Prove $x(t)\geq x(0)e^{\int_{0}^{t} y(s) ds}$?

Question

Given conditions on $x(t)$ and $y(t)$

1. $x(t)$ and $y(t)$ are both infinitely many times differentiable.

2. $x(t)\geq0$ and $y(t)\geq0$ for all $t\geq0$.

3. $x(t)\geq x(0) + $$\int_{0}^{t} y(s)x(s) ds$

Then I need to prove that $x(t)\geq x(0)e^{\int_{0}^{t} y(s) ds}$

I am just confused how to start, anyone give me some ideas so that I could understand how to begin?

Answer 1

\begin{align*} \left(\dfrac{\displaystyle\int_{0}^{t}y(s)x(s)ds}{\exp\left(\displaystyle\int_{0}^{t}y(s)ds\right)}\right)'&=\dfrac{y(t)x(t)-\left(\displaystyle\int_{0}^{t}y(s)ds\right)y(t)}{\exp\left(\displaystyle\int_{0}^{t}y(s)ds\right)}\\ &\geq\dfrac{x(0)y(t)}{\exp\left(\displaystyle\int_{0}^{t}y(s)ds\right)}, \end{align*} so \begin{align*} \dfrac{\displaystyle\int_{0}^{t}y(s)x(s)ds}{\exp\left(\displaystyle\int_{0}^{t}y(s)ds\right)}&\geq x(0)\int_{0}^{t}\dfrac{y(u)}{\exp\left(\displaystyle\int_{0}^{u}y(s)ds\right)}du\\ &=x(0)\int_{0}^{t}y(u)\exp\left(-\int_{0}^{u}y(s)ds\right)du\\ &=x(0)\left(-\exp\left(-\int_{0}^{u}y(s)ds\right)\right)\bigg|_{u=0}^{u=t}\\ &=-x(0)\exp\left(-\int_{0}^{t}y(s)ds\right)+x(0), \end{align*} so \begin{align*} \dfrac{x(t)-x(0)}{\exp\left(\displaystyle\int_{0}^{t}y(s)ds\right)}\geq-x(0)\exp\left(-\int_{0}^{t}y(s)ds\right)+x(0), \end{align*} and we get \begin{align*} x(t)\geq x(0)\exp\left(\int_{0}^{t}y(s)ds\right). \end{align*}