Classify the points at 0 and $\infty$ of $x^7~\frac{d^4y}{dx^4}=y'$

Question

$$\displaystyle x^7~\frac{d^4y}{dx^4}=y'$$

I know that 0 is an irregular singular point. At $\infty$, I'm using the change of variable $x = \frac{1}{t}$ and I don't understand how to differentiate and do the substitution.

Answer 1

Use the chain rule

$$y'_x=\frac {dy}{dx}=\frac {dy}{dt}\frac {dt}{dx}$$ We have that $x=\frac 1t \implies t=\frac 1x \implies \frac {dt}{dx}=-\frac 1 {x^2}=-t^2$ $$\implies y'_x=y'_t\frac {dt}{dx}=-\frac 1 {x^2}y'_t$$ Substitute $\frac 1 {x^2}=t^2$ $$y'_x=-t^2y'_t$$

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$$y''_x=\frac d{dt}(-t^2y'_t)\frac {dt}{dx}$$ Since $\frac {dt}{dx}=-t^2$ $$y''_x=-t^2\frac d{dt}(-t^2y'_t)$$ $$y''_x=t^4y''_t+2t^3y'_t$$

Do the same for $y''', y''''$

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For y''' I got this $$y'''=\frac d{dt}(t^4y''_t+2t^3y'_t )\frac {dt}{dx}$$ Since $\frac {dt}{dx}=-t^2$ $$y_x'''=-t^2\frac d{dt}(t^4y_t''+2t^3y'_t )$$ $$y_x'''=-t^2(t^4y_t'''+4t^3y''_t+2t^3y''_t+6t^2y_t' )$$ Finally $$y'''_x=-(t^6y_t'''+6t^5y_t''+6t^4y_t' )$$