Started to learn ODE, and the second example was of the equation $y^\prime=y^2$
We showed that the group of solutions is $\{y(x)=0 \}\cup \{ y(x)=\frac{1}{c-x} | c \in \mathbb{R} \}$ and mentioned that there are no other solutions.
I would like to prove that fact.
I tried to use a method like the one we used to prove that for the equation $$dy/dx=y$$
the solutions are only $\{y=Ce^x|C>0\}$: what we did there was to define $g = ye^{-x}$ and by looking at the derivative of $g$ we showed that $g$ is constant. But I can't figure how to do it in my case. (I guess I cant define $h(x) = y\cdot (c-x)$ because it is dependent on $c$, right?)
1 Answer
$\begingroup$You can do the same thing here to show that the solution is (at least locally) unique.
So let's fix $x_0\in\Bbb R$ such that $y(x_0) \neq 0$ and an open neighborhood $U$ of $x_0$. By shrinking $U$, we may assume that $y(x) \neq 0$ for any $x \in U$.
We now define $g = x + 1/y$ and proceed to calculate $g'$:$$g' = 1 - \frac 1{y^2} \cdot y' = 0$$ which shows that $g$ is constant on $U$.
To obtain the global result, we may "glue" all the local pieces together.
The above argument already shows that, whenever $U$ is an open set on which $y$ is nonzero, there exists $c$ such that $y(x) = \frac 1{c - x}$ on $U$.
Let $E$ be the set $\{x \in \Bbb R: y(x) = 0\}$. We show that $E$ is either empty or the whole $\Bbb R$.
Assume the contrary. Then there is an open set $U \subset \Bbb R \backslash E$ such that $E \cap \bar U \neq \emptyset$, where $\bar U$ denotes the closure of $U$.
Thus we have $y(x) = \frac 1{c - x}$ on $U$, which extends by continuity to $\bar U$. But this leads to a contradiction for any $x \in E \cap \bar U$.
Therefore either $y = 0$ on the whole $\Bbb R$, or $y$ is never zero on $\Bbb R$, in which case we have $y = \frac 1 {c - x}$ for some $c$.
$\endgroup$ 2
