Solutions of $(\frac{d}{dt}x)y-(\frac{d}{dt}y)x=0$ I would like to find all the possible solutions of the following differential equation, with two variables: $$(\frac{d}{dt}x)y-(\frac{d}{dt}y)x=0.$$
I have the feeling that I should expect infinitely many solutions, however I don't really know how to approach the problem of finding their general expression. 
I'm sure about the trivial one $x=0,y=0$ and about the following
\begin{align*}
x&=(\frac{d}{dt}x),\\
y&=(\frac{d}{dt}y)
\end{align*}
Did I miss "some" of them?
 A: Here is an interesting solution that everybody has missed. Assuming $x=f(t)$ and $y=g(t)$, we can have
$$ f(t) = \begin{cases} 0 & t \leq 0
\\ t^2 (1-t)^2 & 0 \leq t \leq 1
\\ 0 & 1 \leq t \end{cases} $$
$$ g(t) = \begin{cases} 0 & t \leq 1
\\ (1-t)^2 (2-t)^2 & 1 \leq t \leq 2
\\ 0 & 2 \leq t \end{cases} $$
Here's another fun one that keeps $x$ and $y$ proportional, but changes the proportion:
$$ f(t) = t^2 $$
$$ g(t) = \begin{cases} t^2 & t \leq 0
\\2 t^2 & t \geq 0\end{cases} $$
Here's a pretty horrible variation that shows we can't expect the set of transitions to be pleasant, based on the "topologist's sine curve": define
$$ h(t) = 1 - \cos\left( \frac{2 \pi}{t} \right) $$
Then $h(t)$ has the proprety that $h(t) = h'(t) = 0$ whenever $1/t$ is an integer. Using this,we can construct everywhere differentiable functions
$$ f(t) = \begin{cases} 0 & t \leq 0
\\ \left\lfloor \frac{1}{t} \right\rfloor^{-4} h(t)  & t > 0
\end{cases} $$
$$ g(t) = \begin{cases} 0 & t \leq 0
\\ \left\lfloor \frac{1}{t} \right\rfloor^{-3} h(t) & t > 0
\end{cases} $$
where $\lfloor x \rfloor$ is the floor function. Note that the ratio $y/x$ increasing without bound as $t \to 0^+$! You could do even worse things!
A: On an interval where $y\ne0$, you have 
$$
0=x'y-xy'=\frac{x'y-xy'}{y^2}=\left(\frac xy\right)'.
$$
So $x/y=c$, a constant, and $x=cy$. Now you are free to choose $y$ freely (any differentiable function), and take $x$ to be a scalar multiple. 
Note that roles are reversible, so the same game can be played on intervals where $x\ne0$. And, on intervals where both are zero, you already know what they are.
A: You have your equation:
$$\frac{dx}{dt} y- \frac{dy}{dt} x=0 \tag{1}$$
You can rearrange this to give:
$$\frac{dx}{dt} y=\frac{dy}{dt} x$$
We rearrange the $y$ and $x$ terms towards the side with their associated derivatives, so that we can integrate both sides.
$$\frac{1}{x} \frac{dx}{dt}=\frac{1}{y}\frac{dy}{dt}$$
You can now cancel out $dt$ and see that you get an equation only in terms of $y$ and $x$. Therefore, you can now find a function for $y(x)$.
$$\int\frac{1}{y}~dy=\int\frac{1}{x}~dx \tag{2}$$
If we integrate both sides for $x>0$ and $y>0$ we get:
$$\ln{y}=\ln{x}+C$$
We can now exponentiate both sides:
$$e^{\ln{y}}=e^{\ln{x}+C}$$
$$y=x\cdot e^C$$
Since $e^C$ is arbitrary, we can define a new constant $k=e^C$.
Thus, our solution is:
$$\boxed{y=kx}$$
Now, do the same integration from $(2)$ for other cases.
A: We can rewrite this equation in the form
$$
\frac{1}{x} \frac{dx}{dt} = \frac{1}{y} \frac{dy}{dt}
$$
If we integrate both sides with respect to $t$ by applying the chain rule (i.e. making $u$-substitutions), we find
$$
\ln(x(t)) = \ln(y(t)) + C_0
$$
where $C_0$ is an arbitrary constant.  Exponentiating both sides, we have
$$
x(t) = C y(t)
$$
where $C$ is a new constant.
A: It is separable problem. A mnemonic solution is to multiply both sides by $dt$:
$$
y\;dx = x\;dy
$$
And then re-arrange:
$$
\frac{dx}{x} = \frac{dy}{y}
$$
and integrate:
$$
\ln{x} = \ln{y}+C=\ln{Ay}\quad A>0
$$
and from there: $x = Ay$ (I am omitting some details, such as the case of $x<0$, but these can be trivially reconstructed). Also, there is a special case of $x=0,y=0$.
PS. A stricter solution requires understanding that expression $y\;dx-x\;dy$ defines differential form and the fact that $y\;dx-x\;dy=0$ means that the solution vector field is perpendicular to $(y,-x)$
A: $$
\frac{y}{x} = \frac{dy}{dx}
$$
Variable separable ; After transposing it integrates to :
$$ \ln{y} = \ln{x}+ \ln{C}, $$
$$ y = C\,x, $$  which are all straight lines through  the origin.
