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Math Help - Euler equation

  1. #1
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    Euler equation

    Hi. I have this problem, which says: The equation x^2y''+pxy'+qy=0 (p and q constants) is called Euler equation. Demonstrate that the change of variable u=\ln (x) transforms the equation to one at constant coefficients.

    I haven't done much. I just normalized the equation: y''+\displaystyle\frac{p}{x}y'+\displaystyle\frac{  q}{x^2}y=0

    Then P(x)=\displaystyle\frac{p}{x} and Q(x)=\displaystyle\frac{q}{x^2}

    What should I do now? I thought instead of doing x=e^u, then y''+\displaystyle\frac{p}{e^u}y'+ \displaystyle\frac{q}{e^{2u}} y=0 may be this is the right way, cause it seems more like following the problem suggestion.
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  2. #2
    Forum Admin topsquark's Avatar
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    Quote Originally Posted by Ulysses View Post
    Hi. I have this problem, which says: The equation x^2y''+pxy'+qy=0 (p and q constants) is called Euler equation. Demonstrate that the change of variable u=\ln (x) transforms the equation to one at constant coefficients.

    I haven't done much. I just normalized the equation: y''+\displaystyle\frac{p}{x}y'+\displaystyle\frac{  q}{x^2}y=0

    Then P(x)=\displaystyle\frac{p}{x} and Q(x)=\displaystyle\frac{q}{x^2}

    What should I do now? I thought instead of doing x=e^u, then y''+\displaystyle\frac{p}{e^u}y'+ \displaystyle\frac{q}{e^{2u}} y=0 may be this is the right way, cause it seems more like following the problem suggestion.
    x = e^u

    Thus
    dx = e^u~du

    \frac{d}{dx} = e^{-u}\frac{d}{du}

    So
    y' = \frac{d}{dx}y = e^{-u} \frac{d}{du}y

    Now to find y'' we apply the differential operator twice:
    y'' = \frac{d}{dx} \left ( \frac{d}{dx} \right ) y = \left ( e^{-u} \frac{d}{du} \left ( e^{-u} \frac{d}{du} \right ) \right ) y

    Can you take it from here? (Remember to use the product rule!)

    -Dan

    Edit: This is just about the last step. Once we get this we can plug it into the equation, which then simplifies enormously.
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  3. #3
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    I'll try, thank you!
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  4. #4
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    I'm a bit confused with so many operators

    y'' = \frac{d}{dx} \left ( \frac{d}{dx} \right ) y = \left ( e^{-u} \frac{d}{du} \left ( e^{-u} \frac{d}{du} \right ) \right ) y=\left ( e^{-u} \left ( -e^{-u}\frac{d}{du}+e^{-u}\frac{d^2}{d^2u} \right ) \right ) y
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  5. #5
    Forum Admin topsquark's Avatar
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    Quote Originally Posted by Ulysses View Post
    I'm a bit confused with so many operators

    y'' = \frac{d}{dx} \left ( \frac{d}{dx} \right ) y = \left ( e^{-u} \frac{d}{du} \left ( e^{-u} \frac{d}{du} \right ) \right ) y=\left ( e^{-u} \left ( -e^{-u}\frac{d}{du}+e^{-u}\frac{d}{du} \right ) \right ) y
    Close:
    y'' = \frac{d}{dx} \left ( \frac{d}{dx} \right ) y = \left ( e^{-u} \frac{d}{du} \left ( e^{-u} \frac{d}{du} \right ) \right ) y=\left ( e^{-u} \left ( -e^{-u}\frac{d}{du}+e^{-u}\frac{d^2}{du^2} \right ) \right ) y

    So
    y'' = -e^{-2u}\frac{dy}{du} + e^{-2u}\frac{d^2y}{du^2}

    and
    y' = e^{-u}\frac{dy}{du}

    Just to be clear the expressions on the LHS, y' and y'', are in terms of x ie y'(x) and y''(x) whereas the RHS is describing the derivatives in terms of u.

    Now change x into e^{u} in the differential equation and plug in your new expessions for y' and y''. What does the new equation look like?

    -Dan
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  6. #6
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    \frac{d^2y}{du^2}+\frac{dy}{du}(p-1)+qy=0 nice, so it has constant coefficients in u, right?

    Thank you very much sir
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  7. #7
    Forum Admin topsquark's Avatar
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    Quote Originally Posted by Ulysses View Post
    \frac{d^2y}{du^2}+\frac{dy}{du}(p-1)+qy=0 nice, so it has constant coefficients in u, right?

    Thank you very much sir
    Yup.
    -Dan
    Last edited by topsquark; May 1st 2011 at 03:53 PM. Reason: Joke was a little bit too far.
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