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Let $\dot{x} = f(t,x)$, such that $f:\mathbb{R}^2 \rightarrow \mathbb{R}$ is continuous. Prove that if for every solution to the equation, $x(t)$ and for every $c \in \mathbb{R}$, $x(t + c)$ is also a solution, then $f$ is autonomous.

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You mean $x(t+c)$ is also a solution, not $x(t+x)$. –  Robert Israel Aug 31 '12 at 17:44
    
I fixed what I presume was a typo. –  copper.hat Aug 31 '12 at 17:50
    
Yes, it was a typo. Thank you :) –  Hila Aug 31 '12 at 17:52
    
@Hila, I think your notation is confusing you. Following Robert's notation below, a clearer statement (of your problem above) might be that if $x$ satisfies the equation at time $t$, then so does $x_c$. Then look for $s$ such that $x_c(s) = x(t)$ and see what that 'says' about $f$. –  copper.hat Aug 31 '12 at 17:53
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Let's assume that the existence and uniqueness theorem holds throughout the domain $\Omega:={\rm dom}(f)\subset{\mathbb R}^2$. Consider two points $(t_0,x_0)$, $(t_0+c,x_0)\in\Omega$. There is a unique solution $\phi:\ t\mapsto\phi(t)$ of the initial value problem $$\dot x(t)=f\bigl(t, x(t)\bigr)\ ,\quad x(t_0)=x_0\ ,$$ valid in some $t$-interval $I$ with midpoint $t_0$. By assumption the function $$\psi:\quad t\mapsto \psi(t):=\phi(t-c)\ ,$$ defined in an interval $I'$ with midpoint $t_0+c$, is also a solution of the differential equation; furthermore $\psi(t_0+c)=\phi(t_0)=x_0$. It follows that $\psi$ is the solution of the initial value problem $$\dot x(t)=f\bigl(t, x(t)\bigr)\ ,\quad x(t_0+c)=x_0\ .$$ Therefore we have $$f(t_0+c,x_0)=f\bigl(t_0+c,\psi(t_0+c)\bigr)=\dot\psi(t_0+c)=\dot\phi(t_0)=f(t_0,x_0)\ .$$ This shows that $f$ is constant on horizontal lines in $\Omega$.

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Hint: What differential equation does $x_c(t) = x(t+c)$ satisfy?

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Isn't it $\dot{x} = f(t + c, x)$? And if I follow @copper.hat's comment from above, can I say that $f(t + c, x) = f(t, x)$ for every $c$? –  Hila Aug 31 '12 at 18:08
    
Not exactly; you have $\dot{x_c} = f(x_c,t)$, which is $\dot{x}(t+c) = f(x(t+c),t)$. Shifting time by $-c$ gives $\dot{x} = f(x,t-c)$ (abusing notation slightly). –  copper.hat Aug 31 '12 at 18:38
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