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Math Help - transform the Cauchy-Euler equation

  1. #1
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    transform the Cauchy-Euler equation

    ok in this problem im supposed to use the substitution x = e^t to transform the given Cauchy-Euler equation to a differential equation with constant coefficients.

    x^2y'' +9xy' -20y=0

    im not sure how to start this problem

    maybe dy/dx= (dy/dt)(dt/dx) = (dy/dt)e^t

    so then could i say d2y/dx^2 = (e^t)d2y/dt^2<---- ?

    maybe then i could substitute back into the equation? i dunnno..

    any help would be appreciated thanks in advance.
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  2. #2
    MHF Contributor FernandoRevilla's Avatar
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    x=e^t\Rightarrow t=\log x \Rightarrow y'(x)=x'(t)\cdot{(1/x)=x'(t)e^{-t}}

    Then,

    y''(x)=\ldots=(y''(t)-y'(t))e^{-2t}


    P.D. Another method to solve the equation is to try solutions of the form y=x^k .
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  3. #3
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    ok so i saw my mistake in taking the second derivative and this is what i have so far



    am i on the right track?
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  4. #4
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    this problem is supposed to reduce down to m^2 + 8m -20 = 0???
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  5. #5
    A Plied Mathematician
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    For Cauchy-Euler equations, I've always preferred FernandoRevilla's PD in post # 2. You get the characteristic equation immediately by plugging that ansatz into the DE.
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  6. #6
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    The problem with simply taking y= x^r, as both Fernando Revilla and Ackbeet suggest, is that it not clear what to do with "special cases". For example, consider the differential equation x^2y''- xy'+ y= x+ ln(x). Taking y= x^r the equation reduces to (r^2- r)x^r- rx^r+ x^r= (r^2- 2r+ 1)x^r= 0 which gives the characteristic equation r^2- 2r+ 1= (r- 1)^2= 0. Okay, one solution is y= x^{1} but what is the other, independent, solution we need? And what should we do about that right side, x+ ln(x)?

    If we know that the substitution x= e^t changes the Euler-Cauchy equation to the constant-coefficients equation with the same characteristic equation, we know that the solution to the assosciated homogeneous equation, in terms of t, is Ce^{t}+ Dte^{t}, which tells us that the corresponding solution to the original equation is Cx+ Dxln(x). Also, the right hand side, in terms of t, is e^t+ t, which, by "undetermined coefficients" has particular solution \frac{1}{2}t^2e^{t}+ t+ 2 so that the entire general solution to the original equation is
    y(x)= Cx+ Dxln(x)+ \frac{1}{2}(ln(x))^2x+ ln(x)+ 2.

    Of course, we could learn variations of all those rules for Cauchy-Euler equations but it is much simpler to know how to convert such an equation to a constant-coefficients equation.
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  7. #7
    Forum Admin topsquark's Avatar
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    Quote Originally Posted by slapmaxwell1 View Post
    x^2y'' +9xy' -20y=0
    My notation is probably going to make several members cringe, but this is the way I learned it...

    As was suggested let
    \displaystyle x = e^t \implies dx = e^t~dt \implies \frac{d}{dx} = e^{-t} \frac{d}{dt}

    So looking at your original equation as
    \displaystyle x^2 \frac{d}{dx} \left ( \frac{d}{dx} \right ) y + 9x \frac{d}{dx}y - 20y = 0<br />

    gives us
    \displaystyle e^{2t} \cdot e^{-t} \frac{d}{dt} \left ( e^{-t} \frac{d}{dt} \right ) y + 9e^t \cdot e^{-t} \frac{d}{dt}y - 20y = 0

    So,after simplifying:
    \displaystyle e^{2t} \cdot \left ( -e^{-2t} \frac{dy}{dt} + e^{-2t} \frac{d^2y}{dt^2} \right ) + 9 e^t \cdot e^{-t} \frac{dy}{dt} - 20y = 0

    And finally:
    \displaystyle \frac{d^2y}{dt^2} + 8 \frac{dy}{dt} - 20y = 0

    and you can take over from here.

    -Dan
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  8. #8
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    ok im still a little confused, i dont see how this is true?




    i get how to find dy/dx, but not the second derivative...



    i keep getting a 1/x not 1/x^2...but im just using the chain rule
    Last edited by slapmaxwell1; March 19th 2011 at 08:00 PM.
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  9. #9
    Forum Admin topsquark's Avatar
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    Quote Originally Posted by slapmaxwell1 View Post
    ok im still a little confused, i dont see how this is true?




    i get how to find dy/dx, but not the second derivative...



    i keep getting a 1/x not 1/x^2...but im just using the chain rule
    One problem stems from leaving that x there.

    Example:
    x = e^t

    so
    \displaystyle \frac{d^2y}{dx^2} = x^2 \left ( \frac{d^2y}{dt^2} - \frac{dy}{dt} \right )

    should be
    \displaystyle \frac{d^2y}{dx^2} = e^{-2t} \left ( \frac{d^2y}{dt^2} - \frac{dy}{dt} \right )


    If the problem is that you need a full derivation
    \displaystyle \frac{d}{dx} = e^{-t} \frac{d}{dt}

    So the second derivative of y with respect to x becomes:
    \displaystyle \frac{d}{dx} \left ( \frac{dy}{dx} \right ) = \left [ e^{-t} \frac{d}{dt} \right ] \left [ e^{-t} \frac{d}{dt} \right ]y = \left [ e^{-t} \frac{d}{dt} \right ] \left [ e^{-t} \frac{dy}{dt} \right ]

    Now use the product rule
    \displaystyle = e^{-t} \left [ -e^{-t} \frac{dy}{dt} + e^{-t} \frac{d^2y}{dt^2} \right ]

    -Dan
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  10. #10
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    ok i see it, i was taking the derivative wrong. i went and reviewed implicit differentiation....thanks you guys!
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