Use the fast time $t$ and slow time $T=\epsilon^2t$, and write
$$
u=u_0+\epsilon u_1+\epsilon^2u_2+\cdots,
\qquad
u_0=A(T)e^{it}+\overline A(T)e^{-it}.
$$
There is no $O(\epsilon)$ resonant forcing. Solving
$$
(\partial_t^2+1)u_1=-\sin t\,u_0
$$
gives a convenient particular solution
$$
u_1=-\frac{iA}{6}e^{2it}
+\frac{i\overline A}{6}e^{-2it}
+\frac i2(\overline A-A).
$$
At $O(\epsilon^2)$, the coefficient of $e^{it}$ in the forcing must vanish by the <solvability condition in the method of multiple scales>. This gives
$$
-2iA_T-fA+\frac16A-\frac14\overline A=0,
$$
or
$$
A_T=-\frac i2\left(\frac16-f\right)A
+\frac i8\overline A.
$$
The two slow exponents satisfy
$$
s^2=\frac1{64}-\frac14\left(\frac16-f\right)^2.
$$
The <second instability tongue of a weak Mathieu oscillator> has real $s$ when $-1/12<f<5/12$. At either endpoint the repeated zero exponent permits a linearly growing slow solution, so boundedness for every initial condition requires the strict stable ranges
$$
\boxed{f<-\frac1{12}\quad\hbox{or}\quad f>\frac5{12}.}
$$

Solution

Past exam of the mathematics course of the University of Cambridge / 2025 / iii / Paper 336 / 2 / a / Solution

For $u''+[1+\epsilon\sin t+\epsilon^2f]u=0$, the second parametric-instability tongue is $-1/12<f<5/12$ to leading order. Outside its closure every leading slow amplitude is oscillatory.

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