$$x'=1+t-x \; \; x(0)=0 \\ x'+x=1+t \\ xe^t=te^t+C \Rightarrow x(t)=t+Ce^{-t} \\ x(0)=0 \Rightarrow C=0 \\ x(0)=x_0 \Rightarrow C=x_0 \\ x(t,0,x_0)=t+x_0e^{-t} \\ \forall \epsilon>0 \; \; \exists \delta>0: |x_0-0|<\delta \Rightarrow |x_0e^{-t}+t-t|<\epsilon? \\|x_0-0|<\delta \Rightarrow |x_0-0|e^{-t}<\epsilon!$$
My question is regarding the last line. If $\epsilon=\delta$ would the inequality hold if $t<0?$ It is nowhere implied whether $t$ is smaller or larger than $0$. Does anybody know why it is supposed that it is greater than $0$ ?
From $$xe^t=te^t+C$$with $x(0)=0$ we obtain $C=0$ and $$x(t)=t$$ which is not stable since $$\lim_{t\to\infty}x(t)=\lim_{t\to\infty}t=\infty$$