How to solve $\frac{d^{2}y}{dx^2} + \frac{1}{x} \frac{dy }{dx} =0$? Let $y\in C^{2}(\mathbb R)$ (twice continuously differntiable function). We consider the ODE as follows:
$$\frac{d^{2}y}{dx^2} + \frac{1}{x} \frac{dy }{dx} =0$$
($y$ is function of $x$)
My naive  question is: (1) How to solve the above ODE?  (2) How to make this question rigorous? (What can we say about solution, etc....)   
 A: Write the ODE as
$$xy''+y'=0$$
Where prime dentoes derivative wrt to $x$. Then notice the left hand side is simply the derivative of a product, so
$$(xy')'=0$$
By implication therefore
$$xy'=C$$
Where $C$ is a constant; thus
$$dy=\frac{C dx}{x}$$
A: HINT:
Try to make change of variables 
$$
\begin{aligned}
u = \frac{dy}{dx} & \implies \frac{du}{dx}=\frac{d^{2}y}{dx^{2}} \implies 
\boxed{\frac{d^{2}y}{dx^2} + \frac{1}{x} \frac{dy }{dx} =0 \iff
\frac{d u}{dx} + \frac{1}{x} u =0}
\end{aligned}
$$
Note that the resulting differential equation 
$$
\frac{d u}{dx} + \frac{1}{x} u = 0
$$
is separable, so you can easily integrate it and find its solution $u = u(x)$, and consequently $ y = y(x)$.
A: First note that $\color{blue}{y_1=1}$ is a solution, if you are looking for a second solution $y_2$ for an homogeneous ODE $$y''+p(x)y'+q(x)=0...(1)$$ you can use the formula
$$y_2=y_1\int\frac{e^{-\int p(x)dx}}{y_1^2}dx$$
The general solution for $(1)$ can be written as $y(x)=c_1y_1+c_2y_2$ where $c_1$ and $c_2$ are constants.
A: In general this is the Cauchy equation. The general method is to use the transformation $y=x^m$ for some integer $m$. Though for the present case if you simply take $p=dy/dx$ you'll get $$d(\ln p)/dx+1/x=0$$ which can be easily solved to obtain $p x=C\implies dy=C\frac{dx}{x}\implies y=C\ln x+D$.
