Showing there's no exotic solutions to this (kind of) differential equation We want to solve $$ty'(t) = y$$ for all once-differentiable real functions.
My attempt:
It looks like a separable ODE, so first let's try solving it over an interval that doesn't contain $0$, it should be easy then to extend it to 0 by continuity and differentiability. Unfortunately, we also have to divide by $y$ which could be $0$. If we assume our solution doesn't pass through $0$ at $t\neq 0$, it is easy to see the solutions are of form $$y = Ct, C\neq 0.$$
Another trivial solution would be of the above form with $C=0$. Now we'd need to prove whether these are the only solutions, or if there is some other solution that passes through $0$ at some $t_0 \neq 0$.
What i've tried:
Since $y(t_0)=0$ then plugging in the DE, $y'(t_0)=0$. However now i have no idea how to proceed. Any hints? Is there a general technique for this kind of problem?
 A: First, let us say what exactly a solution is.
Definition A differential equation $F(x,y,y')=0$ with $x$ on some open interval $I$ of $\mathbb{R}$ is said to have a solution $y(x)$ on $I$ if

*

*$y(\cdot)$ is differentiable function on $I$;

*$F(x,y(x),y'(x))=0$ for every $x\in I$.

Now, we handle with the equation $xy'=y$ that is naturally defined on the whole line $\mathbb{R}$.

*

*With $x\in I_1=(0,+\infty)$: (i) clearly $y=0$ is a solution; (ii) when $y\not= 0$, the separation technique gives $\frac{dy}{y}=\frac{dx}{x}$, $\ln|y|=\ln|x|+\ln|C_1|$, $|y|=|C_1x|$, so $y=C_1 x$, $C_1\not=0$; (iii) combining (i), (ii) and note that $y=0$ corresponding to $C_1=0$. In summary, its solutions is $y=C_1 x$, $C_1\in\mathbb{R}$, $x>0$ (the right curve).

*With $x\in I_2=(-\infty,0)$: similarly, its solutions is $y=C_2 x$, $C_1\in\mathbb{R}$, $x<0$ (the left curve). Note that $C_1$ and $C_2$ are independent to each other.

Next, there is natural question: Is there any solution defined on the whole line $\mathbb{R}$ ? If yes, then that solution curve $y(x), x\in\mathbb{R}$ must be archived by gluing a left curve and a right curve
$$
y(x)=
\begin{cases}
C_1 x, \quad x> 0\\
C_2 x, \quad x< 0.\\
\end{cases}
$$
It is easy to see that $C_1=C_2$ (otherwise there is no derivative of $y$ at $x=0$). Therefore the solutions of $xy'=y$ defined on the whole line $\mathbb{R}$ is $y=C x$ for all real constant $C$.
More comment: Most of textbooks ignore gluing solutions  in the first course on ODEs (for beginners' convenience I suppose). In some situation gluing solutions is a must, for example on the problem of finding conjugacy of two systems of differential equations (course on Dynamical Systems).
A: In the form $y'=y/t$, the RHS is Lipschitz in $y$ off a neighborhood of $t=0$ so the solutions to IVPs started at $t_0 \neq 0$ are $y=Ct$ on either $(-\infty,0)$ or $(0,\infty)$ depending on the sign of $t_0$. Here $C$ could be zero; the equation is linear so no division by $y$ is required in this derivation.
Additionally the form of the equation without the division by $t$ forces $y=0$ at $t=0$, which agrees with the solutions on either side. So you can extend to $t=0$. To get differentiability at $t=0$ you must assume the slopes were the same from the left and right. Thus all solutions are of the form $y=Ct$.
A: If you'll excuse a physicist's lack of rigor, I teach my students to solve this by writing $y'(t)$ as $dy / dt$ and splitting up the differential. This yields:
$$\frac{dy}{y} = \frac{dt}{t}$$.
Performing indefinite integrals of both sides, we obtain
$$\log y = \log t + C.$$
This result, now that we've obtained it, seems to work fine everywhere, including at $t=0$.
Then, somewhat differently from your result, and reusing the label $C$ to denote a different (nonzero) constant,
$$y(t) = C t$$.
Again in an unrigorous way, it seems as though we couldn't possibly have any solutions other than ones of this form. Our only real assumption was that $y \ne 0, t \ne 0$, so another solution could exist only if $y=0$. Either this is our old case with $C=0$, or this solution must turn into our old solution at $t=0$, where the two solutions intersect. But in this latter case, $y'(t)$ is ill-defined at $t=0$.
