We have the differential equation $y''+ \frac{p}{x} y'+ \frac{q}{x^2}y=0, x>0$ and we set $z=\log x$.

Then $y'=\frac{dy}{dx}=\frac{dy}{dz} \frac{dz}{dx}=\frac{1}{x} \frac{dy}{dz}$

$y''=\frac{d^2y}{dx^2}=\frac{d}{dx}\left( \frac{dy}{dx} \right)=\frac{d}{dx}\left( \frac{1}{x} \frac{dy}{dz} \right)=-\frac{1}{x^2}\frac{dy}{dx}+\frac{1}{x} \frac{d}{dx}\left( \frac{dy}{dz}\right)$

How can we find $\frac{d}{dx}\left( \frac{dy}{dx}\right)$?

We have the differential equation $y''+ \frac{p}{x} y'+ \frac{q}{x^2}y=0, x>0$ and we set $z=\log x$.

Then $y'=\frac{dy}{dx}=\frac{dy}{dz} \frac{dz}{dx}=\frac{1}{x} \frac{dy}{dz}$

$y''=\frac{d^2y}{dx^2}=\frac{d}{dx}\left( \frac{dy}{dx} \right)=\frac{d}{dx}\left( \frac{1}{x} \frac{dy}{dz} \right)=-\frac{1}{x^2}\frac{dy}{dx}+\frac{1}{x} \frac{d}{dx}\left( \frac{dy}{dz}\right)$

How can we find $\frac{d}{dx}\left( \frac{dy}{dx}\right)$?
@DanielFischer @quid Do you maybe have an idea?

We have the differential equation $y''+ \frac{p}{x} y'+ \frac{q}{x^2}y=0, x>0$ and we set $z=\log x$.

Then $y'=\frac{dy}{dx}=\frac{dy}{dz} \frac{dz}{dx}=\frac{1}{x} \frac{dy}{dz}$

$y''=\frac{d^2y}{dx^2}=\frac{d}{dx}\left( \frac{dy}{dx} \right)=\frac{d}{dx}\left( \frac{1}{x} \frac{dy}{dz} \right)=-\frac{1}{x^2}\frac{dy}{dx}+\frac{1}{x} \frac{d}{dx}\left( \frac{dy}{dz}\right)$

How can we find $\frac{d}{dx}\left( \frac{dy}{dx}\right)$?

I thought that it would be as follows:
$y_1(z)=e^{r_1 z} \Rightarrow y_1(\log x)=e^{r_1 \log x} \Rightarrow y_1(\log{e^x})=e^{r_1 \log{e^x}} \Rightarrow y_1(x)=e^{r_1 x}$.
Where am I wrong?

Given that I know that $f'(x) = f(x)$ and that $f(0) = 1$, how do I prove that
$$ f(1) = e = \lim_{n \to \infty} \left( 1 + \frac{1}{n}\right)^n $$?