Let $y(x, \lambda)$ denote the solution for $0 \leqslant x<\infty$ of

$$\frac{d^{2} y}{d x^{2}}-\left(x+\lambda^{2}\right) y=0$$

subject to the conditions that $y(0, \lambda)=a$ and $y(x, \lambda) \rightarrow 0$ as $x \rightarrow \infty$, where $a>0$; it may be assumed that $y(x, \lambda)>0$ for $x>0$. Write $y(x, \lambda)$ in the form

$$y(x, \lambda)=\exp (z(x, \lambda))$$

and consider an asymptotic expansion of the form

$$z(x, \lambda) \sim \sum_{n=0}^{\infty} \lambda^{1-n} \phi_{n}(x),$$

valid in the limit $\lambda \rightarrow \infty$ with $x=O(1)$. Find $\phi_{0}(x), \phi_{1}(x), \phi_{2}(x)$ and $\phi_{3}(x)$.

It is known that the solution $y(x, \lambda)$ is of the form

$$y(x, \lambda)=c Y(X)$$

where

$$X=x+\lambda^{2}$$

and the constant factor $c$ depends on $\lambda$. By letting $Y(X)=\exp (Z(X))$, show that the expression

$$Z(X)=-\frac{2}{3} X^{3 / 2}-\frac{1}{4} \ln X$$

satisfies the relevant differential equation with an error of $O\left(1 / X^{3 / 2}\right)$ as $X \rightarrow \infty$. Comment on the relationship between your answers for $z(x, \lambda)$ and $Z(X)$.