I'm investigating how you get from this:
$$ z\frac{d^2y}{dz^2}+(1-a)\frac{dy}{dz}+a^2z^{2a-1}y=0 $$
to this:
$$ \frac{d^2y}{dx^2}+y=0 $$
with a change of variable:
$$ z=x^{1/a}, (x\ge0) $$
I'm not quite 'getting' the substitution in the derivative...
In a normal situation where we have $ \frac{dy}{dx} $ and a function $ y(x) $, we basically perform $ \frac{d}{dx}y(x) $ to see how $ y $ changes with respect to $ x $.
Now, with the variable substitution are we now saying that $ \frac{dy}{dz} $ is changed to $ \frac{d}{dz}y(z) = \frac{d}{dx}y(x^{1/a}) $.
This leaves me with a derivative that can be solved using the Chain Rule?
So, $$ \frac{d}{dx}y(x^{1/a}) = \frac{dy}{dx}\frac{x^{\frac{1}{a}-1}}{a}$$
I would then just proceed to take the second derivative and substitute $ \frac{dy}{dx} $ , $ \frac{d^2y}{dx^2} $ and $ y $ into the equation original equation above.
Is my understanding correct on this please? Thankyou.
Note, that by the chain rule, we have $$ \frac{dy}{dz} = \frac{dy}{dx} \cdot \frac{dx}{dz} $$ and, taking another derivative, by product and chain rules, $$ \frac{d^2y}{dz^2} = \frac{d^2 y}{dx^2} \cdot \left(\frac{dx}{dz}\right)^2 + \frac{dy}{dx} \cdot \frac{d^2 x}{dz^2} $$ As $x = z^a$, we have $$ \frac{dx}{dz} = az^{a-1}, \quad \frac{d^2 x}{dz^2} = a(a-1)z^{a-2} $$ that is \begin{align*} \frac{dy}{dz} &= \frac{dy}{dx} \cdot az^{a-1}\\ \frac{d^2y}{dz^2} &= a^2z^{2(a-1)} \frac{d^2y}{dx^2} + a(a-1)z^{a-2} \frac{dy}{dx}\\ \end{align*} Giving \begin{align*} z\frac{d^2y}{dz^2} + (1-a)\frac{dy}{dz} + a^2z^{2a-1}y &= a^2z^{2a-1}\frac{dy^2}{dx^2} + a(a-1)z^{a-1}\frac{dy}{dx} - a(a-1)z^{a-1}\frac{dy}{dx} + a^2z^{2a-1}y\\ &= a^2z^{2a-1}\left(\frac{d^2y}{dx^2} + y\right) \end{align*}