Solve $ y^2y’’ = y’, y(0) = 1, y’(0) = 1 $ Solve the ODE with the initial conditions:
$$
y^2y’’ = y’, y(0) = 1, y’(0) = 1
$$

I did the substitution:
$$
y’ = z 
$$
$$
y’’ = z’ = \frac{dz}{dx} = \frac{dz}{dy}\frac{dy}{dx} = z \frac{dz}{dy}
$$
Putting in the ODE:
$$
y^2z \frac{dz}{dy} = z \Rightarrow y^2 \frac{dz}{dy} = 1 \Rightarrow \frac{dz}{dy} = \frac{1}{y^2} \Rightarrow dz = \frac{1}{y^2}dy
$$
By integration we get:
$$
z = -\frac{1}{y} + C \Rightarrow y’ = - \frac{1}{y} + C
$$
Using the initial conditions we get:
$$
1 = -\frac{1}{1} + C \Rightarrow C = 2
$$
Therefore:
$$
y’ = -\frac{1}{y} + 2
$$
Now I am not sure how to continue.

$y^2y''=y'$
Write as
$y''=\frac{y'}{y^2}$
Integrate both sides
$y'=-1/y +C$
$y'(0)=1,y(0)=1\to C=2$
$y'=-1/y+2$
$y=\frac{1}{2} \left(W\left(e^{4 x+1}\right)+1\right)$
Where $W(z)$ is Lambert $W$ function which solves $we^w=z$
The implicit form is
$x=\frac{1}{4} (2 y+\log (2 y-1)-1)$
