... Is it possible to make the substitution $y(x)= z(x)+\sqrt{x}$ and represent the solution y(x) of the following initial value problem for all $x \geqslant1$ in terms of the new function z(x)? \begin{cases} y'=x-y^2 \\\\ y(0)=0 \end{cases} In other words, naming Y(x) the solution of the the initial value problem above , and Z(x) that of the following problem, supposing that solving the first problem we have found the value Y(1) ,is it true that $Z(x)=Y(x)$-$\sqrt{x}$ for $x \geqslant1$ ? \begin{cases} z'=-z^2-2 \sqrt{x}\cdot z-\frac{1}{2\sqrt{x}}\\ z(1)=Y(1)-1 \end{cases} I ask this question because I have found that Z(x) must tend to 0 as x tends to infinity for any -1<z(1)<0 and I'm not sure that this implies that $Y(x)-\sqrt{x}$ tends to 0. I think that the unicity of the solutions Y(x) and Z(x) is enough to state that that Z(x)=Y(x)-$\sqrt{x}$, because if Y(x) satisfies the first problem then Y(x)-$\sqrt{x}$ satisfies the second one. Is this argument correct? Thanks in advance for the answer.
...
For large $x$ and small $z$ the equation is dominated by the terms $z'=-2\sqrt{x}z$ with the solution $z=Ce^{-\frac43x^{3/2}}$ which converges rapidly to zero.
Seeing the equation as Riccati equation and setting $y=\frac{u'}{u}$ leads to $u''=xu$ which has the Airy functions as solution. Additionally, the second order WKB approximation gives for $x\gg 1$ $$ u\approx x^{-\frac14}(Ae^{\frac23x^{3/2}}+Be^{-\frac23x^{3/2}}) $$ which leads back to $$ y(x)\approx-\frac1{4x} +\sqrt{x}\frac{Ae^{\frac23x^{3/2}}-Be^{-\frac23x^{3/2}}}{Ae^{\frac23x^{3/2}}+Be^{-\frac23x^{3/2}}} =\sqrt{x}-\frac1{4x}-\sqrt{x}\frac{-2Be^{-\frac43x^{3/2}}}{A+Be^{-\frac43x^{3/2}}} $$