theEric
  • theEric
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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theEric
  • theEric
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
jamiebookeater
  • jamiebookeater
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At vero eos et accusamus et iusto odio dignissimos ducimus qui blanditiis praesentium voluptatum deleniti atque corrupti quos dolores et quas molestias excepturi sint occaecati cupiditate non provident, similique sunt in culpa qui officia deserunt mollitia animi, id est laborum et dolorum fuga. Et harum quidem rerum facilis est et expedita distinctio. Nam libero tempore, cum soluta nobis est eligendi optio cumque nihil impedit quo minus id quod maxime placeat facere possimus, omnis voluptas assumenda est, omnis dolor repellendus. Itaque earum rerum hic tenetur a sapiente delectus, ut aut reiciendis voluptatibus maiores alias consequatur aut perferendis doloribus asperiores repellat.

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theEric
  • theEric
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 \(\Large {D=e^{-\int -.10dt}\int (-k)e^{\int -.10dt}dt\\~\\~\\ =-k\ e^{\int .10dt}\int e^{-\int .10dt}dt\\~\\~\\ =-k\ e^{.10t+C_1}\int e^{-.10t+C_2}dt\\~\\~\\ =-k\ e^{.10t+C_1}(-10 e^{-.10t+C_2}+C_3)\\~\\~\\ =-k\ e^{.10t}e^{C_1}(-10 e^{-.10t}e^{C_2}+C_3)\\~\\~\\ =-k\ e^{.10t}C_4(-10 e^{-.10t}C_5+C_3)\\~\\~\\ =-k\ e^{.10t}C_4(-10) e^{-.10t}C_5-k\ e^{.10t}C_4C_3\\~\\~\\ =-k\ e^0C_4(-10) C_5-k\ e^{.10t}C_4C_3\\~\\~\\ =10k\ C_6-k\ e^{.10t}C_7\\~\\~\\ =k\ C_8-k\ e^{.10t}C_7}\)
theEric
  • theEric
Where might I be making a mistake? This doesn't look right... I see that I can simplify: \(\Large\qquad\qquad\qquad = k(C_8-e^{0.10t}C_7)\)
anonymous
  • anonymous
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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theEric
  • theEric
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
anonymous
  • anonymous
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.
theEric
  • theEric
Hmm! Thanks! I'll probably look into this tomorrow when I'm less tired! :) Thank you very much, @SithsAndGiggles !

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