Loan Interest Question

Question

Clarissa wants to buy a new car. Her loan officer tells her that her annual interest is 8%, compounded continuously, over a four-year term. Clarissa informs her loan officer that she can make equal monthly payments of $225. How much can Clarissa afford to borrow?

I tried to solve this problem twice using $$P' = (2/25)P - 2700, P(4) = 0, P(0) =\ ?$$ and $$P' = (2/25)P - 225, P(48) = 0, P(0) =\ ?$$ and ended up with two different answers for $P(0)$. Which setup, if either, is correct?

Given that the problem appears in a differential equations book (in an early section), is it safe to assume that the "equal monthly payments" are actually being paid continuously, rather than discretely at the end of each month, or can the compounding frequency and payment frequency still be operating according to two different "clocks"?

Answer 1

The first setup is correct. There is no further condition needed.

$$P' = 0.08\cdot P - 2700, P(4) = 0$$

Method 1 (LutzL method)

Separation of the variables

$\frac1{0.08\cdot P - 2700} \ dP= dt$

Integrating both sides

$\int \frac1{0.08\cdot P - 2700} \ dP= \int dt$

$\frac1{0.08}\ln(0.08\cdot P - 2700)=t+c$

$\ln(0.08\cdot P - 2700)=0.08\cdot t+c_1$

etc.

Hint 1: $P(t)=C\cdot e^{0.08t}+33750$

Hint 2: After using the condition $P(4)=0$ you should get $C=-24507.53$

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Method 2 (laborious method)

You can solve this first order inhomogeneous equation by the method of variation of constants. If you use this method you firstly have to solve homogeneous differential equation.

$$P' = 0.08\cdot P$$

Separation of the variables

$$\frac1{P}dP = 0.08 \ dt$$

etc.

I leave the remaining work for you. But if you have any questions feel free to ask.

Answer 2

Well, the amount owed after one month would be given by:

$$Pe^{1/150}-225$$

After two months would be:

$$(Pe^{1/150}-225)e^{1/150}-225 = Pe^{2/150}-225\cdot e^{1/150}-225$$

After 48 months would be:

$$e^{8/25}\left(P-225\sum_{n=1}^{48}e^{-n/150}\right)$$

Here, you have a geometric sum:

$$r = e^{-1/150}, \sum_{n=1}^{48}r^n = \dfrac{1-r^{49}}{1-r}-1$$

So, this is:

$$e^{8/25}\left(P-225\sum_{n=1}^{48}e^{-n/150}\right) = e^{8/25}\left(P-225\cdot \dfrac{e^{-1/150}-e^{-49/150}}{1-e^{-1/150}}\right)\le 0$$

So, this gives:

$$P \le 225\left(\dfrac{e^{8/25}-1}{e^{49/150}-e^{8/25}}\right) \approx 9,211.696$$

Answer 3

You have a discrete time process. Assuming payment at the end of the month, as the available cash needs not be included in the loan calculation, the remaining debt at the start of the next month is $$ a_{n+1}=qa_n-r $$ where $q^{12}=1.08$ and $r=225$. This is known to have the closed formula $$ a_n-\frac{r}{q-1}=q^n\left(a_0-\frac{r}{q-1}\right)\implies a_n=q^na_0-\frac{q^n-1}{q-1}r $$ With $a_{48}=0$ this solves to $$ a_0=\frac{r(1-q^{-48})}{q-1}=9266.087... $$