Given the general form of an ODE of order $n$, namely: \begin{equation}\label{1} F\left(t,y(t),y'(t),\ldots,y^{(n-1)}(t),y^{(n)}(t)\right)=0 \end{equation} with: $u:t\in\mathbb{R}\longmapsto y(t)\in\mathbb{R}$ and $ F:\Omega\subseteq \underset{t}{\mathbb{R}}\times \underset{y(t)}{\mathbb{R}}\times \ldots\times \underset{y^{(n-1)}(t)}{\mathbb{R}}\times \underset{y^{(n)}(t)}{\mathbb{R}}\longrightarrow \mathbb{R} $ can we state that the general solution (supposing it exists) is a function of at least $n$ arbitrary constants? Without thinking about theorems of existence and uniqueness, this is a priori argument. First consider the case such that the ODE can be written in normal form, setting $(t,y(t),\ldots,y^{(n-1)}(t)\bigl)=z$: \begin{split} y^{(n)}(t)&=F_0\bigl(t,y(t),\ldots,y^{(n-1)}(t)\bigl)\\ \int y^{(n)}(t) dt&=\int F_0\bigl(t,y(t),\ldots,y^{(n-1)}(t)\bigl) dt\\ y^{(n-1)}(t)&=G_0(z)-c_1=F_1(z,c_1)\\ \int y^{(n-1)}(t) dt&=\int F_1(z,c_1) dt\\ y^{(n-2)}(t)&=G_1(z,c_1)-c_2=F_2(z,c_1,c_2)\\ &\vdots\\ y'(t)&=G_{n-2}(z,c_1,\ldots,c_{n-2})-c_{n-1}=F_{n-1}(z,c_1,\ldots,c_{n-1})\\ \int y'(t)dt&=\int F_{n-1}(z,c_1,\ldots,c_{n-1}) dt\\ y(t)&=G_{n-1}(z,x_1,\ldots,c_{n-1})-c_n=F_n(z,c_1,\ldots,c_n) \end{split} so [it seems that] the solution depends at least on the $n$ integrating constants: \begin{equation} y(t)=F_n(z,c_1,\ldots,c_n) \end{equation} Questions:
- Is it a reasonable proof for the normal case?
- If not, what can we say about it?
- Can we say that the case of an ODE in normal form in presence of Lipschitz regularity, is the case in witch we have exactly $n$ constants?
- In other words, can we state the following Theorem: given an ODE of order $n$ if the solution is not unique, than the solution depends on at least $n$ constants. Morover if the ODE can be written in normal form (and there is Lipschitz regularity) the constants are exactly $n$. , the solution Thanks for collaboration! :D
PS: a necessary hypothesis is that the ODE admists more than one solution a priori, e.g. don't consider:
\begin{equation} y'(t)=\sqrt{-|y(t)|} \end{equation}