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

Separate variables away from the equilibrium solutions. Partial fractions giveThus the constant-coefficient Riccati equation has real solution familyon intervals where the denominator is nonzero. The value $K=0$ includes $y=a$. Setting $K=e^{(b-a)c}$ recovers the printed expression after factoring out $e^{ac}$. For real $c$ that substitution covers positive $K$ only; negative $K$ gives additional real branches between the equilibria, and the constant solutions must not be discarded when separating variables.

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.

When the two roots coincide, $y^{\prime }=-(y-a{)}^2$. Separation gives $y=a+1/(x+c)$, and the equilibrium $y=a$ must be added. Integrating $z^{\prime }=yz$ givesThe 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.

Differentiate $z^{\prime }=yz$ without dividing by $z$:Thus the linearization of a Riccati equation givesIts characteristic polynomial is $(\lambda -a)(\lambda -b)$, explaining both the distinct-root exponentials and the repeated-root factor $A+Bx$. Nonzero solutions reconstruct $y=z^{\prime }/z$ between their zeros; the identically zero $z$ also satisfies the linear equation.