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.

$$y’ = -\frac{1}{y} + 2$$
$$y’ = -\frac{1}{y} + \dfrac {2y}y$$
$$y’ = \frac{2y-1}{y}$$
$$\dfrac {y}{2y-1}dy = dx$$
Then integrate both sides.
$$\int \dfrac {y \, dy}{2y-1} = \int dx$$
Substitute $u=2y-1 \implies du=2dy$
and $y=\dfrac {u+1}2$
$$\dfrac 14 \int \dfrac {u+1}{u}du = \int dx$$
$$\dfrac 14 \int \left (1+\dfrac {1}{u} \right ) du = \int dx$$
