How do I solve $y' = x^2 + y^2$ analytically? I came across this equation:
$$\frac{dy}{dx} = x^2 + y^2$$
$$y(0)=0$$
I found numerical solutions to it using Runge-Kutta methods, but I want to verify my answers by solving it analytically.
At first, I thought the expression $x^2 + y^2$ indicated that I use polar coordinates instead:
$$\frac{d(r\sin{\theta})}{d(r\cos{\theta})}=r^2$$
$$\frac{\sin{\theta}~dr+r\cos{\theta}~d\theta}{\cos{\theta}~dr-r\sin{\theta}~d\theta}=r^2$$
Well, the situation just got more complicated, so nope, this isnt gonna work.
I then tried using Laplace transforms, but there wasnt a simple expression for $\mathcal{L}\{y^2\}$.
I finally tried plugging it into Matlab to get an answer, and I got this:
pretty(ySol(x))

       / 3 \        /   3     \               1/4   3/2            /   3     \
x gamma| - | besselk| - -, #1 | 4i - sqrt(2) 4    pi    x z besseli| - -, #1 | 1i
       \ 4 /        \   4     /                                    \   4     /
---------------------------------------------------------------------------------
             / 3 \        / 1     \            1/4   3/2          / 1     \
      4 gamma| - | besselk| -, #1 | + sqrt(2) 4    pi    z besseli| -, #1 |
             \ 4 /        \ 4     /                               \ 4     /

where

          2
         x  1i
   #1 == -----
           2

Matlab isnt giving me any solution in terms of simple functions. Does this mean that I can't solve this equation analytically? Any help would be appreciated, thanks!
 A: The solution of $$\frac{dy(x)}{dx} = x^2 + y^2(x)$$ is given by
$$y(x)=x \,\frac{c_1
   J_{\frac{3}{4}}\left(\frac{x^2}{2}\right)-J_{-\frac{3}{4}}\left(\frac{x^2}{2}\right)}{c_1
   J_{-\frac{1}{4}}\left(\frac{x^2}{2}\right)+J_{\frac{1}{4}}\left(\frac{x^2}{2}\right)}$$
This makes
$$y(0)=-\frac{2 \Gamma \left(\frac{3}{4}\right)}{c_1 \Gamma \left(\frac{1}{4}\right)}\qquad \text{and}\qquad y'(0)=\frac{2 \pi ^2}{c_1^2 \Gamma \left(\frac{1}{4}\right)^3 \Gamma
   \left(\frac{5}{4}\right)}$$
So, if the condition is $y(0)=0$, then $c_1=\infty$ and the solution is
$$y(x)=x\frac{
   J_{\frac{3}{4}}\left(\frac{x^2}{2}\right)}{J_{-\frac{1}{4}}\left(\frac{x^2}{2}
   \right)}$$
A: One may get  Mclaurin series solutions here. $$y(x)=\sum_{k=0}^{\infty} \frac{y^{(k)}(0)}{k!}x^k,~~~~(*)$$ $k$th derivatives of $y(x)$ at $x=0$ can be caklculated as 
by differentiation $$y'=x^2+y^2~~~(1)$$, repeatedly. we get $y'(0)=0$, then $$y''(x)=2x+2y(x)y'(x),~~~~(2)$$ gives $y''(0)=0$. Next, differentiating w.r.t $x$ $$y^{(3)}(x)=2+2y'^2(x)+2y(x)y''(x),~~~~(3)$$ we get $y'''(0)=2.$ So on and so forth. The first term in $y(x)$ is $\frac{x^3}{3}$ and the latter ones can be calcluated by differentiating (1) succsessively after (3).
