Set up a differential equation where $y$ depends on $x$ and $x = e^u$ The question goes likes this: Show that if $y$ depends on $x$ and $x = e^u$, then $x^2\frac{d^2y}{dx^2} = \frac{d^2y}{du^2} - \frac{dy}{du} $
How I started was: $\frac{dy}{dx} = \frac{dy}{du} \frac{du}{dx} $ and saying if $x=e^u$ , then  $lnx = u$ , hence $\frac{du}{dx}$ is $\frac{1}{x}$. 
$\frac{dy}{dx} = \frac{dy}{du} \frac{1}{x} $ 
$x\frac{dy}{dx} = \frac{dy}{du}$
I am stuck after this step. It seems like I should differentiate with respect to $u$, but I am unsure how to. Thanks in advance!
 A: $$\frac{dy}{dx}=\frac{dy}{du}\frac{du}{dx}=\frac{1}{x}\frac{dy}{du}$$
You got this already.
Now take another derivative with respect to $x$. Notice that you have a product on the right side:
$$\frac{d}{dx}\frac{dy}{dx}=\left(\frac{d}{dx}\frac{1}{x}\right)\frac{dy}{du}+
\frac{1}{x}\left(\frac{d}{dx}\frac{dy}{du}\right)
=-\frac{1}{x^2}\frac{dy}{du}+
\frac{1}{x}\left(\frac{du}{dx}\frac{d}{du}\frac{dy}{du}\right)=$$
$$\frac{d^2y}{dx^2}=-\frac{1}{x^2}\frac{dy}{du}+
\frac{1}{x}\left(\frac{1}{x}\frac{d^2y}{du^2}\right)
$$
Take $x^2$ to the other side and you are done.
A: Taking the second derivative of $y$ with respect to $x$ yields:
$$\frac{d^y}{dx^2}=\frac{d}{dx}\left(\frac{1}{x}\frac{dy}{du}\right)=\frac{1}{x^2}\frac{dy}{du}+\frac{d\frac{dy}{du}}{dx}=\frac{1}{x}\frac{du}{dx}\frac{d}{du}\frac{dy}{du}=-\frac{1}{x^2}\frac{dy}{du}+\frac{1}{x^2}\frac{d^2y}{du^2}$$
A: We have
\begin{eqnarray}
x^2\frac{d^2y}{dx^2}&=&e^{2u}\frac{du}{dx}\cdot\frac{d}{du}\left(\frac{du}{dx}\cdot\frac{dy}{du}\right)=e^{2u}e^{-u}\frac{d}{du}\left(e^{-u}\frac{dy}{du}\right)=e^u\left(-e^{-u}\frac{dy}{du}+e^{-u}\frac{d^2y}{du^2}\right)\\
&=&\frac{d^2y}{du^2}-\frac{dy}{du}.
\end{eqnarray}
