A person places an initial deposit of 25000 in an account with a rate of 5% per year, compounded continuously. The person continuously withdraws 700 per year from the account. Find the value of the account at time t after the initial deposit.

I get this linear DE. Is the question just a matter of solving this equation? $$dA/dt = (0.05)A - 700$$

so $$ln(A-14000)^{20} = t + C$$

I greatly appreciate any help

  • $\begingroup$ You look good so far. Since the initial deposit is 25000, you get that when $t = 0$ you have $A = 25000$. So plug that in to your equation to solve for $C$. $\endgroup$ Sep 25, 2016 at 23:51
  • 1
    $\begingroup$ So yes, you need to solve $dA/dt=0.05A-700,A(0)=25000$. That exponent of $20$ makes me a little bit uneasy. To me you have $\frac{dA}{0.05A-700}=dt$, or equivalently $\frac{dA}{A-14000}=\frac{dt}{20}$. Alternatively you could get the factor of $20$ from the "du" upon integration by substitution. Either way you're going to want to move it to the other side so that you wind up with $Ce^{t/20}=\dots$ $\endgroup$
    – Ian
    Sep 25, 2016 at 23:52
  • $\begingroup$ Ian is right, it makes more sense to put the 20 outside the logarithm like this $20 \ln(A - 14000) = t + C$ $\endgroup$ Sep 25, 2016 at 23:53
  • $\begingroup$ I like questions with some nice background. More fun to do IMO. $\endgroup$ Sep 26, 2016 at 0:06
  • $\begingroup$ Thanks for the help. So we get C = 11000? $\endgroup$
    – toy
    Sep 26, 2016 at 0:32

1 Answer 1


I don't know how you did your calculations, but saw this question at the same time that I had the thought "how do I get the most of my money?", so I did what anyone would do, and dusted off my Calculus book and quickly found that an equation of the form

$\frac{dy}{dx} + p(x)y = q(x)$ (with $x$ as time in years)

is a first order linear non-homogeneous differential equation and that is the same as your equation given that we set

$p(x)=-r=-0.05$. (the rate)


$q(x)=U_{flow}=-700$. (denoting the flow in on the account)

The book had two methods for solving this, and we can use the one with an integrating factor $\mu(x)$ which can be any antiderivative of $p(x)$ so we can set the constant to $0$ giving us

$\mu(x)=\int p(x)dx=-rx$

The solution comes from multiplying both sides by $e^{u(x)}$ such that the left-hand side will be the derivative of $e^{\mu(x)}y$

$\frac{dy}{dx} + p(x)y = q(x)$

$\Leftrightarrow e^{\mu(x)}\left(\frac{dy}{dx} + p(x)y\right)=e^{\mu(x)}q(x)$

$\Leftrightarrow \frac{dy}{dx}\left(e^{\mu(x)}y\right)=e^{\mu(x)}q(x)$

$\Leftrightarrow e^{\mu(x)}y=\int e^{\mu(x)}q(x)dx$

To solve the integral we use a substitution $v=-rx$ and $x=\frac{-v}{r}$

$\int e^{\mu(x)}q(x)dx=\int e^{-rx}U_{flow}dx=\int e^v \frac{-U_{flow}}{r}dv=\frac{-U_{flow}}{r}\int e^vdv$

and this innermost integral is easy to solve. We get that

$\int {e^v}dv=e^v+C = e^{-rx}+C$

where $C$ is a constant that soon can be solved for using our starting money


We now have our function


$\Leftrightarrow y=\frac{-U_{flow}}{e^{-rx}r}(e^{-rx}+C)$

and with $x=0$ we get that


$\Leftrightarrow C = \dfrac{b_{start}\cdot r}{-U_{flow}}-1$

Inserting $C$ into the equation and simplifying, we get our final equation

$y = b_{start}e^{rx}+\frac{U_{flow}}{r}\left(e^{rx}-1\right)$.

You should be able to simply insert whatever values you want here and compute it. If you plot this function, you will see that time is a very big factor in getting rich fast. So, I apologize for answering 3 years later during which you could have made \$1780 and had in your hand \$26780 by now.

(please let me know if I've made any mistake)


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