give DE S(t) = k/r + (So -k/r) exp(rt) where r is the interest rate, So is the i

Question

give DE S(t) = k/r + (So -k/r) exp(rt) where r is the interest rate, So is the initial investment value, k is the amount being drawn from the account.
In part b the question asks Assuming that So and r are fixed, determine the withdrawal rate Ko at which S(t) will remain constant. Which will be (So r) (ie withdraw all the interest earned in a time period and the balance will remain constant. What I can't seem to understand is (c) i k exceeds the value ko found in part (b), the S(t) will decrease and ultimately become zero. Find the time T at which S(t) = 0. The answer is 1/r ln(k/(k-ko)). I just don't see how they get there.

Explanation / Answer

S(t) = k/r + (So -k/r) exp(rt) dS(t)/dt = (So - k/r) r exp(rt) = 0 so k = So*r = ko then dS(t)/dt = (ko - k) exp(rt) if k > ko, dS(t) < 0 find T S(T) = 0 k/r + (ko/r - k/r) exp(rT) = 0 k + (ko - k) exp(rt) = 0 exp(rT) = k/(k - ko) rT = ln[k/(k - ko)] so T = 1/r * ln[k/(k - ko)]

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