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$\frac{\mathrm{d}y}{\mathrm{d}x} = f(y)$ where $f(y)$ is continuous on $|y-a|\leq \epsilon$,and $f(y)=0$ iff $y=a$.

To Proof : For the initial value point on $y=a$,the equation has local unique solution iff $\left|\int_a^{a+\epsilon}\frac{\mathrm{d}y}{f(y)}\right|= \infty$

How to proof Initial value $\Rightarrow$ $\left|\int_a^{a+\epsilon}\frac{\mathrm{d}y}{f(y)}\right|= \infty$ ?

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It's not true. Consider e.g. $\dfrac{dy}{dx} = - y^{1/3}, y(0) = 0$ Even though $\int_0^{\epsilon} - y^{-1/3}\ dy$ is finite, it's obvious that the only solution is $y = 0$.

What is true is this. Suppose $f(y) > 0$ for $a+\epsilon > y > a$ and $\displaystyle\int_{a}^{a+\epsilon} \dfrac{dy}{f(y)} = b < \infty$. Then besides the constant solution $y = a$, there is a solution defined implicitly by $\displaystyle\int_a^y \dfrac{ds}{f(s)} = x$ for $0 < x < b$, with $y = 0$ for $x \le 0$.

Similarly, if $f(y) < 0$ for $a-\epsilon < y < a$ and $\displaystyle \int_{a-\epsilon}^a \dfrac{dy}{f(y)} = c > -\infty$, there is another solution in $c < x < 0$.