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Thread: Resolution of a system of differentgial equations

  1. #1
    Aug 2012

    Resolution of a system of differentgial equations

    Letís consider the function F(K,L) and the following system of partial differential equations:

    ∂lnF/∂K = (1-α)/K (1)
    ∂lnF/∂L = α/L (2)
    I have a solution for the above system of differential equations but am missing some of the details of its development.

    STEP 1
    First, the solution says that based on the fact that ∫1/x= ln(x) +c (where c is a constant of integration), we can write:

    ln F(K,L) = (1 - α) lnK + g(L) + c (1í)
    ln F(K,L) = α lnL + h(K) + cí (2í)
    where g(L) and h(K) are constants of integration that may depend on L and K respectively.
    c and cí are also constants of integration that do not depend on any of the two indicated variables L and K.

    STEP 2
    Next , (1í) and (2í) are combined to get to the following expression:

    ln F(K,L) = (1 - α) lnK + α lnL + C (3)

    Ņcan anyone give me more theoretical background and details to understand STEP 1 and STEP 2? That's to say: how to use integrals to solve the system of differential equations and how to combine (1') and (2') to get to (3)

    The context of the equations is the field of economics (F is a production function, K is capital, L is labor and α is the fraction paid to labor. In fact, I am mostly interested in getting support to understand the mathematical side of the posted issue .

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  2. #2
    MHF Contributor

    Apr 2005

    Re: Resolution of a system of differentgial equations

    The only "theoretical" background involved in step 1 is integration. The equation you give is equivalent to $\displaystyle \partial(ln F)= \frac{1- a}{K}\partial K$
    Those partial derivatives only indicate that there are other variables we have to consider- in this case L. Integrating both sides,
    $\displaystyle ln(F)= (1-a)ln|K|+ f(L)$
    where f(L) is some unknown function of L. Of course, whatever function L it's derivative with respect to K would be 0.

    Now, differentiate both sides of that with respect to L. Since K is independent of L, we have $\displaystyle \partial ln(F)/\partial L= df/dL$. Notice that I have used the ordinary derivative on the right. Because f is a function of L only I can do that. Comparing that to the second equation, $\displaystyle \partial ln F/\partial L= a/L$ and we have the equation df/dL= a/L so that df= a dL/L and integrating with respect to L, f(L)= aln|L|+ C where "C" now really is a constant. Putting that into $\displaystyle ln(F)= (1-a)ln|K|+ f(L)$ from before,
    $\displaystyle ln(F)= (1-a)ln|K|+ aln|L|+ C$
    which we could also write as $\displaystyle F= e^{(1-a)ln|K|+ a ln|L|+ C}= e^{ln|K^{1-a}|}e^{ln|L|^a}e^C$
    $\displaystyle = C'|K^{1-a}L^a|$ where $\displaystyle C'= e^C$.
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