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Thread: Differential Equation and Investment Growth

  1. #1
    Senior Member Vinod's Avatar
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    Differential Equation and Investment Growth

    Hello forumites,
    A large corporation starts at time t=0 to continuously part of its receipts at a rate of P dollars per annum in a fund for future corporate expansion. Assume that the fund earns r percent interest per year compounded continuously. Thus , the rate of growth of the amount A in the fund is given by


    $\frac{dA}{dt}=rA +P$ where A=0 when t=0 . I want to solve this differential equation for A as a function of t.


    Using the above result, I want to find A for the following


    a)P= 100000 dollars, r=6%, and t= 5 years b) P=250000 dollars, r=5%, and t=10 years.

    I also want to find t if the corporation needs 800000 dollars and it can invest 75000 dollars per annum in a fund earning 8% interest compounded continuously.

    Answer: I am working on these problems but still no results. It would be better if any forumite guide me.
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    MHF Contributor MarkFL's Avatar
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    Re: Differential Equation and Investment Growth

    I would put the ODE into standard linear form:

    $\displaystyle \frac{dA}{dt}+(-r)A=P$

    Next, we need to compute the integrating factor:

    $\displaystyle \mu(t)=\exp\left(-r\int\,dt\right)=\,?$
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    Senior Member Vinod's Avatar
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    Re: Differential Equation and Investment Growth

    Quote Originally Posted by MarkFL View Post
    I would put the ODE into standard linear form:

    $\displaystyle \frac{dA}{dt}+(-r)A=P$

    Next, we need to compute the integrating factor:

    $\displaystyle \mu(t)=\exp\left(-r\int\,dt\right)=\,?$
    Hello,
    So,$A=\frac{-P}{r}+C*e^{rt}$
    So C=1666666.66666 and A=58,30,98 when P=100000 r=0.06 t=5 years.
    and C=5000000 A=3243606.35 when P=250000 r=0.05 and t=10 years
    Last edited by Vinod; Jun 24th 2018 at 11:38 PM.
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  4. #4
    MHF Contributor MarkFL's Avatar
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    Re: Differential Equation and Investment Growth

    Continuing where I left off...

    $\displaystyle \mu(t)=\exp\left(-r\int\,dt\right)=e^{-rt}$

    And thus the ODE becomes:

    $\displaystyle \frac{d}{dt}\left(e^{-rt}A\right)=Pe^{-rt}$

    Integrate:

    $\displaystyle e^{-rt}A=-\frac{P}{r}e^{-rt}+c_1$

    $\displaystyle A(t)=-\frac{P}{r}+c_1e^{rt}$

    Now, we are given:

    $\displaystyle A(0)=-\frac{P}{r}+c_1=0\implies c_1=\frac{P}{r}$

    Hence, the solution to the IVP is:

    $\displaystyle A(t)=-\frac{P}{r}+\frac{P}{r}e^{rt}=\frac{P}{r}\left(e^{ rt}-1\right)$

    Can you proceed?
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    Senior Member Vinod's Avatar
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    Re: Differential Equation and Investment Growth

    Quote Originally Posted by MarkFL View Post
    Continuing where I left off...

    $\displaystyle \mu(t)=\exp\left(-r\int\,dt\right)=e^{-rt}$

    And thus the ODE becomes:

    $\displaystyle \frac{d}{dt}\left(e^{-rt}A\right)=Pe^{-rt}$

    Integrate:

    $\displaystyle e^{-rt}A=-\frac{P}{r}e^{-rt}+c_1$

    $\displaystyle A(t)=-\frac{P}{r}+c_1e^{rt}$

    Now, we are given:

    $\displaystyle A(0)=-\frac{P}{r}+c_1=0\implies c_1=\frac{P}{r}$

    Hence, the solution to the IVP is:

    $\displaystyle A(t)=-\frac{P}{r}+\frac{P}{r}e^{rt}=\frac{P}{r}\left(e^{ rt}-1\right)$

    Can you proceed?
    Hello,
    How to find t if the corporation needs 800000 dollars and it can invest 75000 dollars per annum in a fund earning 8% interest compounded continuously.
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    Senior Member Vinod's Avatar
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    Re: Differential Equation and Investment Growth

    Quote Originally Posted by Vinod View Post
    Hello,
    How to find t if the corporation needs 800000 dollars and it can invest 75000 dollars per annum in a fund earning 8% interest compounded continuously.
    I got the answer 8.0168 years in hp48 series calculator.
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  7. #7
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    Re: Differential Equation and Investment Growth

    Vinod, try something easier:
    if interest is 8% compounded annually, what will be the value after 8 years?
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  8. #8
    Senior Member Vinod's Avatar
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    Re: Differential Equation and Investment Growth

    Quote Originally Posted by DenisB View Post
    Vinod, try something easier:
    if interest is 8% compounded annually, what will be the value after 8 years?
    Is it $797747.07?
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  9. #9
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    Re: Differential Equation and Investment Growth

    Right!
    OK: what if the rate compounds daily (use 1 year = 360 days)?
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  10. #10
    Senior Member Vinod's Avatar
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    Re: Differential Equation and Investment Growth

    Quote Originally Posted by DenisB View Post
    Right!
    OK: what if the rate compounds daily (use 1 year = 360 days)?
    Hello,
    It would be $807253.35. If the rate compounded continuously, the amount after 8 years would be $807280.79.
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  11. #11
    Newbie dkugg's Avatar
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    Post Re: Differential Equation and Investment Growth

    We can model the growth of an initial deposit with respect to the interest rate with differential equations. If represents time, then the rate of change of the initial.
    . Examples of ordinary differential equations:

    dydx = u(y)

    or

    dydx = u(x, y)
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  12. #12
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    Re: Differential Equation and Investment Growth

    Quote Originally Posted by Vinod View Post
    Hello,
    It would be 807253.35.
    If the rate compounded continuously, the amount after 8 years would be 807280.79.
    807280.79 - 807253.35 = 27.44
    That's 27.44 / 8 = 3.43 average per year.

    My point was to show you "continuously" is NOT that big a deal!!
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