Separable ODE and singular solutions

In most introductory ODE textbooks we can find the following definition:

A separable first-order ODE is the one of the form $$y'=g(x)h(y)$$ and if $h(y)\neq0$, then the general solution is found by integration (using chain rule). Next, we must find every $y_0$ such that $h(y_0)=0$ and the constant function $y=y_0$ satisfies the above ODE (called singular solutions).

I'm trying to learn this method in a rigorous way by using the existence and uniqueness of solution theorems and adding restrictions over $g$ and $h$, but can't deal with the following:

Does there exist a solution $f$ defined in some $A\in \mathbb{R}$ such that there exists $x_0\in A$ such that $h(f(x_0))=0 $ but $f$ is not constant? I mean, if $f(x_0)=y_0$, then $h(y_0)=0$, so we can't divide by $h(y)$ and integrate using separable equation method. I know that $f(x)=y_0$ is a solution, but what can we say about a non-constant $f/f(x_0)=y_0$?

Also, can you recommend books explaining this kind of solving-ODE's methods but in an absolutely rigorous way?

EDIT: This edit is made after the bounty (I thought it was implicit in the spirit of the question, but maybe I couldn't state it properly due to my lack of expertise in english), but I'd really want to know what further hypothesis need to be given in order to assure that every singular solution must be constant.

Any help is highly appreciated. Thanks and regards

Solution 1:

Yes, you might have solutions that are not constant and such that $h(y)=0$ somewhere.

Here's an example I ran across on a textbook some years ago: \begin{equation} \tag{1} \frac{dy}{dx}=2x(1-y^2)^{1\over 2}. \end{equation} You have the constant solutions $y=\pm 1$. If you carelessly separate variables you get $$ \int\frac{dy}{(1-y^2)^{1\over 2}} = \int 2x\, dx $$ from which you obtain the "general integral" $$ \tag{!!} y(x)=\sin(x^2+C). $$ This is the solution I found on the textbook. Problem is, it is wrong. Take for example $C=0$. The function $\sin(x^2)$ is oscillating, so its first derivative changes sign infinitely often for $x\ge 0$. And as you can see from (1), the derivative $\frac{dy}{dx}$ must be nonnegative on the whole half-line $[0, \infty)$.

To understand what is going on let us observe the graph of $\sin(x^2)$ on the phase field of (1):

The black line matches with the phase field until it touches the critical level $y=1$. Then it goes amiss. The correct solution is the following:

As soon as the solution touches the critical level, it merges with the constant solution. The general solution to the Cauchy problem \begin{equation} \begin{cases} y'=2x(1-y^2)^{1\over 2},& x> 0\\ y(0)=y_0\in (-1, 1] \end{cases} \end{equation} is the following: \begin{equation} y(x)= \begin{cases} \sin(x^2+\arcsin y_0), & x<\sqrt{\frac{\pi}{2}-\arcsin(y_0)} \\ 1, & x\ge \sqrt{\frac{\pi}{2}-\arcsin(y_0)} \end{cases} \end{equation} For $y_0=-1$ you have infinite solutions. One is the constant one $y(x)=-1$, the others are pictured below:

The analytical expression of those solutions is the following: \begin{equation} y_{x_0}(x)= \begin{cases} -1, & x<x_0 \\ \sin\left(x^2-x_0^2-{\pi\over 2}\right), & x_0\le x < \sqrt{x_0^2+\pi} \\ +1, & \sqrt{x_0^2+\pi}\le x. \end{cases} \end{equation} The parameter $x_0\ge 0$ is the abscissa of the point in which the solution leaves the critical line $y=-1$.