Change of Variables in differential equation Given the equation $zZ''(z) + Z'(z) + \alpha^2Z(z) = 0$ use the change of variables $x = \sqrt{\frac{z}{a}}$ where $a$ is a constant to map the problem to the differential equation $Z''(x) + \frac{1}{x}Z'(x) + \gamma^2 Z(x) = 0,$ where $\gamma = 2\alpha \sqrt{a}$
Attempt:
I think I am struggling to get the chain rule correct for this particular case. Here is my attempt:  $$\frac{d}{dz} = \frac{d}{dx} \frac{dx}{dz} = \frac{1}{2\sqrt{za}}\frac{d}{dx} = \frac{1}{2xa} \frac{d}{dx}$$ This means by a similar argument $$\frac{d^2}{dz^2} = \frac{1}{4x^2a^2}\frac{d^2}{dx^2}$$
I am not really sure where to go from here. 
Thanks.

$$ \frac{dZ}{dz} = \frac{dZ}{dx}\frac{dx}{dz}=\frac{dZ}{dx}\frac{1}{2ax}$$
$$ \frac{d^{2}Z}{dz^{2}}=\frac{d}{dz}[\frac{dZ}{dx}\frac{1}{2ax}], $$
but $dz=2axdx$, so
$$ \frac{d}{dz}[\frac{dZ}{dx}\frac{1}{2ax}] = \frac{1}{2ax}\frac{d}{dx}[\frac{dZ}{dx}\frac{1}{2ax}] = \frac{1}{4a^{2}x^{2}} \frac{d^{2}Z}{dx^{2}} -\frac{1}{4a^{2}x^{3}} \frac{dZ}{dx}. $$
Now 
$$ z\frac{d^{2}Z}{dz^{2}} + \frac{dZ}{dz} + \alpha^{2}Z =0$$ becomes
$$ \frac{d^{2}Z}{dx^{2}} + \frac{1}{x}\frac{dZ}{dx} + 4a\alpha^{2}Z=0. $$
Cheers,
Paul Safier
