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Math Help - Differentiation Crisis

  1. #1
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    Differentiation Crisis

    Hi I'm stuck with a problem, I'm fine with differentials but the way this is worded confuses me!

    Valuation of a storage facility gives equation:

    V'' - rV = cP where c and r are constants

    V: R->R is a function of the variable P and it wants me to find the general solution V (P).

    I'm going round in circles here, if V is a function of P then am I right in saying V'' = d^2V/dP^2 ??
    but then where do I go, do they want V = P something?

    please help!
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  2. #2
    Grand Panjandrum
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    Quote Originally Posted by mathgrad View Post
    Hi I'm stuck with a problem, I'm fine with differentials but the way this is worded confuses me!

    Valuation of a storage facility gives equation:

    V'' - rV = cP where c and r are constants

    V: R->R is a function of the variable P and it wants me to find the general solution V (P).

    I'm going round in circles here, if V is a function of P then am I right in saying V'' = d^2V/dP^2 ??
    but then where do I go, do they want V = P something?

    please help!
    This is a standard second order linear ordinary differential equation.

    We first look for the general solution of:

    <br />
V'' - rV = 0<br />

    For this take a trial solution V(P)=e^{\lambda P}, then as:

    <br />
V'' - rV =0<br />

    we have:

    <br />
\lambda^2 V(P)-rV(P)=0<br />

    so if V(P) \not \equiv 0

    <br />
\lambda^2-r=0<br />

    and so \lambda= \pm \sqrt{r}

    So the general solution to V'' - rV =0 is:

    <br />
V(P)=A e^{\sqrt{r}P}+B e^{-\sqrt{r}P}\ \ \ \ \dots(1)<br />

    where A and B are arbitary constants.

    Now to get the general solution to V'' - rV = cP we take any solution of this equation and add on the right hand side of (1) above. An obvious solution is V(P)=cP, so the general solution is:

    <br />
V(P)=cP + A e^{\sqrt{r}P}+B e^{-\sqrt{r}P}\ \ \ \ \dots(2)<br />

    RonL
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