Question

Reread Example 3 and then discuss why we can conclude that the interval of definition of the explicit solution of the IVP (the blue curve in Figure 2.4.1) is $\left(-1,1\right)\mathbf{.}$

Short answer

It cannot pass -1 or 1 because the function would be undefined there.

Step-by-step solution

The solution of the initial condition

The solution$y^2=\frac{(c+cos[2\left]\right(x\left)\right)}{1-x^2}$ is well defined for $x\ne -1,x\ne 1$. The initial condition starts between -1 and 1, so the solution would remain in $⟨-1,1⟩$. It cannot pass -1 or 1 because the function would be undefined there.

Final proof

It cannot pass -1 or 1 because the function would be undefined there.