Past exam of the mathematics course of the year 1 of the natural sciences course of the University of Cambridge / 2006 / Paper 2 / 9 / b / Solution

For the printed family, put $D(x)=e^{a(x+c)}-e^{b(x+c)}$. Its logarithmic derivative is $D^{\prime }/D=y$, so $z^{\prime }=yz$ integrates toThis includes $z=0$ and is valid on each interval on which the corresponding $y$ is finite. For the full real family, write $z=A{e}^{ax}+B{e}^{bx}$ and $y=z^{\prime }/z$, 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.