just wondering if anyone can help out here.
I'm solving the following ODE:
$$\frac{\textrm{d}^{2}y_{2}}{\textrm{d}x^{2}}-xy_{2} =\frac{\textrm{d}}{\textrm{d}x}(xy_{1}),$$
where
$$y_{1}=\int\textrm{Ai}(x)\,\textrm{d}x.$$
The solution I have is
$$y_{2}=C_{1}\textrm{Ai}(x)+C_{2}\textrm{Bi}(x)-\pi\left[\textrm{Ai}(x)\int\textrm{Bi}(x)\frac{\textrm{d}}{\textrm{d}x}(xy_{1})\,\textrm{d}x -\textrm{Bi}(x)\int\textrm{Ai}(x)\frac{\textrm{d}}{\textrm{d}x}(xy_{1})\,\textrm{d}x\right].$$
Using integration by parts gives
$$y_{2}=C_{1}\textrm{Ai}(x)+C_{2}\textrm{Bi}(x)+\pi\left[\textrm{Ai}(x)\int xy_{1}\textrm{Bi}'(x)\,\textrm{d}x -\textrm{Bi}(x)\int xy_{1}\textrm{Ai}'(x)\,\textrm{d}x\right].$$
If I consider only the first of the 'tricky' integrals (if I can do one I can do both) I have that
$$\int xy_{1}\textrm{Bi}'(x)\,\textrm{d}x=\frac{x^{2}y_{1}\textrm{Bi}'(x)}{2}-\int\frac{x^{2}y_{1}'\textrm{Bi}'(x)}{2}\,\textrm{d}x-\int\frac{x^{2}y_{1}\textrm{Bi}''(x)}{2}\,\textrm{d}x.$$
Or, equivalently
$$\int xy_{1}\textrm{Bi}'(x)\,\textrm{d}x=\frac{x^{2}y_{1}\textrm{Bi}'(x)}{2}-\int\frac{x^{2}\textrm{Ai}(x)\textrm{Bi}'(x)}{2}\,\textrm{d}x-\int\frac{x^{2}y_{1}\textrm{Bi}''(x)}{2}\,\textrm{d}x.$$
The second term on the right-hand side has a closed form solution. Hence
$$\int xy_{1}\textrm{Bi}'(x)\,\textrm{d}x=\textrm{stuff that I know}-\int\frac{x^{2}y_{1}\textrm{Bi}''(x)}{2}\,\textrm{d}x.$$
I can't really see where else to go from here. I have tried parts on the remaining 'problem' term but that hasn't proved fruitful.
Am I simply going down a rabbit hole? Is a nice closed form solution even obtainable in this case?