The <Riccati equation> $b_s+b^2=-4$ with $b(0)=0$ has solution
$$
b(s)=-2\tan(2s).
$$
Since $\lambda_s/\lambda=-b$, the initial condition $\lambda(0)=1$ gives $\lambda(s)=\sec(2s)$. Consequently
$$
t=\int_0^s\lambda(\sigma)^2d\sigma=\frac12\tan(2s),
\qquad
s=\frac12\arctan(2t).
$$
Finally $\gamma_s=-\omega$. In physical time the solution is therefore
$$
\lambda(t)=\sqrt{1+4t^2},
\qquad
b(t)=-4t,
\qquad
\gamma(t)=-\frac\omega2\arctan(2t).
$$

Solution

Past exam of the mathematics course of the University of Cambridge / 2023 / iii / Paper 154 / 1 / 6 / Solution

 Solution

ID: past-exam-of-the-mathematics-course-of-the-university-of-cambridge/2023/iii/paper-154/1/6/solution