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i have this exercice Let the problem $$y'=f(y) g(x) , y(x_0)=y_0$$ if $f(y_0)=0$ what are the conditions on $f$ for the probleme admits a unique solution thank's

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Use the Picard-Lindeloef's Theorem to give a sufficient condition... –  Alex Apr 19 '13 at 22:24
    
$f(y)$ has to be Lipschetz continuous by Picard-Lindeloef's theorem. –  Mhenni Benghorbal Apr 19 '13 at 22:45
    
We use Picard theorem to $f(y) g(x)$ not just to $f(y).$ or not? –  lili Apr 19 '13 at 22:48
    
@lili: Read the theorem carefully, if $y'=G(x,y)$, then we require $G$ to be Lipschitz continuous in $y$ and continuous in $x$. –  Mhenni Benghorbal Apr 19 '13 at 23:12
    
ok! $f(y)$ must be lipschitz . But why lipschitz continuous? if $f$ is lipschiz, so she is continuous! –  lili Apr 20 '13 at 9:30
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1 Answer

Let

$$f(y)g(x)=h(x,y) $$

If both $h(x,y)$ and $\displaystyle\frac{\partial h}{\partial y}(x,y) $ are continuous over some rectangle $R$ then for $(x_{0},y_{0})\in R$ the solution exists and it is unique

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