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Can anybody help please

  1. May 10, 2009 #1
    I have got a general solution for the equation

    xy'' - y' + 4x^3y = 0, x > 0

    by converting to the normal version of the self adjoint form and solving with an auxiliary equation I have

    y = Acos(2x) + Bsin(2x)

    It then asks to select two independent solutions and verify the wronskian satisfies Abel's identity.

    Can I simply set A = 0, B = 1 for y1 and A = 1, B = 0 for y2

    I did this and I got 1 for the wronskian and kx for Abel's which doesn't prove it!!!

    Can anybody tell me where I'm going wrong please
  2. jcsd
  3. May 10, 2009 #2


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    Science Advisor

    If y= A cos(2x)+ B sin(2x), y'= -2A sin(2x)+ 2B cos(2x), and y"= -4Acos(2x)- 4Bsin(2x).
    Putting those into the equation gives:
    xy'' - y' + 4x^3y = -4Axcos(2)- 4Bxsin(2x)+ 2Asin(2x)- 2Bcos(2x)+ 4x^3Acos(2x)+ 4x^3Bsin(2x) which is definitely NOT equal to 0. y= Acos(2x)+ Bsin(2x) does NOT satisfy the equation.

    Yes, dividing the equation by 1/x2 puts it into self-adjoint form but what did you do from there?
  4. May 10, 2009 #3
    I put it into the normal version of the self adjoint form.

    Where d/dx{p(x)dy/dx}+q(x)y=0

    t = integral of 1/p(x) = x^3/3

    dt/dx = x^2

    dy/dx = dy/dt.dt/dx = dy/dt.x^2


    Then solved using the auxiliary equation. It probably is my working that is wrong, however the question I really needed answering is; if I get a solution to the differential equation, to get two linearly independent solutions from the general solution can I simply assign values to the constants.

    Thank you for your response
  5. May 10, 2009 #4
    I get m=+-sqrt(-4)


    y = exp(alpha*x){Acos(beta*x)+Bcos(beta*x)}

    where alpha is real and beta is imaginary


    alpha is = 0

    beta is = 2

    y = Acos(2x)+Bsin(2x)
  6. May 10, 2009 #5
    my values were for t not x, I forgot to substitute them back in

    y = Acos(2t) + Bsin(2t) where t = x^2/2

    it works now

    Thanks for your help
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