1. Not finding help here? Sign up for a free 30min tutor trial with Chegg Tutors
    Dismiss Notice
Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

Differential Equation Problem bank account

  1. Mar 13, 2008 #1
    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. jcsd
  3. Mar 13, 2008 #2


    User Avatar
    Homework Helper

    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.
  4. Mar 13, 2008 #3
    You can separate the variables and then integrate both sides of the equation.
    Last edited: Mar 13, 2008
  5. Mar 14, 2008 #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.
  6. Mar 14, 2008 #5


    User Avatar
    Homework Helper

    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.
  7. Mar 15, 2008 #6


    User Avatar
    Staff Emeritus
    Science Advisor

    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
Know someone interested in this topic? Share this thread via Reddit, Google+, Twitter, or Facebook

Have something to add?

Similar Discussions: Differential Equation Problem bank account