TITLE: Differential equation change of variables
QUESTION [7 upvotes]: I have a differential equation $$xy''(x) +(n+1-x)y'(x) + ay(x)=0.$$
If I set $x=r^t$ then how to plug in this and how to use change of variable to get the differential equation for $r$ instead of $x,$ i.e. the following equation:
$$\frac{dy}{dx}=\frac{dy}{dr}\frac{dr}{dx}$$

REPLY [13 votes]: By chain rule
$$\frac{dy}{dx}=\frac{dy}{dr}\frac{dr}{dx}$$
$$\frac{d^2y}{dx^2}=\frac{d^2y}{dr^2}\bigg(\frac{dr}{dx}\bigg)^2+\frac{dy}{dr}\frac{d^2r}{dx^2}$$
where
$$x=r^t\Rightarrow dx=tr^{t-1}dr\Rightarrow\frac{dr}{dx}=\frac 1{tr^{t-1}}$$
$$dx^2=t(t-1)r^{t-2}dr^2\Rightarrow\frac{d^2r}{dx^2}=\frac 1{t(t-1)r^{t-2}}$$
By plugging into original equation
$$r^t \Bigg(\frac{d^2y}{dr^2}\bigg(\frac 1{tr^{t-1}}\bigg)^2+\frac{dy}{dr}\frac 1{t(t-1)r^{t-2}} \Bigg)+(n+1-r^t)\frac{dy}{dr}\frac 1{tr^{t-1}} + ay=0$$
---- Addition for chain rule ----
$$\frac{d}{dx}\bigg(\frac{dy}{dx}\bigg)=\frac{d}{dx}\bigg(\frac{dy}{dr}\frac{dr}{dx}\bigg)$$
$$\frac{d^2y}{dx^2}=\frac{d}{dx}\bigg(\frac{dy}{dr}\bigg)\frac{dr}{dx}+\frac{dy}{dr}\frac{d}{dx}\bigg(\frac{dr}{dx}\bigg)$$
$$\frac{d^2y}{dx^2}=\frac{d^2y}{dr^2}\frac{dr}{dx}\frac{dr}{dx}+\frac{dy}{dr}\frac{d^2r}{dx^2}$$