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.
 
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