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Math Help - another diffeq modeling problem..

  1. #1
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    another diffeq modeling problem..

    A young person with no initial capital invests k dollars per year at an annual rate of return r. Assume that
    the investments are made continuously and that the return is compounded continuously.

    determine the sum S(t) accumalted at any time t.

    Can someone please help me set this one up?
    Im very confused on how to really start the process of modeling the problem.
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  2. #2
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    The modelling aspect is to consider the time rate of change of investiment. There are two components. The investment amount k and the interest r*S. This leads to the d.e.

    \frac{dS}{dt} = rS + k

    Which can be solved for S(0) = 0 as the boundary condition as follows...

    \frac{dS}{S + k/r} = r\,dt

    \ln(S + k/r) = rt + C

    Now at t = 0, S = 0, therefore C = \ln(k/r), so

    \ln(rS/k + 1) = rt

    \frac{r}{k} S + 1 = \exp(rt)

    S = \frac{k(\exp(rt) - 1)}{r}.

    Note this is assuming r \ne 0. If r = 0 then the d.e. simplifies and the solution is simply S = kt, as you would expect with no interest.

    Hope this helps.
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  3. #3
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    Quote Originally Posted by Rincewind View Post
    The modelling aspect is to consider the time rate of change of investiment. There are two components. The investment amount k and the interest r*S. This leads to the d.e.

    \frac{dS}{dt} = rS + k

    Which can be solved for S(0) = 0 as the boundary condition as follows...

    \frac{dS}{S + k/r} = r\,dt
    <------------this step.

    \ln(S + k/r) = rt + C

    Now at t = 0, S = 0, therefore C = \ln(k/r), so

    \ln(rS/k + 1) = rt

    \frac{r}{k} S + 1 = \exp(rt)

    S = \frac{k(\exp(rt) - 1)}{r}.

    Note this is assuming r \ne 0. If r = 0 then the d.e. simplifies and the solution is simply S = kt, as you would expect with no interest.

    Hope this helps.

    can u explain the bolded step, it seems you divided by r and solved. Not sure why, is that to get it into a form we can use?
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  4. #4
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    Quote Originally Posted by p00ndawg View Post
    can u explain the bolded step, it seems you divided by r and solved. Not sure why, is that to get it into a form we can use?
    I did that so that when I integrate the LHS there wont be a term hanging around. If you want to you can not do that and it will sort itself out as follows...

    \frac{dS}{rS + k} = dt

    integrating

    \frac{1}{r}\ln(rS + k) = t + C

    now S(0) = 0 therefore C = (1/r)ln(k) so substituting

    \frac{1}{r}\ln(rS + k) = t + \frac{1}{r}\ln(k)

    \ln(rS + k) = rt + \ln(k)

    \ln(rS + k) - \ln(k) = rt

    \ln(rS/k + 1) = rt

    as before,

    Does this explain it?
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  5. #5
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    Quote Originally Posted by Rincewind View Post
    I did that so that when I integrate the LHS there wont be a term hanging around. If you want to you can not do that and it will sort itself out as follows...

    \frac{dS}{rS + k} = dt

    integrating

    \frac{1}{r}\ln(rS + k) = t + C

    now S(0) = 0 therefore C = (1/r)ln(k) so substituting

    \frac{1}{r}\ln(rS + k) = t + \frac{1}{r}\ln(k)

    \ln(rS + k) = rt + \ln(k)

    \ln(rS + k) - \ln(k) = rt

    \ln(rS/k + 1) = rt

    as before,

    Does this explain it?
    perfect thank you.
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