= Solution
For the printed family, put $D(x)=e^{a(x+c)}-e^{b(x+c)}$. Its <logarithmic derivative> is $D'/D=y$, so $z'=yz$ integrates to
$$
\boxed{z=C_1\bigl[e^{a(x+c)}-e^{b(x+c)}\bigr].}
$$
This includes $z=0$ and is valid on each interval on which the corresponding $y$ is finite. For the full real family, write $z=Ae^{ax}+Be^{bx}$ and $y=z'/z$, with $(A,B)\ne(0,0)$. The pure exponentials include the two equilibrium values of $y$; arbitrary relative coefficient signs include the branches missing from real $c$ in the printed expression.
Back to article page
