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I have the following question: Let $I_{1}, I_{2} \subset \mathbb{R}$ be intervals and let $f:I_{1} \rightarrow \mathbb{R}$ be a continous function and let $g:I_{2} \rightarrow \mathbb{R}$ be a continously differentiable function. Consider the ODE $x'(t) = f(t)g(x(t))$, $x(t_{0}) = x_{0}$ and $t_{0} \in I_{1}$, $x_{0} \in I_{2}$ and $g(x_{0}) = 0$. How can one show that this equation has a unique constant solution $x(t) = x_{0}$? Hope for answers. Thanks in advance.

eric

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If $g(x_0)=0$, then $t \mapsto x_0$ is a constant solution. Are you sure that $g^{-1}(\{0\}) = \{x_0\}$? –  Siminore Jun 27 '12 at 14:00
    
Existence is clear: plug it in and you see that $x(t) = x_0$ satisfies the equation. Uniqueness follows from the Picard-Lindelof Theorem. –  Willie Wong Jun 27 '12 at 14:05
    
how does this follow from the Picard-Lindelof Thm.? –  eric Jun 27 '12 at 14:09
    
I dont know that $f$ and $g$ are lipschitz continous ? –  eric Jun 27 '12 at 14:10
    
There is something strange in your question, maybe I can't understand it. Let $x(t)=x_1$ be a constant solution. Then $x(t_0)=x_0=x_1$, so there is one and only one constant solution. But this has little to do with the ODE, since it follows from the initial condition. –  Siminore Jun 27 '12 at 14:14

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