Suppose the given equation is: $$r^2\frac{\text d^2f}{\text{d}r^2}+r\frac{\text{d}f}{\text{d}r}+(\lambda r^2-m^2)f = 0$$
My text demonstrates the following:
Let $$\text{z = }\sqrt{\lambda }r$$
So that we may have
$$z^2\frac{\text d^2f}{\text{d}z^2}+z\frac{\text{d}f}{\text{d}z}+(z^2-m^2)f = 0$$
But this looks very strange for
$$z^2\text{$\neq $ }r^2$$
What am I missing here?
Similar to Fourier trigonometric series solution of the heat equation, the general solution for Bessel equation can be written in terms of series of Bessel functions.
The final solution will be written as an (infinite) sum over $m$ of Bessel Functions of the first kind and Bessel Functions of the second kind.
Now, if you are confused about change of variables, then consider following substitution: $$ z : =\sqrt{\lambda}r \implies r = \frac{z}{\sqrt{\lambda}}, \quad \frac{dz}{dr} = \sqrt{\lambda},\\ r \leftrightarrow z \iff f(r) \leftrightarrow f(z) \implies\\ \frac{d f}{dr} = \frac{d f}{dz} \frac{d z}{dr} = \sqrt{\lambda} \, \frac{d f}{dz}, \\ \frac{d^2 f}{dr^2} = \frac{d }{dr} \left( \frac{d f}{dr} \right) = \frac{d }{dr} \left( \sqrt{\lambda} \, \frac{d f}{dz} \right) = \frac{d }{dz} \left( \sqrt{\lambda} \, \frac{d f}{dz} \right) \cdot \frac{d z}{dr} = \lambda \frac{d^2 f}{dz^2} $$ Substituting everything into original equation, we get $$ r^2\frac{\text d^2f}{\text{d}r^2}+r\frac{\text{d}f}{\text{d}r}+\left(\lambda r^2-m^2\right)\,f = 0 \iff \\ \left( \frac{z}{\sqrt{\lambda}}\right)^2\frac{\text d^2f}{\text{d}r^2}+\frac{z}{\sqrt{\lambda}}\frac{\text{d}f}{\text{d}r}+\left(\lambda \left( \frac{z}{\sqrt{\lambda}}\right)^2-m^2\right)\,f = 0 \iff \\ \left( \frac{z}{\sqrt{\lambda}}\right)^2 \lambda \frac{\text d^2f}{\text{d}z^2}+\frac{z}{\sqrt{\lambda}}\sqrt{\lambda}\frac{\text{d}f}{\text{d}z}+\left(\lambda \left( \frac{z}{\sqrt{\lambda}}\right)^2-m^2\right)\,f = 0 \implies \\ z^2\frac{\text d^2f}{\text{d}z^2}+z\frac{\text{d}f}{\text{d}z}+\left( z^2-m^2\right)\,f = 0 . $$