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Thread: Euler equation

  1. #1
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    Euler equation

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

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

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

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

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

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

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

    Thus
    $\displaystyle dx = e^u~du$

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

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

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

    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

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

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

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

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

    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 $\displaystyle e^{u}$ 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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    $\displaystyle \frac{d^2y}{du^2}+\frac{dy}{du}(p-1)+qy=0$ 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
    $\displaystyle \frac{d^2y}{du^2}+\frac{dy}{du}(p-1)+qy=0$ 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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