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When does the IVP

$$\begin{cases}\dot{y}=y^2 \\ y(0)=1,\end{cases}$$

with $(x,y)\in \Bbb R\times\Bbb R$ have a unique solution?

For

$$\begin{cases}\dot{y}=f(x,y) \\ y(0)=y_0 \end{cases}$$

when do we have a unique solution, no solution, or infinitely many solutions? i want to ask the first IVP have a unique solution in which of the following intervals.. a)($-\infty$,$\infty$) b)($-\infty$,1) c)(-2,2) d)(-1,$\infty$).and how the answer is given option b . please tell me how it comes

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    $\begingroup$ The $\belongs$ command does not seem to be rendering the way you want it to. Do you want the symbol, $\in$? Also by $(R,R)$, do you mean $\mathbb{R}^2$? $\endgroup$ Sep 25, 2013 at 4:08
  • $\begingroup$ yes. i meant this $\endgroup$
    – abc
    Sep 25, 2013 at 4:09
  • $\begingroup$ Tri \in ${}{}{}$ $\endgroup$
    – leo
    Sep 25, 2013 at 4:10
  • $\begingroup$ What do you mean by solution? Do you mean a function $x \mapsto y(x)$ defined for $x \in (-\epsilon, \epsilon)$ for some small $\epsilon$? This is usually called a 'local solution'. Or do you mean a solution $y : \mathbb{R} \to \mathbb{R}$, that is, defined for all 'time'? You should look up Picard iteration, which is a typical method for establishing local uniqueness. $\endgroup$ Sep 25, 2013 at 4:19

2 Answers 2

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One can solve $\dot y=y^2$ where $y\ne0$ noting that $\dot y/y^2$ is the derivative of $-1/y$. Integrating this yields that the unique local solution on a neighborhood of $0$ is such that $1/y(x)=-x+1/y(0)$. By inspection, if $y(0)=1$, the maximal solution is $y(x)=1/(1-x)$ for $x$ in $(-\infty,1)$.

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  • $\begingroup$ what do you mean by maximal solution.and thanks $\endgroup$
    – abc
    Sep 26, 2013 at 7:14
  • $\begingroup$ The solution defined on the maximal interval around x=0 on which a solution exists. $\endgroup$
    – Did
    Sep 26, 2013 at 7:19
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The IVP has a uniaue solution by Picard-Lindelof Theorem.

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  • $\begingroup$ please someone help $\endgroup$
    – abc
    Sep 26, 2013 at 6:55
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    $\begingroup$ @Mhenni how do you apply the theorem here $f(x,y)=y^2$ is not lipschitz! $\endgroup$
    – Mathronaut
    Sep 8, 2014 at 7:17

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