Let y be a solution of the following equation on $\mathbb{R}$: $y'(t)= \sin(ty(t))$
I want to proof that: y is constant $\iff y(0)=0$
I started by proofing: y is constant $\Rightarrow~ y(0)=0$.
Let y be constant, so $y(t)=c$ for $c \in \mathbb{R}$ and $ \forall t \in \mathbb{R}: y'(t)=0$. This implies $\forall t \in \mathbb{R}: 0=sin(t\cdot c) \Rightarrow \forall t \in \mathbb{R}:t\cdot c= \pi \cdot k : k \in \mathbb{Z} $.
Now assume $y(0)\neq0 \Rightarrow c \neq 0$ and choose $t=\frac{\pi}{2c}$. Therefore $\frac{\pi}{2c}\cdot c=\pi\cdot k \Rightarrow \frac{1}{2}=k$, which is a contradiction to $k \in \mathbb{Z}$.
Now for the other implication. y is constant $\Leftarrow$ y(0)=0.
Let $y(0)=0$. I want to show that $\forall t \in \mathbb{R}: y'(t)=0$ or that $\forall t \in \mathbb{R}: y(t)=0$.
And there is also my question I would like a hint on which you think is easier to show, because I tried both being either assuming that there exists $t_0$ for which $y(t_0)\neq0$ or $y'(t_0)\neq0$. As I am new to ODEs and my course starts the upcoming semester I already wanted to try some problems and still want to solve this problem kind of on my own but everyone who can hint something is welcome. :)
