How did we derive $\frac{d}{dx}(\frac{dy}{dt}) = \frac{1}{x}\frac{d^2y}{dt^2}$

Question

So I saw this differential equation $$x^2y^{\prime \prime} - xy^{\prime} + y = \ln(x)$$ let $x=e^t$ or $t = \ln x$ $$\frac{dy}{dx} =\frac{dy}{dt}\frac{dt}{dx}= \frac{1}{x}\frac{dy}{dt}$$ $$\frac{d^2y}{dx^2} = \frac{1}{x}\frac{d}{dx}\left(\frac{dy}{dt}\right) + \frac{dy}{dt}\left(-\frac{1}{x^2}\right)$$ $$\frac{d^2y}{dx^2} = \frac{1}{x}\left(\frac{d^2y}{dt^2}\frac{1}{x}\right) + \frac{dy}{dt}\left(-\frac{1}{x^2}\right)$$

so here how did we derive $\frac{d}{dx}(\frac{dy}{dt}) = \frac{1}{x}\frac{d^2y}{dt^2}$, even the textbook I used this example from has no derivation in it

Answer 1

This can easily be resolved by not using this terrible mess of notation which is unfortunately very common. So what is actually going on is we are considering two functions, namely $y$, which is the solution to the ODE, and $t$, defined by

$$t(x)=\ln x.$$

When we then write "$y(t)$", what we really mean is the function $y\circ t$. This is what's going on when we make a change of variables: we compose our function with another function. What the equation you're confused about is really saying then is that

$$(y'\circ t)'(x)=\frac{1}{x}(y''\circ t)(x)$$

(you could alternatively write $y''(t(x))$ instead of $(y''\circ t)(x)$ if that makes it clearer for you). But this them becomes a really easy to see consequence of the chain rule, which tells us that

$$(y'\circ t)'=t'\cdot(y''\circ t),$$

and so as

$$t'(x)=\frac{1}{x},$$

we get the result above.

Answer 2

Chain rule:

$$\frac{d\frac{dy}{dt}}{dx} = \frac{d\frac{dy}{dt}}{dt} \cdot \frac{dt}{dx} = \frac1x \frac{d^2y}{dt^2}$$