Ode 2nd degree with a polynomial as a coefficient

[bloodlines]

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!

 

[jackmell]

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]

 

[bloodlines]

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.

 

[bloodlines]

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!