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

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.

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    $\begingroup$ If $x=\frac1t$, then $y(x)=y\left(\frac1t\right)$, and by the chain rule $\frac{\mathrm dy}{\mathrm dt}=\frac{\mathrm dy}{\mathrm dx}\frac{\mathrm dx}{\mathrm dt}$, or $\frac{\mathrm dy}{\mathrm dx}=-t^2\frac{\mathrm dy}{\mathrm dt}$. Do this a few more times to find a similar relation between the fourth-order derivatives. $\endgroup$ – user170231 Oct 15 '18 at 2:38
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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 $\frac 1 {x^2}=t^2$ $$y'_x=-t^2y'_t$$


$$y''_x=\frac d{dt}(-t^2y'_t)\frac {dt}{dx}$$ 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 $y''', y''''$


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 $$y'''_x=-(t^6y_t'''+6t^5y_t''+6t^4y_t' )$$

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    $\begingroup$ Can I ask you to verify y'''? I got $y'''_x = t^6 y'''_t - 6t^5 y''_t - 12t^4 y'_t$ $\endgroup$ – p3ngu1n Oct 15 '18 at 3:21
  • $\begingroup$ Be carefull because all the terms should be negative since you multiply by $-t^2$ @p3ngu1n $\endgroup$ – LostInSpace Oct 15 '18 at 3:55

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