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 If a continuous cash flow of P dollars per year is withdrawn from the fund, then the rate of decrease of A is given by the differential equation $\frac{dA}{dt}=rA-P$

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.

2. 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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sample problem in investment sing differential

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