Any solution of second order differential equation has atmost a countable number of zeroes ?
So what i have for this proof is
let $S=\{a\in \mathbb{R}/y(a)=0\}$
suppose $S$ is uncountable then
$S\subseteq \bigcup_{n=0}^{\infty} [0,m]$ ( i don't is this correct or not if this so there is contradiction i think im missing so many things)
let $S^*=S\cap[n,n+1)$
and $S^*$ is bounded and by Bolzan's Weierstrass theorem $S^*$ has limit point $x_n\to x$ and then $x\in S$
and since $y(a)=0$ then $y'(a)=0$ then Picard's theorem in it has unique solution
so solutions are one at most countable
i really sorry if i missed things can someone explained properly please and is this only applicable for homogeneous ODE?
thank you so much for your time.....
No, this is not true. Choose a twice continuously differentiable function $f$ with an uncountable set of zeros - then $f$ is a solution of the equation $y'' - f'' = 0$.
This, on the other hand, is true if the equation has constant coefficients: $y'' + ay' + by + c = 0$ with $a,b,c \in \mathbb R$. The proof is done in two steps.
First, assume that $c=0$. The general solution in this case is either a linear combination of a sine and a cosine, or of two real exponentials, or of an exponential and another one multiplied by $x$ (the variable of $y$), depending on the sign of $a^2-4b$. In all these three cases the general solution will have at most countably many zeros.
Second, if $c \ne 0$, use the method of the variation of constants after having performed the previous step.