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Homework Help: 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


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    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


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    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


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    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 by a moderator: Mar 15, 2008
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