Comparison principle for differential equations

Question

I am trying to solve example 3.8 in the book Nonlinear systems by Hassan Khalil and I have been unable to figure out how they got the answer for $\frac{\mathrm dv(t)}{\mathrm dt}$ as $-2x^2(t)$. I will be grateful if someone can please explain it to me. I have attached a screenshot of the problem.

Answer 1

The term $-2x^4(t)$ is negative(non-positive actually) , so $-2x^2(t) - 2x^4(t) \leq -2x^2(t)$ since adding a negative quantity always reduces what you have first. Since $v(t) = x^2(t)$, it follows that $\dot{v}(t) \leq -2v(t)$, and by the comparison lemma for linear systems, $v(t) \leq u(t)$ and the bound on $x(t)$ follows.