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Let us say we have the following ODE: $$x^7\frac{d^4y}{dx^4} - y' = 0$$

To classify the singular point at $\infty$, we need of course to make a substitution: $x = \frac{1}{t}$, and then evaluate the point $t = 0$. My question is: how can we transform this differential equations in terms on $x$ and $t$? It seems like a very elaborate computational task. Is there an easy way to do this???

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I think there is no other way around it but to use the chain rule: \begin{align} \frac{d y}{d x} &= \frac{d y}{d t}\frac{d t}{d x} = -\frac{1}{x^2} \frac{d y}{d t} = -t^2 \frac{d y}{d t}\\ \frac{d^2 y}{d x^2} &= \frac{d}{d t} \left(\frac{d y}{d x}\right) \frac{d t}{d x} = -t^2 \frac{d}{d t}\left(-t^2 \frac{d y}{d t}\right) =2 t^3 \frac{d y}{dt} + t^4 \frac{d^2 y}{d t^2}\\ \frac{d^3 y}{d x^3} &= -t^2 \frac{d}{d t}\left(2 t^3 \frac{d y}{dt} + t^4 \frac{d^2 y}{d t^2}\right) = -6 t^4 \frac{d y}{d t} -6 t^5 \frac{d^2 y}{d t^2} - t^3 \frac{d^3 y}{d t^3}\\ \frac{d^4 y}{d x^4} &= -t^2 \frac{d}{dt} \left(-6 t^4 \frac{d y}{d t} -6 t^5 \frac{d^2 y}{d t^2} - t^3 \frac{d^3 y}{d t^3}\right)\\ &= 24t^5 \frac{d y}{d t} + 36 t^6 \frac{d^2 y}{d t^2} + 12 t^7 \frac{d^3 y}{d t^3} + t^8 \frac{d^4 y}{d t^4} \end{align}

I didn't doublecheck my calculations, and I can't vouch I didn't made a mistake. The conceptual work is correct though. You should do the math yourself.

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