Proof
Let\(\chi\left(t\right)=\ \Theta\left(t\right)e^{-\Phi\left(t\right)}+\int_{0}^{t}{b\left(s\right)e^{-\Phi\left(s\right)}}\text{ds}\)
Then \(\chi\) is differentiable, and an application of the chain rule shows that
\(\dot{\chi}\left(t\right)=\left(\dot{\Theta}\left(t\right)-\Theta(t)\varphi\left(t\right)\right)e^{-\Phi\left(t\right)}+b\left(t\right)e^{-\Phi\left(t\right)}=(\dot{\Theta}\left(t\right)-\varphi\left(t\right)\Theta(t)-b\left(t\right))e^{-\Phi\left(t\right)}\ \leq 0\)
The differential inequality (6) means in particular,\(\chi\left(t\right)\leq\chi\left(0\right)\ ,\ \ \ \forall t\in\left[0,t\right.\ )\)
and the claim follows.