I have to solve $\dfrac{dy}{dx} = \dfrac{y}{x}$. So I set $v = \dfrac{y}{x}$ and so $$ \dfrac{dy}{dx} = v $$ Then by product rule $x\dfrac{dv}{dx} + v = v$ and so $x\dfrac{dv}{dx} = 0$. But then that means there is no unique solution to the differential equation; am I wrong in my reasoning?
Wolframalpha said the solution was $y(x) = cx$.
Approach 1
$$\dfrac{dy}{y} = \dfrac{dx}{x}$$
Integrate each side and simplify, you get:
$$y = c x$$
Approach 2
This is your approach, we get:
$$v' x = 0 \rightarrow v = c$$
From the initial substitution, you have:
$$v = \dfrac{y}{x} \rightarrow y = v x = cx$$
In other words, both methods yield the same family of curves and your reasoning is correct.
${\frac{dy}{dx}}={\frac{y}{x}}$
${\frac{dy}{y}}={\frac{dx}{x}}$
Integrating
${\ln}y={\ln}x+C$,
where $C$ is an integration constant.
${\ln}y={\ln}x+{\ln}a$
${\ln}y={\ln}(ax)$
$y=ax$
