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