Differentiate a substitution before replacing every dependent term

$dy/dx=dz/dx-1$. Since the original equation requires $y\ne0$, substitution gives $[dz/dx-1]/(z-x)-1=(x-2)/(z-x)$. Multiplying by $z-x$ gives $dz/dx-1-(z-x)=x-2$, hence $dz/dx=z-1$.

For $z\ne1$, $\ln|z-1|=x+A$, hence $z-1=\pm e^Ae^x=Be^x$. The equilibrium $z=1$ is restored by allowing $B=0$. Thus $z=1+Be^x$, B real.

Replacing $z=y+x$ gives $y=1-x+Be^x$. This is the general family on intervals where $1-x+Be^x\ne0$, as required by the original denominators. The teacher contrasts this general separable solution with the particular direct-integration solution in Example 6 and the particular substituted solution in Example 7.