Can I conclude that, if $v=0$ somewhere, then $v=0$ everywhere? In solving a homework problem, I have encountered the following DE ($v$ is a function of $x$).
$$v' = \frac{1}{x}\frac{1 + x^2}{1-x^2} v, \quad x \in (0,1)$$
I'd like to split the problem into two cases.


*

*$v=0$ somewhere

*$v=0$ nowhere


For Case 2, I can solve the problem by taking $v$ to the LHS and choosing an appropriate substitution.
However, Case 1 is giving me a bit of grief. I want to show that, assuming the hypothesis of Case 1, the only solution is the zero function.
Question A. Can I conclude that, if $v=0$ somewhere, then $v=0$ on the entire interval?
Question B. Is there a general principle that works for all DE's of the form $$v'(x) = f(x)v(x),\; x \in I$$
where $f$ is locally Lipschitz continuous and $I \subseteq \mathbb{R}$ is an open interval, allowing us to conclude that if $v=0$ somewhere, then $v=0$ everywhere?
Please keep answers as non-technical as possible; I have received answers to similar questions in the past, and had a lot of trouble understanding them.
 A: To solve such kind of equation you need to solve it on the intervals where $x(1-x^2)\neq 0$ which are $(-\infty,-1)$, $(-1,0)$, $(0,1)$ and $(1,+\infty)$. If you need a solution on $\mathbb{R}$, you have to try to find a $\mathcal{C}^1$ function that is a solution on each sub-intervals ie the form of a solution $f$ on $\mathbb{R}$ will be
$f(x)=\begin{cases}
f_1(x) &\text{on } (-\infty,-1) \\
f_2(x) &\text{on } (-1,0) \\
f_3(x) &\text{on } (0,1) \\
f_4(x) &\text{on } (1,+\infty) \\
\end{cases}$
where $f_1$,$f_2$,$f_3$ and $f_4$ are solution on $(-\infty,-1)$, $(-1,0)$, $(0,1)$ and $(1,+\infty)$ respectively. You don't to consider if $v=0$ or not. You have a solution directly by integration : $v(x)=\exp(\int_{x_0}^x \frac{1+t^2}{t(1-t^2)}dt)$.
A: If $v\ne0$, the equation is equivalent to
$$
\frac{\mathrm{d}}{\mathrm{dx}}\log(v)=\frac1x+\frac1{1-x}-\frac1{1+x}
$$
Integrate to get
$$
\log(v)=\log\left(\frac{Cx}{1-x^2}\right)
$$
so that
$$
v=\frac{Cx}{1-x^2}
$$
In this particular case, $v(0)=0$, yet $v$ is not necessarily $0$ everywhere.
Since $f(x)=\frac{v'}{v}=\frac{\mathrm{d}}{\mathrm{dx}}\log(v)$, if $v(x_0)=0$ at some point and $v$ is not identically $0$ in a neighborhood of $x_0$, $\log(v)$, and therefore, $\frac{\mathrm{d}}{\mathrm{dx}}\log(v)$, must blow up at $x_0$.
Thus, if $f$ is bounded in a neighborhood of $x_0$ and $v(x_0)=0$, then $v$ is identically $0$ in that neighborhood.
