We want to show that the solution to the initial value problem $\dot x = (x-3)(1-3t^2)\cos(10000(4x-t))$, $x(1) = 4$ is bounded by $3+e^{t-t^3} < x(t) < 3 + e^{t^3-t}$ when $t> 1$
What I tried:
Firstly I attempted to prove the upper bound. If $t > 1$ then $1-3t^2$ is negative and so $\dot x \leq(x-3)(3t^2-1) \leq (3t^2-1)x$
So $x(t)$ does not increase as fast as $e^{t^3-t}$, which trivially means it doesn't increase as fast as $3 + e^{t^3-t}$
What now? Is this the right direction?
If we had $x'=(x-3)(3t^2-1)$, then $$\frac{1}{x-3}x'= 3t^2-1$$ Integrating both sides with respect to $t$, $$\int\frac{1}{x-3}dx=\int(3t^2-1 )dt$$ $$\log{(x-3)}= t^3-t+c$$ $$x-3= Ae^{t^3-t}$$ $$x=3+ Ae^{t^3-t}$$ Where $A=e^c$. We have the inital condition $x(1)=4$, so $A=1$. $$\therefore x=3+ e^{t^3-t}$$ We can similarly find the lower bound by using $x' \ge (3-x)(3t^2-1)$