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I am having trouble interpreting the meaning of this differential equation model for interest on an account. The problem is as follows:

Assume you have a bank account that grows at an annual interest rate of r and every year you withdraw a fixed amount from the account (denoted w). Assuming continuous compounding interest and continuous withdrawal, the described account follows the differential equation:

${\frac{dP}{dt} = rP - w}$

Where P(t) is the amount of money in the account at time t.

I am confused as to why we subtract the entire amount ${w}$ in the equation? My interpretation (which is obviously incorrect) is that ${\frac{dP}{dt}}$ is the change in the amount of money in the account with respect to time for any given time ${t}$. If this is so, wouldn't the above equation mean that at every instant ${t}$ we are adding ${rP}$ to the account and subtracting the entire annual deduction ${w}$ so that at the end of an entire year we've deducted more than ${w}$ from the account?

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  • $\begingroup$ FYI: I edited your question to correct a misspelling in the title. $\endgroup$
    – wltrup
    Jul 30 '15 at 0:01
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$r$ is a rate of interest, meaning a percentage per unit time. So say $r$ is 5% per year, the instantatneous rate of growth of the account is 5% of the amount in the account per year: $rP$.

Similarly you have to interpret $w$ as the withdrawal rate per unit time, for example, $1,000 per year.

You are right that the equations are meaningless if yo take $w$ to mean that every instant you withdraw the finite amount $w$. But that is not what is meant.

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  • $\begingroup$ That makes sense. I was having trouble dealing conceptually with how units fit into the equation. I feel like I am understanding calculus theoretically and I get most of the proof problems right, but as soon as it comes to application problems where there's units and modelling and estimation I can't make heads or tails of it. $\endgroup$ Jul 29 '15 at 22:44
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$\frac{dP}{dt}$ measures the rate of increase of $P$ in currency units per year

At any instant the total in the account is increasing at a rate of $rP$ per year and decreasing at a rate of $w$ per year.

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There is a small problem with the whole setup and maybe I can clarify the issue, by pointing out this small error when constructing the differential equation. Although I am not sure how banks do it, the differential equation should be valid for any amount of time passing: $t$ is measured in years, but fractional values are still allowed. So, an effective annual return is assumed to work like that: $$P(1)=P(0)\times(1+r)\,,$$ not taking withdrawals into account. But waiting just one week and looking at the account needs a different $r$, i.e. this would not be $$P(1/52)\ne P(0)\times\left(1+\frac{r}{52}\right)\,,$$ as you could iterate that to something inconsistent. 52 weekly returns need a different rate constant than one yearly return: $$P(52\times1/52)=P(0)\times\left(1+\frac{r}{52}\right)^{52} \ne P(0)\times(1+r)$$ So, depending on how much smaller the given period is, the return percentage needs to be a bit smaller than $r$. The correct (effective, infitesimal) rate constant $k$, valid for tiny steps, should be calculated as: $$\log(1+r)=\log\left(\lim_{n\to\infty} (1+k/n)^n\right) = k\,,$$ this is because the limit turns out to be an exponential: $\exp(k)$ (hence we take the logarithm).

As a side note: for small $r$, that turns out to be: $$r \ll 1: \quad k\approx \frac{r}{1+r}\,,$$ so, definitely not $r$: $k\ne r$.

Given Example (with $r=0.05$): By taking the natural logarithm of $(1+0.05)$ we get $k=0.04879$ (slightly less than $r$).

Solving the differential equation, with known $k$ $$\dot P = kP\,,\quad P(0)=P_0$$ we get $$P(t)=P_0 \exp(kt)\,,$$ which, after one year, yields $P(1)=P_0\times 1.05\,,$ as we wanted it to do.

Now, annual accumulated withdrawals are tricky as they can occur at precise moments, or spread out. But, if we assume an almost steady (constant) rate of—say—daily withdrawals: $$365\times\left(P(t+1/365) - P(t)\right)=-w\,,$$ which amount to $w$, it begins to look like an approximated derivative. Assuming small time steps between withdrawals, we get: $$\Delta_t \ll 1:\quad \dot P \approx \frac{P(t+\Delta_t) - P(t)}{\Delta_t} = -w$$ So, no surpises, just a linear rate constant: any amount of time can be multiplied with this rate to get the aggregated withdrawals for that time length. Combining the two actions we get: $$\frac{dP}{dt}=-kP-w\,,\quad P(0)=P_0\,,$$ where $k=\log(1+r)\,.$

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$w$ is the (continuous, constant) rate of withdrawal, and $rP$ is the (continuous, proportional) rate of interest accrual.

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