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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"?

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  • $\begingroup$ Are there other examples or tasks in the book that would justify this continuous transfer of repayments? It could still be an example of a discrete dynamical process, using more obvious examples like nuclear decay or chemical reactions for the switch to continuous dynamics. $\endgroup$ – LutzL Jul 4 at 15:19
  • $\begingroup$ @LutzL Yes! This is Exercise 6 from Section 3.3 at academia.edu/32329978/… The paragraph before Exercise 11 seems to imply that the previous ten assumed continuous deposits/withdrawals/payments, but it also seems to conflate this with the continuity of compounding. For example, Exercise 15 sounds symmetric in every way except compounding period, but my current hypothesis is that 15a secretly envisions continuous payments, while 15b secretly envisions monthly payments. $\endgroup$ – user10478 Jul 5 at 15:55
  • $\begingroup$ The wording in Exercise 14 is slightly more viable for this interpretation, with the grammatical distinction, "deposit every year," versus "deposited at the end of each compounding period," but I still find it far from clear. $\endgroup$ – user10478 Jul 5 at 15:57
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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 $C=-24507.53$


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.

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    $\begingroup$ You can also directly go to the separation of variables, $\frac{dP}{0.08P-2700}=dt$ integrates as well as the homogeneous equation. $\endgroup$ – LutzL Jul 4 at 16:30
  • $\begingroup$ @LutzL Good catch. I will include this fact into my answer. $\endgroup$ – callculus Jul 4 at 19:26
  • $\begingroup$ How is Method 1 the LutzL method? Isn't LutzL's answer a difference equation? Also, perhaps I should have mentioned that I'm not looking for an explanation of how to solve a separable or linear ODE. I'm good with that, but having trouble understanding the modeling aspect. $\endgroup$ – user10478 Jul 5 at 15:59
  • $\begingroup$ The LutzL method is referring to his comment below my answer. Read it. // Can you please say what do you not understand? $\endgroup$ – callculus Jul 5 at 16:02
  • $\begingroup$ @user10478 If you want to adjust your second approach you have to divide the interest rate by 12 (12 month=1 year): $$\boxed{\large{P'=\frac{0.08}{12}\cdot P-225, P(48)=0}}$$ $\endgroup$ – callculus Jul 5 at 16:16
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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:

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

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  • $\begingroup$ The difference between my answer and LutzL's answer is that I assume the 8% annual interest rate is not the annual yield, which would be $e^{0.08} \approx 1.083287$. $\endgroup$ – InterstellarProbe Jul 3 at 20:45
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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 $$ a_0=\frac{r(1-q^{-48})}{q-1}=9266.087... $$

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