To illustrate the problem that I'm having, consider the following ODE:
$$y' = \frac{y}{1+x}$$
To solve this ODE, my professor had done the followings:
Assuming $y(x) \not = 0$, and $x \not = -1$ (actually, he never stated these), then we can arrange the ODE as
$$\frac{dy}{y} = \frac{dx}{1+x} \\
\ln|y| = \ln|x+1| + \ln|c| \\
y = c(x+1). \quad \text{(The General solution)}$$
And later, while introducing some initial values to this ODE, he gave the case where $y(-1) = 0$, and claimed that since when $x = -1$, for any values of $c\in \mathbb{R}$, we get different solutions from the "general solution", there exists infinitely many solutions satisfying this initial condition.
However, (even if we skip how we found $y = c(x+1)$) to verify that $y = c(x+1)$ is indeed a solution (for some $c$) to this ODE, when we plug it in to the ODE, in order to cancel the terms $1+x$ in the denominator and in the numerator, we need to assume that $x \not =-1$, otherwise it would also mean we are calcelling zero's, which is not valid since, in the algebra that we are doing, zero do not have any multiplicative inverse.Therefore, any solution of the form $y = c_0(x+1)$ cannot contain the the point $x = -1$, so how can we say that "there are infinitely many solution of this ODE satisfying the initial condition $y(-1) = 0$" ?
With my logic, there exists no such solution satisfying this initial condition, so which one of us is right, and why?
A Further Question:
As we have done it in this particular case, to reach some solutions of a given ODE, we need to do some arrangements to the form of the ODE, such as collection same variables to the same side, but this generally requires some assumptions, such as $y \not= 0$, so in such cases, the result that we get should not be applicable to the cases, for example, in the above case, it there was any other point other than $x = -1$ where $y = 0$, say $y(x_1) = 0$ the solutions that we got from the equation $y = c(x+1)$ should not contain that point $(x_1, 0)$ also, right ?
Best Answer
I fully agree that, already when writing an ODE in the form $$ y' = \frac{y}{1+x}, \tag{1} $$ you're already implicitly assuming that $x \neq -1$, otherwise the fraction is not defined. However, since we're dealing with an ODE (and hence implicitly assume that $y$ is continuously differentiable), the concept of a limit is intimately involved.
So, let's look at equation $(1)$ on $I := \mathbb{R} \backslash \{-1\}$. It's easy to that on $I$, the function $\eta_c(x) = c(1+x)$ is a solution of $(1)$, for any $c \in \mathbb{R}$. Now, even though $\eta_c$ is not defined at $x=-1$, the limit $\lim_{x \to -1} \eta_c(x)$ exists. That is, both left and right limits (to $x=-1$) of exist and are equal. Furthermore, by the same reasoning, the limit of the derivative ($x \to -1$) also exists.
Therefore, it's a small step to extend $\eta_c$ to $\mathbb{R}$ by including the definition $\eta_c(-1) := \lim_{x \to -1} \eta_c(x)$. This new extended function -- let's call it $y_c$ -- is continuously differentiable on the entire real line. Since the procedure sketched above is straightforward and unambiguous, people tend to call $y_c : \mathbb{R} \to \mathbb{R}, x \mapsto c (1+x)$ 'the solution' to $(1)$, without worrying too much about the point $x=-1$.
Be that as it may, there's still something going on at $x=-1$, of course. To see what's the problem, I think it's instructive to look at the slightly rewritten ODE $$ (1+x) y'=y. \tag{2} $$ It's clear that on $I$, equations $(1)$ and $(2)$ are equivalent, so they give rise to the same solutions. However, if we are looking for continuously differentiable solutions to $(2)$ on the entire real line, we take the limit $x \to -1$ on both sides of equation $(2)$. As we have assumed $y$ to be continuously differentiable on $\mathbb{R}$, both the limit $\lim_{x \to -1} y(x)$ and the limit $\lim_{x \to -1} y'(x)$ exist (and are finite), so we obtain $$ 0 \times y'(-1) = y(-1), $$ i.e. $y(-1) = 0$. Hence, we can't choose the function value at $x=-1$; every continuously differentiable solution to $(2)$ obeys $y(-1)=0$. Of course, you can still take your initial condition $y(x_0)=y_0$ anywhere else (i.e. $x_0 \in I$), and you'll get a unique, continuously differentiable solution to $(1)$.
To summarise: equation $(2)$ is singular at $x=-1$. This has the effect that every continuously differentiable solution to $(2)$ obeys $y(-1)=0$.