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Math Help - Investment word problem w/differential equations

  1. #1
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    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 If a continuous cash flow of P dollars per year is withdrawn from the fund, then the rate of decrease of A is given by the differential equation \frac{dA}{dt}=rA-P

    where A=A_0 when t=0

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

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

    (c) Find 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 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 A' - rA = -P and my P(x)=-r so the IF is e^{\int-rdt}=e^{-rt}

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

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

    From there...

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

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

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

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

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

    but then part B asks me to input t as one of the variables and t doesn't show up in this new soln so I'm not sure where to go from here.
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  2. #2
    Junior Member
    Joined
    Nov 2007
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    Can anyone help me with this please? I only need help with the A_0 part, not the entire part B & C.
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