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

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SUMMARY

The differential equation y' = √(y² + x² + 1) with the initial condition y(0) = 0 has a solution that is defined for all x ∈ ℝ. The solution y(x) satisfies the condition y(x) ≥ sinh(x) for all x ≥ 0. This conclusion is derived from analyzing the behavior of the function and its derivatives, confirming that the solution remains bounded and adheres to the specified inequality.

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Malmstrom
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Consider the problem

[tex]y'=\sqrt{y^2+x^2+1}[/tex]
[tex]y(0)=0[/tex]

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

[tex]y'=\sqrt{y^2+x^2+1}[/tex]
[tex]y(0)=0[/tex]

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

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