Given the system of equations $$\begin{cases} x'=x(x-1) \\ y'=-2xy+y \end{cases}$$ I am asked to calculate the general solution $\phi(t,0,x_0,y_0)$ that is, the general solution with initial conditions $\phi(0)=(x_0,y_0)$
Since I didn't know how to do it, I checked the solution and this was the first thing I saw:
$$\frac{x'}{x(x-1)}=1\implies \ln\bigg(\frac{|x-1|}{|x|}\bigg) - \ln\bigg(\frac{|x_0-1|}{|x_0|}\bigg)=t$$
Where does $$\ln\bigg(\frac{|x_0-1|}{|x_0|}\bigg)$$ come from?
I would have done $$\frac{x'}{x(x-1)}=1\implies \ln\bigg(\frac{|x-1|}{|x|}\bigg) - =t+c$$ and from here it's easy to find the explicit solution $x$.
I am studying dependence of solutions respect initial conditions and parameters of ordinary differential equations.
$$\frac{x'}{x(x-1)}=1$$ can be written as $$\frac{\frac{dx}{dt}}{x(x-1)}=1$$ or $$\frac{dx}{x(x-1)}=dt$$ We can separate the left hand side:$$\frac{dx}{x-1}-\frac{dx}x=dt$$ When you integrate bot sides, you get $$\ln|x-1|-\ln|x|=t+c$$ Finally, using the properties of the logarithm $$\ln\left|\frac{x-1}x\right|=t+c$$ Plugging in $t=0$ and $x=x_0$ yields $$\ln\left|\frac{x_0-1}{x_0}\right|=c$$ Therefore $$\ln\left|\frac{x-1}x\right|=\ln\left|\frac{x_0-1}{x_0}\right|+t$$