Welcome to our community

Be a part of something great, join today!

JustCurious' question at Yahoo! Answers regarding a mathematical model

  • Thread starter
  • Admin
  • #1

MarkFL

Administrator
Staff member
Feb 24, 2012
13,775
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.
 
  • Thread starter
  • Admin
  • #2

MarkFL

Administrator
Staff member
Feb 24, 2012
13,775
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:

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

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

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

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

\(\displaystyle \ln\left|\frac{rL-k}{rL_0-k} \right|=rt\)

Converting from logarithmic to exponential form, we have:

\(\displaystyle \frac{rL-k}{rL_0-k}=e^{rt}\)

Solving for $k$, there results:

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

Plugging in the desired data:

\(\displaystyle r=\frac{1}{10},\,L_0=8000,\,t=3,\,L=0\)

we obtain:

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

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