# Continuous Interest with Continuous Withdrawal

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

• 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 $C$. Sep 25, 2016 at 23:51
• 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 $Ce^{t/20}=\dots$
– Ian
Sep 25, 2016 at 23:52
• Ian is right, it makes more sense to put the 20 outside the logarithm like this $20 \ln(A - 14000) = t + C$ Sep 25, 2016 at 23:53
• I like questions with some nice background. More fun to do IMO. Sep 26, 2016 at 0:06
• Thanks for the help. So we get C = 11000?
– toy
Sep 26, 2016 at 0:32

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)

and

$$q(x)=U_{flow}=-700$$. (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

$$b_{start}=y(0)=25000$$.

We now have our function

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

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

and with $$x=0$$ we get that

$$b_{start}=\frac{-U_{flow}}{r}(1+C)$$

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

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.