<s>[INST] Solving a second order nonlinear differential equation 
Solve $$y’’ + 2y = 4 -\frac{1}{\sqrt{y}}$$ with initial conditions $y(0) = 1$ and $y’(0) = 2$.

I am trying different substitutions, but none seem to work.
 [/INST] $$y'' + 2y - 4 +\frac{1}{\sqrt{y}}=0$$
$$2y''y'+4yy'-8y'+\frac{2y'}{\sqrt{y}}=0$$
$$(y')^2+2y^2-8y+4\sqrt{y}=c_1$$
$y(0)=1$ and $y'(0)=2 \quad;\quad 4+2-8+4=c_1=2$
$$y'=\pm\sqrt{-2y^2+8y-4\sqrt{y}+2}$$
$$dx=\pm\frac{dy}{\sqrt{-2y^2+8y-4\sqrt{y}+2}}$$
$$x=\pm\int \frac{dy}{\sqrt{-2y^2+8y-4\sqrt{y}+2}}+c_2$$
$y(0)=1$
$$x=\pm\int_1^y \frac{d\xi}{\sqrt{-2\xi^2+8\xi-4\sqrt\xi{}+2}}$$
$$x=\pm\int_1^{\sqrt{y}} \frac{2\zeta\:d\zeta}{\sqrt{-2\zeta^4+8\zeta^2-4\zeta+2}}$$
This is the solution for the inverse function $x(y)$.
Further calculus appears arduous, involving elliptic integral and inverse.
Better use numerical calculus directly to solve the original ODE.
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