First remove the small damping term by writing
$$
y=e^{-(1-e^{-2\varepsilon t})/4}v.
$$
The transformed <linear ordinary differential equation> is
$$
v_{tt}+\left[e^{-2\varepsilon t}+\varepsilon^2e^{-2\varepsilon t}-\frac{\varepsilon^2}{4}e^{-4\varepsilon t}\right]v=0.
$$
The leading frequency is $\omega=e^{-\varepsilon t}$. Its <WKB approximation> has amplitude $\omega^{-1/2}$ and phase $\int_0^t\omega\,ds=(1-e^{-\varepsilon t})/\varepsilon$. The <initial conditions> select
$$
\boxed{y_{\rm WKB}=e^{\varepsilon t/2-(1-e^{-2\varepsilon t})/4}\sin\left(\frac{1-e^{-\varepsilon t}}{\varepsilon}\right).}
$$
This leading expression has $y(0)=0$ and $y_t(0)=1$. Its <WKB approximation for a slowly varying oscillator> requires $|\omega_t|/\omega^2=\varepsilon e^{\varepsilon t}\ll1$, so it fails around $t_*=\varepsilon^{-1}\log(\varepsilon^{-1})$.

To resolve the <Bessel transition for an exponentially decaying oscillator>, shift to $T=\varepsilon t-\log(\varepsilon^{-1})$, put $Z=e^{-T}$ and rescale $y=\varepsilon^{-1/2}e^{-1/4}Y(T)$. The exact transformed equation is
$$
Y_{TT}+e^{-2T}Y+\varepsilon^2e^{-2T}Y_T=0.
$$
For fixed $T$ its leading form is $Y_{TT}+e^{-2T}Y=0$. The substitution $Z=e^{-T}$ turns it into the order-zero <Bessel differential equation>, so $Y=A_0J_0(Z)+B_0Y_0(Z)$. In the overlap $1\ll Z\ll\varepsilon^{-1}$, matching the large-argument <Bessel functions> to $Z^{-1/2}\sin(\varepsilon^{-1}-Z)$ gives, with $\theta=\varepsilon^{-1}-\pi/4$,
$$
\boxed{y\sim\varepsilon^{-1/2}e^{-1/4}\sqrt{\frac\pi2}\left[\sin\theta J_0(Z)-\cos\theta Y_0(Z)\right].}
$$
Here $J_0$ and $Y_0$ are the <Bessel function of the first kind> and <Bessel function of the second kind>; the symbol $Y_0(Z)$ in this question is unrelated to the leading inner function in the preceding question.

At late times $T\gg1$, the small-argument expansions give
$$
\boxed{y\sim\varepsilon^{-1/2}e^{-1/4}\sqrt{\frac\pi2}\left[\sin\theta+\frac2\pi\cos\theta\bigl(T+\log2-\gamma\bigr)\right],}
$$
where $\gamma$ is the <Euler--Mascheroni constant>. The solution becomes asymptotically linear rather than maintaining the exponentially growing WKB envelope. Its leading late-time slope is $\varepsilon^{1/2}e^{-1/4}\sqrt{2/\pi}\cos\theta$. These are leading asymptotic coefficients as $\varepsilon\to0$: near a zero of $\cos\theta$, higher-order phase corrections determine the small actual slope. The formula is not an absolute-error estimate uniform to arbitrarily late times. The <exact late slope of an exponentially damped oscillator>, obtained from a <Kummer function> and a <Wronskian>, provides a separate check even near those exceptional phases.

Solution

Past exam of the mathematics course of the University of Cambridge / 2018 / iii / Paper 336 / 3 / i / Solution

{title2=$\lim_{t\to\infty}y_t=e^{-1/2}M(-1/(2\varepsilon^2),1,1/2)$}

For $y_{tt}+e^{-2\varepsilon t}(\varepsilon y_t+y)=0$, $y(0)=0$, $y_t(0)=1$, set $r=e^{-2\varepsilon t}/2$. The exact equation becomes $ry_{rr}+(1-r)y_r+y/(2\varepsilon^2)=0$. Let $m(t)=M(-1/(2\varepsilon^2),1,r)$, using the <Kummer function>. It is regular at $r=0$ and tends to one. The <Wronskian> $W=my_t-m_ty$ obeys $W_t=-\varepsilon e^{-2\varepsilon t}W$, with $W(0)=M(-1/(2\varepsilon^2),1,1/2)$. Integration gives the displayed exact late slope. The other local solution grows at most logarithmically in $r$, so $m_ty\to0$ and $W\to y_t$. This provides a check on a <WKB approximation> even near phases where its leading predicted slope vanishes.

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