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Math Help - bondary value problem

  1. #1
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    Feb 2008
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    bondary value problem

    u=x and v= - 1/3x^2 are two solutions of the differential equation
    x^2y'' + 2xy' - 2y = 0
    such that u(0)=0 and 2v(1)+v'(1)=0
    solve the boundary value problem
    x^2y'' + 2xy' - 2y = x^2 y(0)=0 and 2y(1) + y'(1) = 0
    if u and v are solutions does that mean y_1=u and y_2=v
    whats the difference between u(0)=0 and 2v(1)+v'(1)=0 and y(0)=0 and 2y(1) + y'(1) = 0
    they have just replaced y=u=v. I will be able to solve this using green's function once I know what are y_1 and y_2 equal to?
    Thanks for any help
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  2. #2
    Super Member
    Joined
    Aug 2008
    Posts
    903
    Quote Originally Posted by charikaar View Post
    u=x and v= - 1/3x^2 are two solutions of the differential equation
    x^2y'' + 2xy' - 2y = 0
    such that u(0)=0 and 2v(1)+v'(1)=0
    solve the boundary value problem
    x^2y'' + 2xy' - 2y = x^2 y(0)=0 and 2y(1) + y'(1) = 0
    if u and v are solutions does that mean y_1=u and y_2=v
    whats the difference between u(0)=0 and 2v(1)+v'(1)=0 and y(0)=0 and 2y(1) + y'(1) = 0
    they have just replaced y=u=v. I will be able to solve this using green's function once I know what are y_1 and y_2 equal to?
    Thanks for any help
    I think you should have written the two particular solutions as:

    y_1=x

    y_2=-1/3 \frac{1}{x^2}

    It's not hard to use reduction of order with y_1 to obtain the general solution of the non-homogeneous equation:

    y(x)=c_1 \frac{1}{x^2}+c_2 x+1/4 x^2

    Not sure though what you wish to do with the boundary information you gave.
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