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Math Help - Finished - but is it Right (ODE)??

  1. #1
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    Finished - but is it Right (ODE)??

    Here's my working for this ODE problem - could someone venture an opinion on the result please??

    Before I start 'topsquark' from this forum was kind enough to offer the suggestion in reply to my earlier post to make a substitution x = e^t which confused me a little since he called 'x' the dependent variable, so perhaps someone may have a thought on that advice too??

    Anyways, here's my attempt

    x^2*y" + 4x*y' - 4y = 0 need to find 'p' where y = x^p satisfies the ODE. My analysis:
    This is a 2nd order ODE of form f(x, y', y") so I'm using reduction of order process. I've found one solution by trial and error (y = 4x)
    Form of 2nd solution is:
    y(x) = xv , y'(x) = v + xv' , y"(x) = 2v' + xv"
    sub
    x^2(2v' + xv") + 4x(v + xv') - 4x(v) = 0
    x^3*v" + 6x^2*v' = 0
    Let p = v' and p' = v" ----> x^3p' + 6x^2*p = 0
    (1st order ODE) separate the variables
    x^3*dp/dx = -6x^2*p ----> 1/p dp = -6/x dx ---> integral 1/p dp = imtegral -6/x dx
    lnp = -6lnx + C = -6lnx + lnK
    (lnp - lnK)/ lne^-6 = lnx = lne^t = t
    t = ln(p/K)/ lne^(-6) ---> -6t = ln(p/K) ---> e^(-6t) = p/K
    p = K*e^(-6t) = K(e^t)^(-6) = K*x^(-6)
    p = K*x(-6)
    Correct?????? How about the x = e^t advice?????
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  2. #2
    is up to his old tricks again! Jhevon's Avatar
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    Quote Originally Posted by dave d View Post
    Here's my working for this ODE problem - could someone venture an opinion on the result please??

    Before I start 'topsquark' from this forum was kind enough to offer the suggestion in reply to my earlier post to make a substitution x = e^t which confused me a little since he called 'x' the dependent variable, so perhaps someone may have a thought on that advice too??

    Anyways, here's my attempt

    x^2*y" + 4x*y' - 4y = 0 need to find 'p' where y = x^p satisfies the ODE. My analysis:
    This is a 2nd order ODE of form f(x, y', y") so I'm using reduction of order process. I've found one solution by trial and error (y = 4x)
    Form of 2nd solution is:
    y(x) = xv , y'(x) = v + xv' , y"(x) = 2v' + xv"
    sub
    x^2(2v' + xv") + 4x(v + xv') - 4x(v) = 0
    x^3*v" + 6x^2*v' = 0
    Let p = v' and p' = v" ----> x^3p' + 6x^2*p = 0
    (1st order ODE) separate the variables
    x^3*dp/dx = -6x^2*p ----> 1/p dp = -6/x dx ---> integral 1/p dp = imtegral -6/x dx
    lnp = -6lnx + C = -6lnx + lnK
    (lnp - lnK)/ lne^-6 = lnx = lne^t = t
    t = ln(p/K)/ lne^(-6) ---> -6t = ln(p/K) ---> e^(-6t) = p/K
    p = K*e^(-6t) = K(e^t)^(-6) = K*x^(-6)
    p = K*x(-6)
    Correct?????? How about the x = e^t advice?????
    what you did is overkill. this is an Euler equation
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