I'm solving $y''=2y(y')$
So using the substitution $P=y'$ and $P\frac{dP}{dy}=y''$
That yields to a separable equation then by integrating i end up with another separable equation,
$$\frac{dy}{dx}=y^2+C_1$$
So by integrating again my final answer is,
$$\frac1{C_1}\arctan\left(\frac{y}{C_1}\right)+C_2=x$$
But the final answer should be, $\arctan\left(\frac{y}{C_1}\right)+C2=x$
Can I combine the $1/C1$ to the $C1$ at at the $\arctan$?
The given ODE implies that $$(y'-y^2)'=y''-2yy'=0\ .$$ It follows that $$y'=y^2+C$$ for a constant $C\in{\mathbb R}$. We have to distinguish three cases:
(i)$\quad C=0\ :\quad$ The ODE $y'=y^2$ has the "ordinary" solutions $$y(x)={1\over c-x}\qquad (c\in{\mathbb R})$$ and the special solution $y(x)\equiv0$.
(ii)$\quad C=\lambda^2>0\ :\quad$ The ODE $y'=y^2+\lambda^2$ is solved by the functions $$y(x)=\lambda\tan\bigl(\lambda(x-c)\bigr)\qquad(c\in{\mathbb R})\ .$$ (iii)$\quad C=-\lambda^2<0\ :\quad$ The ODE $y'=y^2-\lambda^2$ has the two constant solutions $y(x)=\pm\lambda$ whose graph partitions the $(x,y)$-plane into three zones. The solutions in the central zone $-\lambda<y<\lambda$ are given by $$y(x)=\lambda\tanh\bigl(\lambda(c-x)\bigr)\qquad(c\in{\mathbb R})\ .$$ I leave the two other zones to you. You will obtain $\coth$ functions there.