Is the Solution to y'=\sqrt{y^2+x^2+1} Defined for All x and Greater Than sinhx?

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Consider the problem

y'=\sqrt{y^2+x^2+1}
y(0)=0

Prove that the solution is defined for all x \in \mathbb{R} and that y(x) \geq \sinh (x) \forall x \geq 0
 
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Malmstrom said:
Consider the problem

y'=\sqrt{y^2+x^2+1}
y(0)=0

Prove that the solution is defined for all x \in \mathbb{R} and that y(x) \geq \sinh (x) \forall x \geq 0
Is this homework?
 
Not really, I just don't know where to start from.
 

Thread 'Volterra equation of the second kind'

There is the following linear Volterra equation of the second kind $$ y(x)+\int_{0}^{x} K(x-s) y(s)\,{\rm d}s = 1 $$ with kernel $$ K(x-s) = 1 - 4 \sum_{n=1}^{\infty} \dfrac{1}{\lambda_n^2} e^{-\beta \lambda_n^2 (x-s)} $$ where $y(0)=1$, $\beta>0$ and $\lambda_n$ is the $n$-th positive root of the equation $J_0(x)=0$ (here $n$ is a natural number that numbers these positive roots in the order of increasing their values), $J_0(x)$ is the Bessel function of the first kind of zero order. I...
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