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

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

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.

• 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. – user170231 Oct 15 '18 at 2:38

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

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