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I want to prove that the I.V.P. $$\frac{dy}{dx}=y^{2},y(0)=1$$ does not have unique solution on $\mathbb{R.}$ I tried as follows

solution is given by $y=\frac{1}{c-x}$ but initial condition gives that $c=1$ i.e. solution is given by $y=\frac{1}{1-x}$ which is defined on $(-\infty ,1)$ or on $(1,\infty )$. Now how to combine $\frac{1}{c-x}$ and $\frac{1}{1-x}$ to obtain many solutions? Like $$f(z) = \left\{ \begin{array}{rcl} \frac{1}{1-x} & \mbox{for} & x< 1 \\ \frac{1}{c-x} & \mbox{for} & x\geq1 \ &\end{array}\right.$$ But in this way i can discontinuity at $1.$ Please suggest me exact way. Thanks a lot.

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    Your IVP has a unique solution. – 2017-01-06
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    it has unique solution in bounded neighbourhood of $0$ but not on whole $\mathbb{R}$ – 2017-01-06
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    You showed it yourself, $y(x)=1/(1-x)$ on $(-\infty,1)$. That's it. – 2017-01-06

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The maximal interval in which, there exists a solution $y$ satisfying $y(0)=1$ is $(-\infty,1)$. this solution is given by

$\forall x\in(-\infty,1)\;\;\; y(x)=\frac{1}{1-x}$.