# Thread: Investment word problem w/differential equations

1. ## Investment word problem w/differential equations

Ok, here's the problem:

Investment. Let A(t) be the amount in a fund earning interest at the annual rate of r, compounded continuously. If a continuous cash flow of P dollars per year is withdrawn from the fund, then the rate of decrease of A is given by the differential equation $\displaystyle \frac{dA}{dt}=rA-P$

where $\displaystyle A=A_0$ when $\displaystyle t=0$

(a) Solve this equation for A as a function of t.

(b) Use the result of part (a) to find A when $\displaystyle A_0$=$2,000,000, r=7%, P=$250,000, and t=5 years

(c) Find $\displaystyle A_0$ if a retired person wants a continuous cash flow of $40,000 per year for 20 years. Assume that the person's investment will earn 8%, compounded continuously. Ok, so C is simple enough once A and B is done since you basically just plug in$\displaystyle t=20, r=0.08, A=$40,000$ for the variables. Part A & B is what's getting me. Here's what I did.

I rearranged it into a first-order linear DE so $\displaystyle A' - rA = -P$ and my $\displaystyle P(x)=-r$ so the IF is $\displaystyle e^{\int-rdt}=e^{-rt}$

Then I multiplied the original equation on both sides by the IF to get

$\displaystyle e^{-rt}(A'-rA)=-Pe^{-rt}$

From there...

$\displaystyle \int \frac{d}{dt}(e^{-rt}*A) = \int (-Pe^{-rt})dt$

$\displaystyle e^{-rt}A= \frac{P}{r}e^{-rt} + C$

$\displaystyle A= \frac{P}{r} + Ce^{rt}$

What's tripping me up is $\displaystyle A=A_0$ only when $\displaystyle t=0$ so I'm not sure how to proceed.

I can set $\displaystyle t=0$ and plug in $\displaystyle A_0$ for $\displaystyle A$ with the above general soln to get $\displaystyle A_0= \frac{P}{r} + C$

but then part B asks me to input t as one of the variables and $\displaystyle t$ doesn't show up in this new soln so I'm not sure where to go from here.

2. Can anyone help me with this please? I only need help with the $\displaystyle A_0$ part, not the entire part B & C.

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