Solving a second degree non-homogeneous ODE with variation of parameters

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.

• How can Your solution be zero? This isn't correct. Could You provide the equation? – Logic_Problem_42 May 4 '18 at 9:07
• @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) – kathi_h May 4 '18 at 9:22
• The dimension of the solution space of any second order homogeneous differential equation is always two, – Logic_Problem_42 May 4 '18 at 9:25
• @kathi_h : Why don't you definitively write the ODE in black and white ? – JJacquelin May 4 '18 at 10:00
• @JJacquelin sorry guys, I wrote the equation now. – kathi_h May 4 '18 at 19:55

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.