Simple Nonlinear Differential Equation I have tried in vain to search for a closed form solution (involving known functions) to the simple first order equation $\dfrac{dy}{dx}=x^2+y^2, y(0) = 0$. Can anyone help with a solution that does not have the form of an infinite series?
I tried to solve this using maple and here's my code:

dy:=diff(y(x),x)=x^2+y(x)^2;

dsolve({dy,y(0)=0},y(x));

And it gives me a solution in terms of Bessel function, so I don't think there's a solution in terms of elementary functions.
Putting
$$ y = -\dfrac{d(\ln z)}{dx} = -\dfrac{z'}{z}$$
gives
$$y'=\left(\dfrac{z'}{z}\right)^2-\dfrac{z''}{z},$$
and so $$ y' = y^2 - \frac{z''}{z}$$

So in $y' = y^2 + x^2$ the $y^2$ cancels out leaving $-\dfrac{z''}{z} = x^2$, or $z''=-x^2z$, which according to Wolfram has the solutions 
$z = c \cdot D[-1/2, u]$, where D is the parabolic cylinder function of fractional order $1/2$ and $u = (1+i)x$
 or (-1+i)x

The relation $y = -\dfrac{d(\ln z)}{dx}$ then leads to
$$(1-i)y = \frac{2D[1/2, u]}{D[-1/2, u]}-u,$$
where $u$ can be $(1+i)x$.  However, Wolfram says $2D[1/2, 1] - D[-1/2, 1] =  1.03133....$ and not $0$ as is required for $y(0)=0$ so there also seems to be a problem with this solution!