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Math Help - Cauchy-Euler Equation - what's going on with the coefficients?

  1. #1
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    Cauchy-Euler Equation - what's going on with the coefficients?

    I'm working with the Cauchy-Euler equation... In the direction of


    a \frac{d^2y}{dt^2} + b \frac{dy}{dt} + cy = 0


    ...being transformed into


    Ax^2 \frac{d^2y}{dx^2} + Bx \frac{dy}{dx} + Cy = 0


    by setting

    t = ln(x)


    ... I have been able to prove how


    \frac{dy}{dt} = x \frac{dy}{dx}

    and

    \frac{d^2y}{dt^2} = x^2 \frac{d^2y}{dx^2} + x \frac{dy}{dx}


    Oddly enough, what I'm getting stuck on, is how a, b, c relate to A, B, C.

    It seems to me (please let me know if I'm wrong) that

    b = B

    c = C


    ...but I'm not exactly sure what's going on with a.

    Can someone help me figure out what I'm missing?
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  2. #2
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    Re: Cauchy-Euler Equation - what's going on with the coefficients?

    Quote Originally Posted by Lancet View Post
    I'm working with the Cauchy-Euler equation... In the direction of


    a \frac{d^2y}{dt^2} + b \frac{dy}{dt} + cy = 0


    ...being transformed into


    Ax^2 \frac{d^2y}{dx^2} + Bx \frac{dy}{dx} + Cy = 0


    by setting

    t = ln(x)


    ... I have been able to prove how


    \frac{dy}{dt} = x \frac{dy}{dx}

    and

    \frac{d^2y}{dt^2} = x^2 \frac{d^2y}{dx^2} + x \frac{dy}{dx}


    Oddly enough, what I'm getting stuck on, is how a, b, c relate to A, B, C.

    It seems to me (please let me know if I'm wrong) that

    b = B

    c = C


    ...but I'm not exactly sure what's going on with a.

    Can someone help me figure out what I'm missing?
    Have you substituted your results and compared the resulting DE with the given DE ....?
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    Re: Cauchy-Euler Equation - what's going on with the coefficients?

    Quote Originally Posted by mr fantastic View Post
    Have you substituted your results and compared the resulting DE with the given DE ....?
    Do you mean plug in numbers? I wouldn't think that would be as accurate in determining what was going on.

    Regardless, I was just exposed to this a couple of days ago, so I'm very new to this equation, and am not quite sure how to approach this particular facet of it.
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    Re: Cauchy-Euler Equation - what's going on with the coefficients?

    Quote Originally Posted by Lancet View Post
    I'm working with the Cauchy-Euler equation... In the direction of


    a \frac{d^2y}{dt^2} + b \frac{dy}{dt} + cy = 0


    ...being transformed into


    Ax^2 \frac{d^2y}{dx^2} + Bx \frac{dy}{dx} + Cy = 0


    by setting

    t = ln(x)


    ... I have been able to prove how


    \frac{dy}{dt} = x \frac{dy}{dx}

    and

    \frac{d^2y}{dt^2} = x^2 \frac{d^2y}{dx^2} + x \frac{dy}{dx}


    Oddly enough, what I'm getting stuck on, is how a, b, c relate to A, B, C.

    It seems to me (please let me know if I'm wrong) that

    b = B

    c = C


    ...but I'm not exactly sure what's going on with a.

    Can someone help me figure out what I'm missing?
    One of my first posts in MHF...

    http://www.mathhelpforum.com/math-he...ion-98829.html

    Kind regards

    \chi \sigma
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    Re: Cauchy-Euler Equation - what's going on with the coefficients?

    Quote Originally Posted by chisigma View Post
    One of my first posts in MHF...

    http://www.mathhelpforum.com/math-he...ion-98829.html

    Kind regards

    \chi \sigma
    I see what I was already able to prove, but I don't see how that explains what happens to the coefficients...
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    Re: Cauchy-Euler Equation - what's going on with the coefficients?

    Quote Originally Posted by Lancet View Post
    Do you mean plug in numbers? I wouldn't think that would be as accurate in determining what was going on.

    Regardless, I was just exposed to this a couple of days ago, so I'm very new to this equation, and am not quite sure how to approach this particular facet of it.
    No I don't meant that at all. I meant what I said and said what I meant.

    You have results for dy/dt and d^2y/dt^2. Substitute them into the first DE. Compare the result with the second DE. Hence write down the obvious relationship betwen a, b, c and A, B, C. Did you do that?
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    Re: Cauchy-Euler Equation - what's going on with the coefficients?

    Quote Originally Posted by mr fantastic View Post
    No I don't meant that at all. I meant what I said and said what I meant.

    You have results for dy/dt and d^2y/dt^2. Substitute them into the first DE.


    Do you mean this?

    ax^2 \frac{d^2y}{dx^2} + x \frac{dy}{dx} + bx \frac{dy}{dx} + cy = 0
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    Re: Cauchy-Euler Equation - what's going on with the coefficients?

    Quote Originally Posted by Lancet View Post
    Do you mean this?

    ax^2 \frac{d^2y}{dx^2} + x \frac{dy}{dx} + bx \frac{dy}{dx} + cy = 0
    You have not substituted correctly. The second term should have an a in its coefficient.

    Then you compare the coefficients of d^2y/dx^2, dy/dx and y in this DE with the second DE. You are making hard work of this - I do not see what the difficulty is.
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    Re: Cauchy-Euler Equation - what's going on with the coefficients?

    Quote Originally Posted by mr fantastic View Post
    You have not substituted correctly. The second term should have an a in its coefficient.

    Then you compare the coefficients of d^2y/dx^2, dy/dx and y in this DE with the second DE. You are making hard work of this - I do not see what the difficulty is.
    It's a new subject to me along with a new twist on things, and I'm trying to figure it out on my own. That's what the problem is.

    Okay, that being said, would it then be accurate to say this?


    A = a

    B = a + b

    C = c
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    Re: Cauchy-Euler Equation - what's going on with the coefficients?

    Quote Originally Posted by Lancet View Post
    It's a new subject to me along with a new twist on things, and I'm trying to figure it out on my own. That's what the problem is.

    Okay, that being said, would it then be accurate to say this?


    A = a

    B = a + b

    C = c
    Yes.
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  11. #11
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    Re: Cauchy-Euler Equation - what's going on with the coefficients?

    Quote Originally Posted by mr fantastic View Post
    Yes.
    Thanks!
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