Trying to apply my DiffEq solution.

In summary, the conversation is discussing a differential equation involving a savings account balance, withdrawal amount, and continuous interest rate. The solution involves solving for the balance over time and finding the stability of the solution. In the third question, the conversation discusses using the solved equation to determine how long a given amount of savings will last if it grows at a certain interest rate and a set amount is withdrawn each year. The solution involves using the starting values to determine the value of C3 and then plugging in the values for the interest rate, withdrawal amount, and initially given balance to solve for the time it will take for the savings to run out.
  • #1
Geofram
5
0

Homework Statement



S is the balance of a savings account
W is the amount withdrawn per year.
k is a rate percentage of continuous interest per year

1. Solve the differential Equation above.
2. Draw a phase portrait and assess the solution's stability.
3. Assume you have $1,000,000 for retirement. How long would this savings last if the balance grows at 2% per year and you have to withdraw $50,000 a year to live?


Homework Equations



dS/dt = kS - W

The Attempt at a Solution



To solve this I've isolated like terms:

dS/(kS - W) = dt

ln(kS - W)/k = t + C1

ln(kS - W) = t + C2

kS - W = ekt + eC2

kS = C2ekt + W

S = (C2ekt + W) / k

S = C3ekt + W/k


I'm just unsure about what to plug into find out what happens to the solution for the phase diagram, and also I'm lost on how to solve #3.

For number 3, I would plug in 0 for S because we want to know when it runs out.
k = .02, t = ?, and W = 50,000

I'm just lost on how I get C3. I can plug in the starting values to get:
t = 0 because we're just starting?
1,000,000 = C3e0.02(0) + 50,000 / .02
-1,500,000 = C3e0
C3 = -1,500,000

But if I plug this back in:

0 = (-1,500,000)e0.02t + 50,000/.02
(-2,500,000)/(-1,500,000) = e0.02t
ln(1.6667) = 0.02t
t = 25.54 years?

That actually makes sense...
 
Last edited:
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  • #2
Geofram said:

Homework Statement



S is the balance of a savings account
W is the amount withdrawn per year.
k is a rate percentage of continuous interest per year

1. Solve the differential Equation above.
2. Draw a phase portrait and assess the solution's stability.
3. Assume you have $1,000,000 for retirement. How long would this savings last if the balance grows at 2% per year and you have to withdraw $50,000 a year to live?


Homework Equations



dS/dt = kS - W

The Attempt at a Solution



To solve this I've isolated like terms:

dS/(kS - W) = dt

ln(kS - W)/k = t + C1

ln(kS - W) = t + C2

kS - W = ekt + eC2
You went wrong here.
[tex]e^{t+ C_2}= (e^t)(e^{C_2}[/tex]
the product of the exponentials, not the sum. Also, there should be no "k" multiplying t.

kS = C2ekt + W

S = (C2ekt + W) / k

S = C3ekt + W/k


I'm just unsure about what to plug into find out what happens to the solution for the phase diagram, and also I'm lost on how to solve #3.

For number 3, I would plug in 0 for S because we want to know when it runs out.
k = .02, t = ?, and W = 50,000

I'm just lost on how I get C3.
 
  • #3
Geofram said:
ln(kS - W)/k = t + C1

ln(kS - W) = t + C2

Actually I just forgot to do it in the step I multiplied k out.
It should be:

Geofram said:
ln(kS - W)/k = t + C1

ln(kS - W) = kt + C2

Take a look at what I added, does that seem right? I'm still confused on how to accomplish #2
 

1. How do I know if my DiffEq solution is correct?

There are a few ways to check the accuracy of your DiffEq solution. One way is to compare it to a known solution for the same equation. Another way is to use numerical methods, such as Euler's method, to approximate the solution and compare it to your solution. Additionally, you can check for any physical or mathematical constraints that your solution should satisfy.

2. What if my DiffEq solution diverges or gives unrealistic results?

If your DiffEq solution diverges or gives unrealistic results, it could be due to a few reasons. It could be caused by incorrect initial conditions or boundary conditions, an incorrect differential equation, or a mistake in your calculations. Double-checking your work and making sure all parameters are correct can help identify the issue.

3. Can I use my DiffEq solution for different values of the parameters?

Yes, in many cases, your DiffEq solution can be used for different values of the parameters in the equation. However, it is important to check if the equation and initial conditions still hold true for the new parameters. In some cases, the solution may need to be adjusted or recalculated for different parameter values.

4. How do I interpret the results of my DiffEq solution?

The interpretation of your DiffEq solution will depend on the context of the problem and the specific equation being solved. Generally, the solution will provide a numerical value or a function that describes the behavior of the system over time. It is important to understand the meaning of the variables and parameters in the equation to correctly interpret the results.

5. Can I use my DiffEq solution to make predictions for future behavior?

In some cases, yes, your DiffEq solution can be used to make predictions for future behavior. However, this depends on the accuracy of the solution and the stability of the system. It is important to keep in mind that small errors in the initial conditions or parameters can greatly affect the predictions of the solution. Additionally, the solution may need to be recalculated for longer time periods if the system is not stable.

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