JustCurious' question at Yahoo Answers regarding a mathematical model

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To determine the continuous payment rate required to pay off an $8,000 car loan at a 10% annual interest rate compounded continuously over three years, a differential equation model is utilized. The model is expressed as dL/dt = rL - k, where L is the loan balance, r is the interest rate, and k is the payment rate. By solving the initial value problem and applying integration techniques, the payment rate k is derived. Substituting the known values into the equation yields a required annual payment rate of approximately $3,086.64. This calculation provides a clear financial strategy for managing the loan repayment effectively.
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Here is the question:

Differential modeling help?

A college grad. borrows $8000 dollars for a car. The lender charges interest at an annual rate of 10%. Assuming interest is compounded continuously and that the borrower makes payments continuously at a constant annual rate of k, determine the payment rate k that is required to pay off the loan in 3 years.

I have posted a link there to this topic so the OP can see my work.
 
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Hello JustCurious,

If we let $L$ be the balance in dollars left on the loan at time $t$ in years, and $r$ be the annual interest rate, then from the given information, we may model the loan balance with the following IVP:

$$\frac{dL}{dt}=rL-k$$ where $$L(0)=L_0$$

Separating variables, and switching dummy variables of integration to allow using the boundaries as limits, we obtain:

$$\int_{L_0}^L\frac{1}{ru-k}\,du=\int_0^t\,dv$$

Applying the anti-derivative form of the FTOC, we get:

$$\ln\left|\frac{rL-k}{rL_0-k} \right|=rt$$

Converting from logarithmic to exponential form, we have:

$$\frac{rL-k}{rL_0-k}=e^{rt}$$

Solving for $k$, there results:

$$k=\frac{r\left(L_0e^{rt}-L \right)}{e^{rt}-1}$$

Plugging in the desired data:

$$r=\frac{1}{10},\,L_0=8000,\,t=3,\,L=0$$

we obtain:

$$k=\frac{800}{1-e^{-\frac{3}{10}}}\approx3086.636730808066$$

Hence, we find that the annual rate of repayment is about $3,086.64.
 
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