# Irregular singular point of a DE

In my notes it said that ${d^2y\over dx^2} -xy=0$ has an irregular singular point at $x=\infty$.

It was illustrated by doing a transformation $y(x)=w(t)$ with $t=\frac{1}{x}$. But I don't understand, how did they get

${dy\over dx} =-t^2 {dw\over dt}$ and ${d^2y\over dx^2}=t^4 {d^2w\over dt^2}+2t^3 {dw\over dt}$?

And how did they get

${d^2w\over dt^2} + \frac{2}{t}{dw\over dt}+\frac{1}{t^5}w=0$?

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They used the chain rule. $\frac{\mathrm dy}{\mathrm dx}=\frac{\frac{\mathrm dw}{\mathrm dt}}{\frac{\mathrm dx}{\mathrm dt}}$. –  Guess who it is. May 12 '12 at 16:54
...and that particular equation you're looking at is Airy's equation. –  Guess who it is. May 12 '12 at 16:55

Before we start, let's use $y'', y', w', w''$, etc. just for efficiency's sake.
So, you know that the solution to the differential equation $y'' - xy = 0$ comes with two linearly independent solutions called the Airy $Ai(x)$ and the Airy $Bi(x)$ functions.
We are told that the substitution is $y(x) = w(x)$. Making this substitution, find $y'$ using the chain rule $dy/dx = (dw/dt)(dx/dt)$. This is the same as $y' = x'w'$ which gives us $y' = (-t^2)w'$. Notice that we want to get them equal, which is why the function becomes $-t^2$ and not $-1/t^2$. Now, notice that for example when $y=uv$, where $u,v$ are both functions, by chain rule we have $y'=u'v + v'u$. In the same manner, $y'=(-1/t^2)w'$ becomes $y''=(-1/t^2)'w' + (-1/t^2)w''$. Substituting to maintain equivalence, we see that $y'' = t^4w'' + 2t^3w'$ (check yourself).
Substitute the newly found equivalencies into the original form of $y''-xy = 0$ and you will see that $w'' + (2/t)w' + (1/t^5)w = 0$.
Now, you can surely see that when $x=∞$, t becomes 0. As you probably know well, $P(x)(x-x_0)$ and $Q(x)(x-x_0)^2$ must be singular in order to be irregular singular. Well, once you plug in $t=0$ you can surely see that $Q(x)$ part is still singular, and therefore this function is irregular at $x=∞$