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Say you have a bank account in which your invested money yields 3% every year, continuously compounded. Also, you have estimated that you spend $1000 every month to pay your bills, that are withdrawn from this account.

Create a differential model for that, find its equilibriums and determine its stability.

My problem here is that the \$1000 withdrawal is not continuous on time, it's discrete. The best I could achieve is, if $S(t)$ is the current balance: $\dot S (t) = 0,0025S(t) - 1000$. I'm using $0,0025$ as the interest rate because it yields 3% every year, so it should yield 0,25% every month. But I'm pretty confident that it's wrong. Any help would be highly appreciated! Thanks!

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    $\begingroup$ Perhaps they meant to model the expenditures as continuous as well? If the expenditures are time varying then you do not have enough information to model the problem. $\endgroup$
    – copper.hat
    May 27 '16 at 19:07
  • $\begingroup$ Yes, I think so, but I don't know how to do that. :/ $\endgroup$ May 27 '16 at 19:07
  • $\begingroup$ How is money expended? On a daily basis? Once per month? First day? Last day? $\endgroup$ May 27 '16 at 19:17
  • $\begingroup$ The problem set just says "you every-day purchases sum around $1000 per month". I'm pretty confident that the question wants me to model everything continuously but I'm not sure how to do it. $\endgroup$ May 27 '16 at 19:24
  • $\begingroup$ It seems to me that if $t$ is to be measured in years, then $\dot S(t)=.03S-1000 \lfloor {12t}\rfloor$. $\endgroup$
    – Taylor
    May 27 '16 at 20:11
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Let $x (t)$ be the amount of money in the account at time $t$ (years). Hence, if no money is spent,

$$\dot x = r x$$

where $r = \ln (1.03)$. If $\$1000$ is spent continuously every month, then we have the ODE

$$\dot x = r x - 12000$$

We have an equilibrium point when we have

$$\bar{x} := \frac{12000}{r} \approx \$406,000$$

in the account, as the interest earned per year then equals the amount of money expended per year. If we have more than $\bar{x}$ in the bank, then our wealth is growing. If we have less than $\bar{x}$ in the bank, then our wealth is decaying. Let us verify. Integrating the non-homogeneous ODE above, we obtain

$$x (t) = \bar{x} + (x_0 - \bar{x}) \, \mathrm{e}^{r t}$$

If $x_0 > \bar{x}$, our wealth is growing. If $x_0 < \bar{x}$, our wealth is decaying. If $x_0 = \bar{x}$, our wealth is stationary. Note that $\bar{x}$ is an unstable equilibrium point.

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  • $\begingroup$ Thanks a lot, got it! In your solution, the time is measured in years. If we represent time in months, the correct interest rate should be $r=\frac{\ln (1.03)}{12}$, right? $\endgroup$ May 28 '16 at 14:19
  • $\begingroup$ @RaulGuarini Yes. And the ODE becomes $\dot x = r x - 1000$. $\endgroup$ May 28 '16 at 15:01
  • $\begingroup$ @RaulGuarini Why doesn't $r = .03$ or $\frac{.03}{12}$? $\endgroup$
    – user10478
    Jul 3 '19 at 3:05
  • $\begingroup$ @user10478 If $r = \ln (1.03)$, then $$e^{rt} = \left( \exp (r) \right)^t = \left( \exp(\ln( 1.03 )) \right)^t = (1 + 0.03)^t$$ $\endgroup$ Jul 6 '19 at 10:35
  • $\begingroup$ @RodrigodeAzevedo Hmm, is this because the problem says "yields $3\%$ every year," whereas $r$ would be $.03$ if it said "annual interest rate of $3\%$"? $\endgroup$
    – user10478
    Jul 6 '19 at 11:10
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The ballance must be sufficient to generate $1,000 / month in cash flow

$B(e^{0.0025}-1) = 1000\\ B = \dfrac{1000}{e^{0.0025}-1} = \$399,500$

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The money flow consists of two contributions $$ \dot{S} = \dot{S}_y + \dot{S}_m $$ with the continous contribution $$ \dot{S}_y = a S \quad (*) $$ where $a$ must be adjusted to give the yearly interest rate such that $$ S_y(1\text{y}) = (1 + p) S_y(0\text{y}) $$ for $p = 3\% = 3/100$ and the monthly part $$ \dot{S}_m = -M \sum_{k=1}^\infty \delta(t - k\cdot 1\text{m}) $$ with $M = 1000 \$$.

Determining $a$ from $p$: Equation $(*)$ means $$ S_y(t) = S_y(0\text{y}) e^{at} $$ and $$ S_y(1\text{y}) = S_y(0\text{y}) e^{a \cdot 1 \text{y}} = (1 + p) S_y(0\text{y}) $$ so $a = \ln(1+p)/1\text{y}$. In summary: $$ \dot{S} = S(0\text{y}) \, (1+p)^{t/1\text{y}} - M \sum_{k=1}^\infty \delta(t - k\cdot 1\text{m}) $$

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