For Bessel function $$R_{rr}+\frac1r R_r+(\mu-\frac{n^2}{r^r})R=0$$
We want to transform it into a standard form. Let $t=\sqrt{\mu}r$ , then $$R_r=R_t \frac{dt}{dr}=\sqrt{\mu}R_t$$ and similarily $$R_{rr}=\mu R_{tt}$$ So, we have $$R_{tt}+\frac1t R_t+(1-\frac{n^2}{t^2})R=0$$
A solution of the last equation is $$R(t)=J_n(t)$$ So I think a solution of the first equation should be $$R(\sqrt{\mu}r)=J_n(\sqrt{\mu}r)$$ Which is not the case since the original solution is $$R(r)=J_n(\sqrt{\mu}r)$$
I know there's something wrong above , but I can't see where did I do wrong . Let's consider a simple case.
$\frac{dy}{dx}=2x$ and $2x=t$ , then we have $$t=2x=\frac{dy}{dx}=2\frac{dy}{dt}$$
So a solution of the above equation should be $y(t)=\frac{t^2}{4}$ , then we have $$y(x)=\frac{x^2}{4}$$
However , assume the equation we get was $y(x)=\frac{t^2(x)}{4}$ , then we get $$y(x)=x^2 \,\,\,\,\,!$$ which is the right solution . I know there are some other approaches such as let $\hat{R}(t)=R(\frac{t}{\sqrt{\mu}})$ , but I need some help to show why the approach above seems wrong and how to correct it . Any help would be very apprreciate !