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Personal Wealth Model using DE's

  1. Apr 28, 2008 #1
    [SOLVED] Personal Wealth Model using DE's

    Hey, I am having trouble with this question.

    Most people have an income that comes from 2 sources: salary and personal investments. From this income, 'necessary' expenses (housing, food) are paid, some money is spent on luxuries and the rest is saved (increasing investments).

    Given that income must equal outflow, show the steps in developing the following mathematics model for a person's wealth at any time t:

    dW/dt=(1-p)(s-n+rW)
    Where s= your salary
    W(t)= your wealth (savings), which is a function of time
    r= rate of interest on your wealth (savings)
    n= amount spent on necessities
    p=proportion of your income after necessities that you spend on luxuries.

    Ok what I have to try to figure out is how to differentiate this model (if possible) to get it to the form of:

    W(t)=((s-n)/r)(e^((1-p)rt)-1)


    I'm not sure whether this is sufficient information as I left what appeared to be a lot of irrelevant information out.

    Thanks very much for your help.
     
  2. jcsd
  3. Apr 28, 2008 #2

    Hootenanny

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    Welcome to PF Shadow,

    Firstly, I'm not quite sure what the question is asking. Are you given this,
    Or do you have to derive it?
     
  4. Apr 29, 2008 #3
    We are given that function and are then asked to differentiate it (if possible), and if we can it should equal the second model for W(t), which was also given to us.

    In other words we have to show the steps to obtaining the W(t) function from the original function.

    Sorry if this isn't making much sense, if you want me to explain futher I will.
     
  5. Apr 29, 2008 #4

    Hootenanny

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    No, it makes perfect sense. However, there's one thing that we have to fix up before we can start. The equation which you quoted,
    Is a differential equation for w(t) and therefore we need to integrate it to find an expression for w(t) not differentiate it.

    So what type of differential equation is it?
     
  6. Apr 29, 2008 #5
    Seems like there is not much to derive if all of that is part of the givens. I'm guessing s, r, n, p are assumed to be constant functions so basically you have a DE thats first order and linear in W, so use separation of variables and you'll find W.
     
  7. Apr 29, 2008 #6

    Yes I realised that I need to integrate it last night before I went to sleep.

    I am a bit confused by your question... Do you mean what method should I use to solve it? Like partial fractions or DE's with separable variables? If so I think I should be using partial fractions.
     
  8. Apr 29, 2008 #7
    ok I think I understand that. The thing thats confusing me is when I have to integrate it, there is no 't' to integrate with respect to. Does this matter at all?
     
  9. Apr 29, 2008 #8

    Hootenanny

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    Sounds good to me :approve:. However, there's no need to use partial fractions,

    [tex]\frac{dW}{dt} = (1-p)(s-n+rW) \Rightarrow \frac{1}{s-nr+W} \frac{dW}{dt} =1-p[/tex]

    For [itex]s\neq nr+W[/itex]. Notice that everything is constant except for W. Can you take it from here?
     
    Last edited: Apr 29, 2008
  10. Apr 29, 2008 #9
    umm I have to go to school now, I will give it a shot there and let you know.

    Thanks

    P.S. should it be s-n+rW?
     
  11. Apr 29, 2008 #10

    Hootenanny

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    Of course it should, sorry it was a typo, which I corrected appropriately above.
     
  12. Apr 30, 2008 #11
    I tried to intgrate it by taking dt over to the other side and integrating both sides. But I am not sure how to go about integrating it. For example on the right side does it become t+pt + c, and I don't have a clue how to integrate the left side.
     
  13. May 1, 2008 #12

    Hootenanny

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    You're almost correct, just watch your signs. Now for the LHS we have,

    [tex]\int\frac{dW}{s-nr+W}[/tex]

    Notice that the denominator is a linear function of W, how do you usually integrate one over a linear function?

    HINT:

    [tex]\frac{d}{dW}\left(s-nr+W\right) = 1[/tex]
     
  14. May 2, 2008 #13
    Ummm do you take out the coefficient of what you're integrating with respect to, in this case it would be 1.

    then it would be 1/1 ln(s-nr+W)?

    Also, why does the r move off the W and onto the n? I don't quite get that sorry.
     
  15. May 2, 2008 #14

    Hootenanny

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    Sorry, I've been making a lot of typos recently. The r should remain the coefficient of the W.
     
  16. May 2, 2008 #15
    no that's quite allright. I appreciate your help.

    So was my method of integrating correct?
     
    Last edited: May 2, 2008
  17. May 2, 2008 #16

    Hootenanny

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    Yes it was, but note that your integral will change since W now has a coefficient r.
     
  18. May 2, 2008 #17
    ok so it would be

    1/r ln (s-n+W)=t-pt+c
    ln (s-n+W)=tr-ptr+c
    s-n+W=e^((1-p)rt)*e^c
    W=(A*e^((1-p)rt))-s+n, where A=e^c

    ?
     
  19. May 2, 2008 #18

    Hootenanny

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    You seem to be missing the coefficient of W in your logarithm :wink:
     
  20. May 2, 2008 #19
    I'm not sure I understand what you mean. I don't really understand what I'm doing above, I am just going through rules we learnt in class.

    Are you trying to say that when i integrate it should be 1/r ln (s-n+rW)?
     
  21. May 2, 2008 #20

    Defennder

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    Yes, that should be correct. I got the same answer.
     
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