How to solve the non-linear differential equation $y''=x-y^2$?

Question

$y''(x)=x-y^2(x)$

I'm particularly interested in solutions when $x>0$.

I've performed asymptotic analysis and reached the conclusion that solutions must behave as $\pm\sqrt{x}$ when $x\rightarrow \infty$ but I don't know what to do next.

I've also tried looking for a solution of the form $y(x)=y_p(x)\pm\sqrt{x}$, but I got to an even more difficult differential equation for $y_p(x)$

I can also deduce the obvious thing that $y''>0$ when $x>y^2$ and $y''<0$ when $x<y^2$, so there are like three regions. My guess is that there will be solutions which tend to the parabola $y=\sqrt{x}$ in every region, but I'm really stuck.

Any insights would be very appreciated.

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EDIT: When I say ''solve'', I mean that I want a feel on how possible solutions behave (e.g., are there any oscillating solutions?, if I have a solution of an IVP where $y^2(x_0)>x_0$, what will that solution do?...)

Also, I would be thrilled if there were solutions expressable as elementary functions.

Now that i think about it, a better phrasing than ''solve this ODE'' would be ''study [the behaviour of solutions of] this differential equation''

Answer 1

If you are interested in the behavior near infinity, you can provide very good approximations through the following method: $$ y=(x^{1/2}+A(x)),\qquad y''=(-\frac{1}{4}x^{-3/2}+A''(x))$$ $$ y^2 = x + 2x^{1/2}A(x) + A^2(x)$$ $$ y^2 + y'' = x + 2 x^{1/2} A(x) -\frac{1}{4}x^{-3/2}+A''(x)+A^2(x) \tag{1}$$ Now set $A(x)$ in order to cancel out the second and third term in the RHS: $$ A(x)=\frac{1}{8x^2}+B(x). $$ You get: $$ y^2 + y'' = x+ \left(2\sqrt{x}+\frac{1}{4x}\right) B(x)+\frac{3}{4x^4}+B''(x)+B^2(x),\tag{2}$$ hence by setting: $$ B(x) = - \frac{3}{8x^{9/2}+x^3}+C(x)$$ you get the quite close approximation:

$$ \sqrt{x}+\frac{1}{8x^2}-\frac{3}{8x^{9/2}+x^3}\leq y(x)\leq \sqrt{x}+\frac{1}{8x^2}\tag{3}$$

that can even refined through successive steps of the method, even if the bounds become more and more involved.