# FO Differential equations and account balance

by fogvajarash
Tags: account, balance, calculus, differential, equations, interest, variables separable
 P: 126 1. The problem statement, all variables and given/known data a. Assume that yo dollars are deposited into an account paying r percent compounded continuously. If withdrawals are at an annual rate of 200t dollars (assume these are continuous), find the amount in the account after T years. b. Consider the special case if r = 10% and y0=$20000 c. When will the account be depleted if y0=$5000? Give your answer to the nearest month. 2. Relevant equations 3. The attempt at a solution I've realized that the rate at which the account balance varies is the following: dy/dt = ry - 200 (where r is the r percent rate, 0.10; and y the amount of money present) However, when i try to obtain the differential equation, I keep getting that the amount of money present is the following: y(T) = 200/r + (y0-200/r)erT This would, mean that the function would never decrease in the case of $20000 and as well for$5000 (meaning it will never be depleted). However, I'm pretty sure that i'm wrong on this one. Could anyone please help me with this? My procedure: 1/(ry-200) dy = 1 dt (integrate both parts) ln(ry-200) 1/r = t + M1 ln(ry-200) = rt + M2 M3ert=ry-200 y = M4ert + 200/r Then, if y(0) = y0: y0 - 200/r = M4 We then plug this result into our equation: y = 200/r + (y0 - 200/r)erT This corresponds to the equation i've been getting. Is my procedure done right?
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 Quote by fogvajarash 1. The problem statement, all variables and given/known data a. Assume that yo dollars are deposited into an account paying r percent compounded continuously. If withdrawals are at an annual rate of 200t dollars (assume these are continuous), find the amount in the account after T years. b. Consider the special case if r = 10% and y0=$20000 c. When will the account be depleted if y0=$5000? Give your answer to the nearest month. 2. Relevant equations 3. The attempt at a solution I've realized that the rate at which the account balance varies is the following: dy/dt = ry - 200 (where r is the r percent rate, 0.10; and y the amount of money present) However, when i try to obtain the differential equation, I keep getting that the amount of money present is the following: y(T) = 200/r + (y0-200/r)erT This would, mean that the function would never decrease in the case of $20000 and as well for$5000 (meaning it will never be depleted). However, I'm pretty sure that i'm wrong on this one. Could anyone please help me with this? My procedure: 1/(ry-200) dy = 1 dt (integrate both parts) ln(ry-200) 1/r = t + M1 ln(ry-200) = rt + M2 M3ert=ry-200 y = M4ert + 200/r Then, if y(0) = y0: y0 - 200/r = M4 We then plug this result into our equation: y = 200/r + (y0 - 200/r)erT This corresponds to the equation i've been getting. Is my procedure done right?
Your withdrawal rates are incorrect; you said that the annual withdrawal rate is 200t, so at t = 1 it is at rate 100, at t = 2 it is at rate 200, etc. In other words, the withdrawal rate varies with t, so your DE is not correct.

In the corrected problem the value of y0 determines whether or not the account will ever be depleted, and when that will happen.
P: 126
 Quote by Ray Vickson Your withdrawal rates are incorrect; you said that the annual withdrawal rate is 200t, so at t = 1 it is at rate 100, at t = 2 it is at rate 200, etc. In other words, the withdrawal rate varies with t, so your DE is not correct. In the corrected problem the value of y0 determines whether or not the account will ever be depleted, and when that will happen.
So this means i can't solve the problem until i have done first order linear DE?

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## FO Differential equations and account balance

 Quote by fogvajarash So this means i can't solve the problem until i have done first order linear DE?
Have you learned yet about using integrating factors for first order linear ODEs with constant coefficients?

Chet
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u
 Quote by fogvajarash So this means i can't solve the problem until i have done first order linear DE?
Presumably you know how to solve an equation of the form du/dt = ru - c for constant c. You can use a trick to reduce your problem to that form: in your equation dy/dt = ry - 200t you have ry - 200t on the right, and you can write this as r(y - (200/r)t) = ru, where u = y - (200/r)t. Now dy/dt = du/dt + 200/r, so the DE is du/dt + 200/r = ru, or du/dt = ru - 200/r = ru - c, and that is a form you already know how to solve.
P: 126
 Quote by Ray Vickson u Presumably you know how to solve an equation of the form du/dt = ru - c for constant c. You can use a trick to reduce your problem to that form: in your equation dy/dt = ry - 200t you have ry - 200t on the right, and you can write this as r(y - (200/r)t) = ru, where u = y - (200/r)t. Now dy/dt = du/dt + 200/r, so the DE is du/dt + 200/r = ru, or du/dt = ru - 200/r = ru - c, and that is a form you already know how to solve.
I finally solved the problem by using the fact that it's a first order linear differential equation and then multiply it by the integrating factor.

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