How to solve $ \frac{d^2y}{dx^2} - \left(1/x\frac{dy}{dx}\right) = 0$ I'm trying to solve:

$ \frac{d^2y}{dx^2} - \left(1/x\frac{dy}{dx}\right) = 0$

I've been following a tutorial and they say the answer is:

$y = \frac{1}{2}Cx^2 + D$

They've used:

$\frac{dy}{dx} = Cx$

How did they get that result??
 A: I think that the equation read as follows:
$\frac{d^2 y}{dx^2} - \left(1/x\frac{dy}{dx}\right) = 0$. Put $z: =y'$ then we have
$z'=\frac{1}{x}z$. The last equation has the general solution $z(x)=Cx$. Hence $y'(x)=Cx$, thus $y(x) = \frac{1}{2}Cx^2 + D$.
A: Let us consider your differential equation:
$$y''(x)-\frac{1}{x}\cdot y'(x)=0.$$
Add $\frac{1}{x}\cdot y'(x)$ on both sides:
$$y''(x)=\frac{1}{x}\cdot y'(x).$$
Divide by $y'(x)$ on both sides:
$$\frac{y''(x)}{y'(x)}=\frac{1}{x}.$$
Integrate with respect to $x$ on both sides:
$$\int \frac{y''(x)\ dx}{y'(x)}=\int \frac{dx}{x}.$$
It follows that
$$\ln\left(y'(x)\right)=C+\ln(x).$$
An expression for $y'(x)$ is given by
$$y'(x)=e^{C+\ln(x)}=e^{C}\cdot e^{\ln(x)}=e^{C}\cdot x.$$
Redefine $e^{C}$ as $C$, since it is an arbitrary constant:
$$y'(x)=C\cdot x.$$
Integrate with respect to $x$ on both sides:
$$\int y'(x)\ dx=\int C\cdot x\ dx.$$
An expression for $y(x)$ is given by
$$y(x)=\frac{1}{2}\cdot C\cdot x^2 + D.$$
A: The equation writes
$$\frac{y''(x)}{y'(x)}=\frac1x$$
or after integration
$$\frac{y'(x)}{y'_0}=\frac x{x_0}.$$
Then after a second integration,
$$\frac{y(x)-y_0}{y'_0}=\frac{x^2-x_0^2}{2x_0}$$
or
$$y(x)=\frac{y'_0}{2x_0}x^2+y_0-\frac{y'_0x_0^2}{2x_0}.$$
