! But thank you! I was just using a formula that was derived with the integrating factor, I think. I don't quite understand it! :P
You're welcome! And by the way, the reason (according to wikipedia) is that we only need *a* solution, not the general solution, to the integral.
Hmm! Thanks! I'll probably look into this tomorrow when I'm less tired! :) Thank you very much, @SithsAndGiggles !

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