Solve\begin {align*} y^{\prime } & =y\left ( x-1\right ) \\ y\left ( 2\right ) & =0 \end {align*}
\(f=y\left ( x-1\right ) \) which is clearly continuous everywhere and so is \(f_{y}\). Hence it is guaranteed that solution exist and unique. Since \(y=0\) at initial conditions, then we can’t divide by \(y\) to separate. So we use the alternative method. At IC the ode itself becomes\[ y^{\prime }=0 \] Hence \[ y=c \] Since \(y\) is constant, then \(y=0\) because it can only have one value due to uniqueness. Therefore the solution is\[ y=0 \] Let now look at the general case to make things more clear.