Solution
= Solution
When the two roots coincide, $y'=-(y-a)^2$. Separation gives $y=a+1/(x+c)$, and the equilibrium $y=a$ must be added. Integrating $z'=yz$ gives
$$
\boxed{y=a+\frac1{x+c},\quad z=C_1(x+c)e^{ax};\qquad\text{or}\qquad y=a,\quad z=C_1e^{ax}.}
$$
The nonconstant pair is restricted to an interval excluding $x=-c$. Equivalently every nonzero linearizing function is $z=(A+Bx)e^{ax}$, with $y=a+B/(A+Bx)$ wherever its denominator is nonzero.