Decide whether given function is the general solution to ODE I have a set of function and must decide whether each of them is a general solution to an ODE of the form : \begin{cases}y'(x)=f(x,y(x))\\y(x_0)=y_0\end{cases}
When the an ODE comes obvious for a function, I know how to prove that it is a general solution simply by giving the ODE and solving it using the methods we've seen in class.
But when a solution isn't obvious, how do I prove that is cannot be solution to an ODE ? (The goal would be to try to prove that it isn't, and if there is a contradiction then there must be some ODE it is the general solution to without necessarily expliciting what ODE it is a general solution of)

The ODEs we've seen in class are of the form :
\begin{split}
y'(x)=g(x)\cdot h\circ y(x)\ \ \Rightarrow\ \ H\circ y(x)&=\int_{y_0}^{y(x)}\frac1{h(t)}dt\\&=\int_{x_0}^xg(t)dt=G(x)\\& (iif\ G(x)\in H(Y)\forall x\in I)
\end{split}
and
\begin{split}
y'(x)=g(x)-h(x)\cdot y(x)\ \ \Rightarrow\ \ y(x)&=e^{-G(x)}(y_0+\int_{x_0}^xh(t)e^{G(t)}dt\\& where\ G(x)=\int_{x_0}^xg(t)dt
\end{split}
We also have seen standard variable changes to help with complex ODEs. With the examples above you can see the $y(x,x_0,y_0)=(3(x-x_0)+y_0^3)^{1/3}$ is a very obvious solution to an ODE. But I think for $y(x,x_0,y_0)=y_0cosh(x-x_0)$ it isn't easy to see what ODE it could be a solution to.

So if I wanna prove that y isn't a general solution or to prove that it is by contradiction assuming it is not, how should I go about it?
 A: I am going to change the notation a bit to make it more similar to the linked Wikipedia article.
Recall that the family of solutions to the ODE
\begin{cases}\tag{1}\dot x(t)=F(x(t),t)\\[3mm]x(t_0)=x_0\end{cases}
constitute a time dependent flow. Writing
$$
\boldsymbol{y}(t)=(x(t),t)
$$
this means in my opinion nothing else than
$$
\begin{cases}\dot{\boldsymbol{y}}(t)=G(\boldsymbol{y}(t))\\[3mm]\boldsymbol{y}(t_0)=(x_0,t_0)\end{cases}\quad\quad\begin{cases}\dot{\boldsymbol{y}}(t+s)=G(\boldsymbol{y}(t+s))\\[3mm]\boldsymbol{y}(t_0+s)=\big(x(t_0+s),t_0+s\big)\end{cases}
$$
where $G(\boldsymbol{y})=F(x,t)\,.$

*

*This seems very trivial. What is however crucial is that the functions $G$ and $F$ are independent of $x_0$ and $t_0$.


*It seems an interesting question to ask if every family of functions consitutes a time dependent flow, or, what is equivalent: constitutes the family of general solutions to an ODE of type (1) where $F$ is independent of $x_0$ and $t_0$.
Your first example is (with a typo corrected)
$$\tag{3a,3b}
x(t)=x(t,t_0,x_0)=\big(3(t-t_0)+x_0^\color{red}{3}\big)^{1/3}\,,\quad\quad
\dot x(t)=\frac{1}{x^2(t)}\,.
$$
The last expression holds independently of $t_0$ and $x_0$ so that
$$F(x,t)=\frac{1}{x^2}$$
does the job of defining the ODE of which (3a) is the family of solutions, hence a time-dependent flow.
Your second example is
$$\tag{4a,4b}
x(t)=x(t,t_0,x_0)=x_0\cosh(t-t_0)\,,\quad\quad \dot x(t)=x_0\sinh(t-t_0)\,.
$$
It seems not possible to express $\dot x(t)$ as a function only of $x(t)$ and $t$ therefore this family does not consitute a time-dependent flow.
It is obviously possible to write
$$
\dot x(t)=\color{red}{x_0}(\sinh\circ\operatorname{arccosh})\Big(\frac{x(t)}{\color{red}{x_0}}\Big)
$$
but it is not possible to get rid of those $\color{red}{x_0}$ that are not part of $x(t)$ already.
A: Ok so I've found way to prove it properly.
We say that $f:D\subset\mathbb{R}^{1+n+m}\rightarrow\mathbb{R}^{1+n+m}$ $(x,y(x),\alpha)\mapsto f(x,y(x),\alpha)$ satisfies the standard requirements for an ODE if it is continuous on D and locally lipschitz with respect to $y$ and :
\begin{equation*}
y'(x)=f(x,y(x),\alpha)
\end{equation*}
Thus when we say we want to know if $\lambda(x,x_0,y_0,\alpha)$ is the general solution to an ODE, what we want to know is if there is a function $f$ that has the properties presented above.
Property (seen in class) :
If $\lambda(x;x_0,y_0,\alpha)$ is the solution to a ODE satisfying the standard requirements then it satisfies the cocycle property :
\begin{equation*}\lambda(x;t,\lambda(t,x_0,y_0,\alpha),\alpha)=\lambda(x,x_0,y_0,\alpha)\end{equation*}
$\forall(x_0,y_0,\alpha)\in D$ and $\forall x,t\in I_{max}(x_0,y_0,\alpha)$. $I_{max}$ being the maximal interval of existence of our ODE.

So showing that $\lambda(x;t,\lambda(t,x_0,y_0,\alpha),\alpha)\neq\lambda(x,x_0,y_0,\alpha)$ will show that it cannot be a general solution to an ODE satifying the standard requirements.
