# Differential Equation Problem bank account

1. Mar 13, 2008

### the7joker7

1. The problem statement, all variables and given/known data

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)

2. Relevant equations

General Differential Equation stuff.

3. 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?

2. Mar 13, 2008

### rock.freak667

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. Mar 13, 2008

### Feldoh

You can separate the variables and then integrate both sides of the equation.

Last edited: Mar 13, 2008
4. Mar 14, 2008

### the7joker7

Here's what I have.
$$\frac{1}{Y - 67000}$$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$$^{.05t}$$e$$^{c}$$

67000 = e$$^{.2}$$e$$^{c}$$

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$$^{-.05*4}$$

And don't really know how to proceed.

5. Mar 14, 2008

### Defennder

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 Common sense would dictate that the money is withdrawn first, but you can show easily that you would need a larger initial sum of money if 3350 is deducted first before you get 5% of the remaining for that year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. Mar 15, 2008

### HallsofIvy

Staff Emeritus
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.

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

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

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).

Last edited: Mar 15, 2008