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:
$$\frac{dy}{dx}=\frac{dy}{dr}\frac{dr}{dx}$$
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$\begingroup$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 ---- $$\frac{d}{dx}\bigg(\frac{dy}{dx}\bigg)=\frac{d}{dx}\bigg(\frac{dy}{dr}\frac{dr}{dx}\bigg)$$ $$\frac{d^2y}{dx^2}=\frac{d}{dx}\bigg(\frac{dy}{dr}\bigg)\frac{dr}{dx}+\frac{dy}{dr}\frac{d}{dx}\bigg(\frac{dr}{dx}\bigg)$$ $$\frac{d^2y}{dx^2}=\frac{d^2y}{dr^2}\frac{dr}{dx}\frac{dr}{dx}+\frac{dy}{dr}\frac{d^2r}{dx^2}$$
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