Deriving a differential equation for a loan/interest problem

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SUMMARY

The discussion focuses on deriving a differential equation for a loan repayment scenario involving continuous interest and decreasing payment rates. The borrower pays back a loan amount S at a continuous interest rate r% per month, with repayments modeled by K₀e^{-at}. The differential equation established is dS/dt = rS - K₀e^{-at}, which simplifies to S' - rS = -K₀e^{-at}. The integration of the equation is crucial for finding the solution, particularly the integral of -K₀e^{-t(a+r)}.

PREREQUISITES
  • Understanding of differential equations and their applications.
  • Familiarity with continuous interest calculations.
  • Knowledge of integration techniques, particularly for exponential functions.
  • Basic concepts of loan amortization and repayment structures.
NEXT STEPS
  • Study the method of integrating exponential functions, specifically ∫ e^(kt) dt.
  • Explore the application of differential equations in financial modeling.
  • Learn about continuous compounding in finance and its mathematical implications.
  • Investigate various repayment models for loans, including decreasing payment structures.
USEFUL FOR

Students studying differential equations, financial analysts, and anyone interested in mathematical modeling of loan repayment scenarios.

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Homework Statement


You borrow money from a friend at a continuous interest rate of r% per month. You want to pay your friend back as quickly as you can at the beginning, but reduce your payment rate over time. You decide to pay off at a continuously decreasing rate given by K₀e^{-at}, in dollars per month.

Write a differential equation that describes how much you owe and solve it.


Homework Equations


None


The Attempt at a Solution


Let S be the amount borrowed -

dS/dt = rS - K₀e^{-at}

S' - rS = -K₀e^{-at}

S'(I(x)) - rS(I(x)) = -K₀e^{-at}(I(x))

Se^{-rt} = -K₀\inte^{-t(a+r)}

Se^{-rt} = ...

This is where I get stuck, I have don't understand how to integrate -K₀\inte^{-t(a+r)}, any hints?
 
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if 'a' and 'r' are constants then you can simply recall that ∫ ekt = (1/k)ekt+ constant.
 

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