$$y'=y^2$$
This differential equation can be solved by dividing both sides with $y^2$ and integrating both sides to obtain $$y=\frac{1}{c_1-x}$$
But if I differentiate both sides instead, I get $$y''=2yy'=2y^3=2y'^\frac{3}{2}$$
This is a Bernoulli DE that can be solved with the substitution $u=y'^\frac{-1}{2}$ and results in
$$y'=\frac{1}{(-x+c_1)^2}$$
$$y=\frac{1}{c_1-x}+c_2$$
Is the method of differentiating both sides of the DE correct?
It resulted in an extra constant and I'm not sure what went wrong.