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Let $f$ be a continuous function from $\Bbb R^3 \to \Bbb R$. By a solution of the differential equation $$f(x,y,\dot{y}) = 0$$ We mean a function $y\colon U \subset \Bbb R \to \Bbb R$ where $u$ is an open interval in $R$ that satisfies $f(x,y(x),\dot{y}(x))= 0$.

Let $f\colon \Bbb R^3 \to\Bbb R$ be a smooth function. Let $f(0,0,0) = 0$. Give a sufficiency condition under which the differential equation $f(x,y,\dot{y})$ admits a solution satisfying the initial condition $y(0) = 0$ in a neighbourhood of $0$.

I think I must use the Implicit Function Theorem but not sure how?

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  • $\begingroup$ To give a sufficiency condition for $f(x,y,\dot{y}$ to admit a soln with the given initial conditions. $\endgroup$ – Arnold Apr 25 '15 at 17:13
  • $\begingroup$ Think of what do you need in order to rewrite an equation $f(x,y,z)=0$ as $z=g(x,y)$. $\endgroup$ – user147263 Apr 25 '15 at 17:37
  • $\begingroup$ I was thinking of the same thing. But nothing comes to mind. $\endgroup$ – Arnold Apr 25 '15 at 17:38
  • $\begingroup$ In order to use the implicit function theorem, you have to know what it says. Do you know the statement of the implicit function theorem, its hypotheses and conclusion? $\endgroup$ – user147263 Apr 25 '15 at 17:39
  • $\begingroup$ If $f(a,b)=0$ and $df$ is invertible at $(a,b)$ then there is a unique $g$ such that $f(g(y),y) =0$ for every $y$ in a neighbourhood of $b$. $\endgroup$ – Arnold Apr 25 '15 at 17:45
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You need to put together two facts:

  • Implicit function theorem. If $\partial f/\partial z$ is nonzero at $(0,0,0)$, then for some neighborhood $U $ of $(0,0,0)$ the set $U\cap \{(x,y,z) : f(x,y,z)=0\}$ can be written as $U\cap \{(x,y,g(x,y)\}$ with a smooth $g$.

  • Existence-Uniqueness theorem. The ODE problem $\dot y =g(x,y)$, $y(0)=0$, has a unique solution provided that $g$ is smooth in a neighborhood of $(0,0)$.

Indeed, the smoothness of $g$ implies the continuity and Lipschitz continuity conditions in the existence-uniqueness theorem.

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