JustCurious' question at Yahoo Answers regarding a mathematical model

  • Context: Undergrad 
  • Thread starter Thread starter MarkFL
  • Start date Start date
  • Tags Tags
    Mathematical Model
Join the discussion
Registration is free. Ask a follow-up in this thread, or start your own.
1 reply · 3K views
MarkFL
Gold Member
MHB
Messages
13,284
Reaction score
12
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.
 
Physics news on Phys.org
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.