Answer
See below
Work Step by Step
Given: $y'=y(y-2)(y-1)$
Rewrite as: $\frac{dy}{dx}=y(y-2)(y-1)\\\rightarrow \frac{dy}{y(y-2)(y-1)}=dx$
Integrate both sides:
$\int \frac{dy}{y(y-2)(y-1)}=\int dx$
To solve this: $\frac{dy}{y(y-2)(y-1)}=\frac{A}{y}+\frac{A}{y-2}+\frac{C}{y-1}=\frac{(A+B+C)y^2+(-3A-B-2C)y+2A}{y(y-2)(y-1)}$
We have the system: $A+B+C=0\\3A+B+2C=0\\2A=1$
then $A=\frac{1}{2}\\
B+C=-\frac{1}{2}\\ B+2C=-\frac{3}{2}\\
\rightarrow C=-1\\
\rightarrow B=\frac{1}{2}$
Hence, $\int \frac{dy}{y(y-2)(y-1)}=\int\frac{dy}{2y}+\int\frac{dy}{2(y-2)}-\int\frac{dy}{y-1}\\
=\frac{1}{2}\int\frac{dy}{y}+\frac{1}{2}\int \frac{dy}{y-2}-\int\frac{dy}{y-1}\\
=\frac{1}{2}\ln(y)+\frac{1}{2}\ln (y-2)-\ln(y-1)\\
=\frac{1}{2}(\ln(y)+\ln(y-2))-\ln(y-1))\\
=\frac{1}{2}\ln(y(y-2))-\ln(y-1)$
Solve this, we get $\frac{1}{2}\ln(y(y-2))-\ln(y-1)=x+c$
where $c_1=0\\c_2=2\\c_3=4\\c_4=8$
