# If $y(z) = C y_0(z) \int_w^z \frac{d\zeta}{y_0(\zeta)^2}$, what limit can we take in $C$ and $w$ to obtain $y(z) \to y_0(z)$?

This is Exercise 6.5 from Miller's Applied Asymptotic Analysis.

The book shows that, given any solution $y_0(z)$ to the equation

$$y''(z)+f(z)y(z)=0, \tag{1}$$

a general solution is given by

$$y(z) = C y_0(z) \int_w^z \frac{d\zeta}{y_0(\zeta)^2}, \tag{2}$$

where $w$ and $C$ are constants.

One solution of $(1)$ which is not represented by $(2)$ for any admissible choice of the constants $C$ and $w$ is $y_0(z)$ itself. On the other hand, $y_0(z)$ can always be obtained from $(2)$ by taking an appropriate limit in the space of parameters $C$ and $w$. Suppose that $y_0(z)$ is a nontrivial solution that vanishes at an ordinary point $z_0$ of the differential equation, and find an appropriate limit, showing how this limit results in the original solution $y_0(z)$.

Hint: Recall that if $z_0$ is an ordinary point of the differential equation and if $y_0(z)$ is a nontrivial solution satisfying $y_0(z_0) = 0$, then $y_0'(z_0) \neq 0$.

Naively it seems like I should be able to take the limit as $(C,w) \to (0,z_0)$ along the path

$$C = \left(\int_w^z \frac{d\zeta}{y_0(\zeta)^2}\right)^{-1},$$

but this doesn't take advantage of the fact that $y_0'(z_0) \neq 0$, so this is probably not what the question is asking for (perhaps the limit should be independent of the path).

I really don't know what I should be looking at. I tried taking the limit as $z \to z_0$ and found that

$$\lim_{z \to z_0} y(z) = - C/y_0'(z_0)$$

and

$$\lim_{z \to z_0} y'(z) = C \lim_{z \to z_0} \left(y_0'(z) \int_w^z \frac{d\zeta}{y_0(\zeta)^2} + \frac{1}{y_0(z)}\right),$$

but I can't really see anything more to deduce about $C$ and $w$ from here.

Any further hints would be appreciated.

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