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Ode 2nd degree with a polynomial as a coefficient

  1. Oct 18, 2011 #1
    Hi friends,

    I have been trying to solve the ode of second degree below with respect to z:

    d2y/dz2=(i/a*z+b)*y

    i is the complex i, a and b are constants

    i ended up with the summation of bessel functions of first end second kind. Then I checked with matlab ode solver it gives no explicit solution. please help me.

    Thank you so much!
     
  2. jcsd
  3. Oct 18, 2011 #2
    Just scrap the i/a thing for now and call it k and consider the equation:

    [tex]y''-(kz+b)y=0[/tex]

    Now, suppose we know the solution to:

    [tex]y''-xy=0[/tex]

    in terms of AiryAi and AiryBi functions.

    Then consider a change of independent variable:

    [tex]u=\frac{b+kz}{k^{2/3}}[/tex]

    and convert the DE to a DE in y as a function of u. That is:

    [tex]\frac{dy}{du}=\frac{dy}{dz}\frac{dz}{du}[/tex]

    then compute the second derivative and make those substitutions and see if you get it in the form:

    [tex]y''-uy=0[/tex]

    Which the solution is C_1 AiryAi(u)+C_2 AiryBi(u)

    or for the original equation:

    [tex]C_1 AiryAi(\frac{b+kz}{k^{2/3}})+C_2 AiryBi(\frac{b+kz}{k^{2/3}})[/tex]
     
  4. Oct 18, 2011 #3
    Thank you so much, I am also trying to solve for

    d2y/dz2=i*y/(a*z^3+b*z^2+c*z+d)

    could you help me with this one too? I appreciate your help.
     
  5. Oct 18, 2011 #4
    By the I got a solution like:

    y=m*[C1*besselj(1,2*m*sqrt(-i)/a)+C2*bessely(1,2*m*sqrt(-i)/a)]

    where m=sqrt(a*z+b)

    is this another form of the same solution (Airy transform to bessel)?

    if you could help with the 3rd degree polynomial, that would be great!

    Thanks a lot!
     
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