Past exam of the mathematics course of the University of Cambridge / 2025 / ii / Paper 4 / 31D (Asymptotic Methods)

Near the endpoint $t=0$,The Watson lemma integrates this local expansion term by term, andTherefore the reciprocal-amplitude Laplace integral has expansionso

Differentiation under the integral sign givesHenceThus

After division by $x$, the equation isThe removal of the first derivative from a second-order differential equation usesWiththe normal-form coefficient isConsequentlyso the constants in the PDF's expression
$1+A{x}^{-1}+B{x}^{-2}$ are

To classify infinity, put $z=1/x$. Sincethe coefficient of $v$ in the transformed equation has a pole of order four at $z=0$. This exceeds the order two allowed at a regular singular point. Hence $x=+\infty$ is an irregular singular point at infinity.

Put $v=e^S$ and $p=S^{\prime }$. The Liouville-Green exponential ansatz turns the normal-form equation into the Riccati equationWriteMatching the constant, $x^{-1}$, and $x^{-2}$ coefficients gives

For the recessive branch,soSince $y=e^{x/2}{x}^{-1/2}v$, this givesThis agrees with the first two terms of part (a),Continuing the Riccati recursion reproduces the factorial asymptotic series, so $U$ is proportional to this recessive branch.

For the dominant branch,and in factsolves the phase equation exactly. It follows thatIndeed, $e^x$ is an exact second solution of the original differential equation.