Solve ODE: $y'(x)=y(x)^2+x^2$

Question

Consider ODE $y'(x)=y(x)^2+x^2$. Using Mathematica with code DSolveValue[{y'[x] == y[x]^2 + x^2, y[0] == a}, y[x], x]// FullSimplify, I see the solution with Bessel and Gamma functions $$y(x)= -\frac{x \left(a \Gamma \left(\frac{1}{4}\right) J_{-\frac{3}{4}}\left(\frac{x^2}{2}\right)+2 \Gamma \left(\frac{3}{4}\right) J_{\frac{3}{4}}\left(\frac{x^2}{2}\right)\right)}{a \Gamma \left(\frac{1}{4}\right) J_{\frac{1}{4}}\left(\frac{x^2}{2}\right)-2 \Gamma \left(\frac{3}{4}\right) J_{-\frac{1}{4}}\left(\frac{x^2}{2}\right)}$$ Does anyone know how to achieve this result?

Answer 1

Let $y=-z'/z$, then $z$ satisfies $z''+x^2z=0$ whose linearly independent solutions in the notation of DLMF 12.2.3 are $W(0,\pm\sqrt2x)$. By DLMF 12.14.13 another linearly independent pair of solutions is $z_\pm(x)=\sqrt xJ_{\pm1/4}(x^2/2)$. Now let $z(x)=cz_-(x)+dz_+(x)$; computations in Mathematica then show that $$y(0)=-\frac{2d\Gamma(3/4)}{c\Gamma(1/4)}$$ from which we may set $c=2\Gamma(3/4)$ and $d=-a\Gamma(1/4)$; simplifying gives the desired result.