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I am solving a second order non-homogeneous ODE and trying to use the variation of parameters. For this, I first solve the associated homogeneous ode, and get two solutions of which one is equal to zero. In this case the Wroskian is equal to zero, and I fail to apply the variation of parameters. What is the approach in this type of problem, i.e. when one of the complementary solutions are zero?

Edit: the equation is as follows:

$$ x^{2}y^{''}(x)+2xy^{'}(x)-(ax^{2}+n(n+1))y(x)=bi_{n}(\sqrt{c}x) $$ and defined on a unit ball. I find the complementary solution as: $$ y_{c}=ki_{n}(\sqrt{a}n), k\in IR $$

where $$ i_{n}(x)=\sqrt{\frac{\pi}{2x}}I_{n+\frac{1}{2}}(x) $$ and where $I_{n+\frac{1}{2}}(x)$ is a modified Bessel function of the first kind.

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    $\begingroup$ How can Your solution be zero? This isn't correct. Could You provide the equation? $\endgroup$ – Logic_Problem_42 May 4 '18 at 9:07
  • $\begingroup$ @Logic_Problem_42 It's a spherical Bessel equation and the second part drops out because of regularity in the center of the domain (unit Ball) $\endgroup$ – kathi_h May 4 '18 at 9:22
  • $\begingroup$ The dimension of the solution space of any second order homogeneous differential equation is always two, $\endgroup$ – Logic_Problem_42 May 4 '18 at 9:25
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    $\begingroup$ @kathi_h : Why don't you definitively write the ODE in black and white ? $\endgroup$ – JJacquelin May 4 '18 at 10:00
  • $\begingroup$ @JJacquelin sorry guys, I wrote the equation now. $\endgroup$ – kathi_h May 4 '18 at 19:55
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HINT :

The solutions of the associated homogeneous ODE : $$x^{2}y^{''}(x)+2xy^{'}(x)-(ax^{2}+n(n+1))y(x)=0$$ are : $$y(x)=c_1\:j_{-n-1}(\sqrt{-a}\:x)+c_2\:y_{-n-1}(\sqrt{-a}\:x)$$ $j_\nu(z)$ and $y_\nu(z)$ are the spherical Bessel functions of the first and second kind respectively. They are independent functions. The Wronskian isn't equal to zero.

Note that the trivial solution $y(x)=0$ (for $c_1=c_2=0$), which isn't an independent function from the two others, has not to be considered in the Wronskian calculus.

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  • $\begingroup$ but shouldn't the second kind be left out, since it blows up when x is zero? mathworld.wolfram.com/… $\endgroup$ – kathi_h May 5 '18 at 7:02
  • $\begingroup$ Of course, this can be done but after the calculus of the Wronskian, not before. If you want eliminate the second kind at first, don't use the Wronskian in trying the variation of parameters method. There is only one parameter left. $\endgroup$ – JJacquelin May 5 '18 at 7:45
  • $\begingroup$ How would you justify the usage of the second kind in the calculation but then dropping it out of the solution? Intuitively, it doesn't seem right because when x=0 it is so in all the places where there is x, hence also in the wronskian. $\endgroup$ – kathi_h May 5 '18 at 7:48
  • $\begingroup$ There is nothing to justify. If you use the Wronskian in order to find a particular solution of the non-homogeneous ODE, you don't take account of any boundary condition. This leads to the general solution. Then, after that, if a boundary condition implies that the second kind of parabolic cylinder function be eliminated either it is possible and the problem has a solution, or it is not possible and the problem has no solution according to the specified boundary conditions. $\endgroup$ – JJacquelin May 5 '18 at 7:56
  • $\begingroup$ thank you @JJacquelin! $\endgroup$ – kathi_h May 5 '18 at 7:58

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