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Math Help - Economics Problem

  1. #1
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    Economics Problem

    So I've never taken differential equations, but this problem popped up in an Econ class so I'm pretty lost.

    Let R(w) = {(-w)[u''(w)]}/u'(w)

    Suppose R(w) = r > 0 for all w.

    Derive an expression for u(w).

    So i began by multiplying both sides by u'(w) and then bringing everything over to the left. I'm not sure where to introduce d/dw. I guess I'm pretty lost. There was a hint: let z(w) = u'(w) and solve the differential equation for z.

    This then becomes:
    R(w) = {(-w)[z'(w)]}/z(w)
    [z(w)]R(w) = -w[z'(w)]
    [z(w)]R(w) + w[z'(w)] = 0

    Then I'm stuck.
    Thanks in advance for the help.
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  2. #2
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    So you have

    -w\dfrac{u''(w)}{u'(w)}=r, where r>0 is a constant. Correct?
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  3. #3
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    correct!!
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  4. #4
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    Right then. I wouldn't use the R(w) notation, because that's essentially an extra variable. Follow the hint (which reduces the order of the DE by one), and you get

    -w\dfrac{z'(w)}{z(w)}=r. You can divide both sides by -w to obtain

    \dfrac{z'(w)}{z(w)}=-\dfrac{r}{w}.

    You can now integrate both sides directly. You obtain

    \ln|z(w)|=-r\ln|w|. Can you continue from here?
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  5. #5
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    do i also have to multiply each side by d/dw ?
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  6. #6
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    No. Right now, I would solve for z(w), and then plug back in what that was. What do you get?
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  7. #7
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    z(w) = e ^ -r ln(w)

    u(w) = the above integrated

    i wish i knew how to type this math notation...
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  8. #8
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    You can simplify the above expression a good deal:

    z(w)=e^{-r\ln|w|}=e^{\ln(|w|^{-r})}=(e^{\ln|w|})^{-r}=w^{-r}.

    Incidentally, I forgot a constant of integration. We would get that back at this step:

    \ln|z(w)|=-r\ln|w|+C, which would translate to the multiplication of an arbitrary constant (equal to e^{C}), which gives us

    z(w)=Aw^{-r}. Now continue?

    Incidentally, you can learn something about LaTeX here.
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