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The nonlinear existence and uniqueness theorem looks at the ODE $y'=f(t,y)$ and requires continuity of the functions $f$ and $\frac{\partial f}{\partial y}$ to apply.

Problem: Use the nonlinear existence and uniqueness theorem to find where in the $ty$-plane the IVP $$y'=\dfrac{1}{\sqrt{1-t^2-y^2}};y(t_0)=y_0$$ is guaranteed to exist.
Solution: We just need to figure out when the functions $f(t,y)=\dfrac{1}{\sqrt{1-t^2-y^2}}$ and $\dfrac{\partial f}{\partial y} = \dfrac{-y}{\sqrt{1-t^2-y^2}}$ are continuous (note they are continuous at the same places). The "bad places" are defined when $1-t^2-y^2=0$ or when $1-t^2-y^2<0$. Rearrangement of these "bad place formulas" yields $$t^2+y^2 \geq 1,$$ or in other words, the "bad places" are outside of the circle of radius $1$ centered at $(0,0)$ in the $ty$-plane. Everywhere else in the plane is "good", i.e. the existence and uniqueness theorem applies to the interior of the unit disk in the $ty$-plane, i.e. any initial condition of the form $y(t_0)=y_0$ must obey $y_0^2+t_0^2 < 1$ will guarantee existence and uniqueness.