an implicit solution is valid to show that the solution exists??

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let us suppose we have a differential equation

$$ \frac{d^{2}y(x)}{dx^{2}}+F(y(x), y')=0 $$

for a knownn function $ F(x) $ which

assume we know that the INVERSE of the solution is

$$ y^{-1}(x)=x+g(x) $$

however $ g(x) $ is a very complicate function, sometimes $ g'(x) >0 $ and other times $ g(x) <0 $

so we do not know an ANALYTIC exact equation

my question is , from the very RIGOROUS mathematic is my solution (1) valid or i should prove that at least the inverse function theorem holds on a certain interval $ (a,b) $ what happens if the inverse function theorem does not hold??

assume also we can NOT use 'Lagrange theorem' to invert the function as a Taylor prower series because the function $ g(x) $ may not be differentiable on certain points or it may be piecewise differentiable

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I'll assume from the form of the splitting that $g(0)=0$ and $g'(0)=0$ or more generally that $g(s)=\mathcal o(s)$? $$ x=y+g(y) $$ is invertible on a neighborhood of $0$ if one assumes that $g$ is Lipschitz with constant $L<1$ on a ball $B(0,R)$. Because then you can consider the iteration $$ y_0=x, \quad y_{n+1}=x-g(y_n) $$ which is contractive and a self-map of $B(0,R)$ for $x\in B(0,r)$ with $r=(1-L)R$ because of
$$ \|y_{n+1}\|\le \|x\|+L\|y_n\|\le r+LR\le R. $$ So one gets an inverse map $y=y(x):B(0,r)\to B(0,R)$ which is Lipschitz as well with constant $\frac1{1-L}$


So you will have to prove that either $g(s)=\mathcal O(s^2)$ or that $|g'(s)|\le L<1$ for $s\approx 0$ or something similar that amounts to the above assumptions.