FO Differential equations and account balance

[fogvajarash]

Homework Statement

a. Assume that y_o 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.

Homework Equations

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)e^rT

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

M₃e^rt=ry-200

y = M₄e^rt + 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)e^rT

This corresponds to the equation I've been getting. Is my procedure done right?

 

[Ray Vickson]

Homework Statement

a. Assume that y_o 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.

Homework Equations

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)e^rT

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

M₃e^rt=ry-200

y = M₄e^rt + 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)e^rT

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.

 

[fogvajarash]

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?

 

[Chestermiller]

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

 

[Ray Vickson]

u

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.

 

[fogvajarash]

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.