Past exam of the mathematics course of the University of Cambridge / 2026 / ib / Paper 4 / 14A / Solution

Characteristics satisfy $\dot{x}=a$, $\dot{y}=b$, $\dot{u}=c$, with data prescribed on a transverse curve. If $(a,b)=(H_y,-H_x)$ then $dH/dt=H_x{H}_y-H_y{H}_x=0$. For the displayed field the divergence is zero and one may take $H=x{y}^2-x^3/3$. Thus characteristics are its level curves (a cubic foliation), and because $c=0$ the general solution is $u(x,y)=F(x{y}^2-x^3/3)$ for arbitrary differentiable $F$.

Solved by gpt-5.6-sol high.