Differential Equation Problem bank account

In summary: C = 10.912 (Can't possibly be right.)In summary, you need $67,000 to be able to afford 4 years of college with a 5% interest rate.
  • #1
the7joker7
113
0

Homework Statement



You have a bank account with Y amount of dollars. The interest on it is 5% a year. Every year, you extract 3,350 dollars to pay for college tuition. What is the smallest amount Y can be to allow you to pay for 4 years of college? (Use Calculus Methods)

Homework Equations



General Differential Equation stuff.

The Attempt at a Solution



Now, I believe an appropriate Differential Equation for this problem would be...

dy/dt = 0.05y - 3,350

t will be 4 and y will be 0, ultimately.

In class today, we got to the point where you divide both numbers by 0.05 so you get...

dy/dt = y - 67000

Or something. Now we don't know where to go. Help?
 
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  • #2
you can just divide by y-67000 and get a variables are separable form

or put the y on the other side and you will get the form for the integrating factor.
 
  • #3
You can separate the variables and then integrate both sides of the equation.
 
Last edited:
  • #4
Here's what I have.
[tex]\frac{1}{Y - 67000}[/tex]dy = .05dt

Take integrals...

ln(Y - 67000) = .05t + C

From here, I tried two different things.

1: ln(-67000) = .2 + C

11.11245 = .2 + C

C = 10.912 (Can't possibly be right.)

2: 67000 = e[tex]^{.05t}[/tex]e[tex]^{c}[/tex]

67000 = e[tex]^{.2}[/tex]e[tex]^{c}[/tex]

67000/1.2214 = e^c

C = ln(67000/1.2214) = Some also very low number that can't possibly be right.

Then I started to try something like...

y(4) = Ae[tex]^{-.05*4}[/tex]

And don't really know how to proceed.
 
  • #5
the7joker7 said:
C = 10.912 (Can't possibly be right.)

(snip)

C = ln(67000/1.2214) = Some also very low number that can't possibly be right.

y(4) = Ae[tex]^{-.05*4}[/tex]
What's the difference between these 2? What value of C are you looking for? Seems to me you got it correct. Where did you get that last line y(4) from? Remember that once you take ln|y - 67000|, you are implying that either y-67000 > 0 or 67000-y>0. Solve them both as two different cases and then find y(0) for them both. Take the smaller y(0) value as the answer, as required by the question.

One last consideration: The question as you have posed it does not specify whether $3350 is deducted from the bank account before or after 5% interest is applied. Common sense would dictate that the money is withdrawn first, but you can show easily that you would need a larger initial sum of money if 3350 is deducted first before you get 5% of the remaining for that year than if you get 5% of your initial sum before 3350 is deducted. This means you only have to check the case for which the college fee is deducted before 5% interest is credited. Check that your answer for y(0) (both answers) satisfies this.
 
  • #6
the7joker7 said:
Here's what I have.
[tex]\frac{1}{Y - 67000}[/tex]dy = .05dt

Take integrals...

ln(Y - 67000) = .05t + C
No, ln|Y- 67000|= 0.05t+ C. In many problems, you know you are working with positive numbers so the absolute value isn't necessary. Here, it is essential.

From here, I tried two different things.

1: ln(-67000) = .2 + C

11.11245 = .2 + C
Actually, ln(-67000) doesn't even exist- however, by forgetting the "-" you have got exactly the right thing!

C = 10.912 (Can't possibly be right.)
Why not? Do you understand what C represents in terms of the orginal problem? And do you understand what it is you are trying to find? If ln|Y- 67000|= 0.05t+ 10.912, then when t= 0, ln|Y- 67000|= ln(67000- Y)= 10.912. How much money must you have started with? Solve that equation for Y

2: 67000 = e[tex]^{.05t}[/tex]e[tex]^{c}[/tex]

67000 = e[tex]^{.2}[/tex]e[tex]^{c}[/tex]

67000/1.2214 = e^c

C = ln(67000/1.2214) = Some also very low number that can't possibly be right.
Again, why are you saying that some "very low number" for C can't be right? C is NOT the original deposit you are looking for.
You should have ln|Y- 67000|= ln(67000- Y) (since Y is always less than 67000) = 0.05t+ C. Taking exponentials of both sides 67000- Y= Ce0.05t, which is the formula you have below. When t= 4, Y= 0 so 67000= Ce0.2 so C= 67000e-0.2 and 67000- Y= 67000e0.05t- 0.2[/itex]. When t= 0, 67000-Y= 67000e-.2 so Y= 67000(1- e-.2).

Then I started to try something like...

y(4) = Ae[tex]^{-.05*4}[/tex]

And don't really know how to proceed.
 
Last edited by a moderator:

What is a differential equation?

A differential equation is a mathematical equation that relates a function to its derivatives. It is used to model many real-world phenomena, such as growth, decay, and change over time.

How is a differential equation related to a bank account?

In the context of a bank account, a differential equation can be used to model the balance of the account over time. This can be helpful in predicting future balances and making financial decisions.

What are the variables in a differential equation for a bank account?

The variables in a differential equation for a bank account are typically the time, t, and the balance, B. The equation may also include other variables such as interest rate, deposits, and withdrawals.

Why are initial conditions important in a differential equation for a bank account?

Initial conditions, such as the starting balance and interest rate, are necessary in a differential equation for a bank account because they provide the starting point for the model. Without these initial conditions, the equation cannot accurately predict the future balance of the account.

How can differential equations be used to optimize a bank account?

By using differential equations, it is possible to determine the optimal interest rate or deposit/withdrawal amounts that will result in the maximum balance over time. These equations can also be used to find the best strategies for paying off debt or making investments.

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