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$f:\mathbb{R}\rightarrow \mathbb{R}$ be a continuous nowhere vanishing function and consider the differential equation $$\frac{dy}{dx}=f(y)$$

$1.$ For each real number $c$ show that the D.E. has a unique , continuously differentiable solutiuon $y=y(x)$ on a neighbourhood of $0$ which satisfy the initial condition $y(0)=c.$

$2.$ Deduce the conditions on $f$ under which the solution $y$ exist for all $x\in\mathbb{R},$ for every initial value $c.$

In book only solution of this problem is not given. We know uniqueness of solution exist if $f$ is Lipschitz function. How we can say that solution is unique in some neighbourhood of $0.$ Similarly i have no result for $2$nd part. Please give me sufficient theorem or results regarding this problem. Thanks in advance.

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    Let $G:\bf R\to \bf R$ defined by $G(y)=\int _0^y {dy\over f(y)}$ . Prove that $G$ is a $C^1$ bijection form $\bf R$ to a certain interval $I$. Let $Y: I\to \bf R$ its inverse. Check that $Y$ is the unique solution of your ODE. When is $I=\bf R$ ? – 2017-01-25
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    Please if possible give solution in details...thanks.. – 2017-01-26

1 Answers 1

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You should have a look at the Cauchy-Lipschitz Theorema and its proof which is making use of Zorn's lemma. Some proofs won't make it obvious.

It covers your entire problem : See https://en.wikipedia.org/wiki/Picard%E2%80%93Lindel%C3%B6f_theorem

Concerning the proof, you will read with great interest the article : https://mathoverflow.net/questions/112199/differential-equations-and-axiom-of-choice

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    YES if possible please provide exact answer with explanation .... – 2017-02-06