I am looking to solve the following integro-differential inequality: $$\dot{V}(t)\leq AV(t)+B\int_{0}^{t} V(\tau)d\tau+C_{2}t+D,$$ where $A$, $B$, $C_2$, and $D$ are known positive constants.
Things to know:
- $V(t)\geq0$,
- $\dot{V}(t)\leq0$,
- $V(0)\leq\overline{V}_{1}$,
- $\dot{V}(0)\leq\overline{V}_{2}$,
where, $\overline{V}_{1}$ and $\overline{V}_{1}$ are known constants.
My initial thought is to differentiate the integro-differential inequality by applying Leibniz's integral rule and the fundamental theorem of calculus: $$\ddot{V}(t)\leq A\dot{V}(t)+B\int_{0}^{t}\dot{V}(\tau)d\tau+C_{2}$$ $$\ddot{V}(t)\leq A\dot{V}(t)+B\left(V(t)-V(0)\right)+C_{2}$$ $$\ddot{V}(t)-A\dot{V}(t)-BV(t)\leq C_{2}-BV(0)$$ To simplify, define $C\triangleq C_{2}-BV(0)$. Then, I get the following linear second order non-homogeneous inequality with constant coefficients: $$\ddot{V}(t)-A\dot{V}(t)-BV(t)\leq C$$
To solve this, I first find the homogeneous solution: $$\ddot{V}(t)-A\dot{V}(t)-BV(t) = 0,$$ which takes on the form $$V_{h}=c_{1}\text{e}^{r_{1}t}+c_{2}\text{e}^{r_{2}t},$$ where, $$r_{1}\triangleq\frac{-A+\sqrt{A^{2}+4B}}{2}$$ and $$r_{2}\triangleq\frac{-A-\sqrt{A^{2}+4B}}{2}.$$ Then the particular solution is determined by assuming it takes on the form $V_p=k$, which yields $$Bk\leq C.$$ Putting the homogeneous and the particular solution together, we get $$V(t)\leq c_{1}\text{e}^{r_{1}t}+c_{2}\text{e}^{r_{2}t}+\frac{C}{B}.$$ From here, we can solve for $C_1$ and $c_2$ with the initial conditions and taking the derivative of the solution: $$V(0)\leq c_{1}+c_{2}+\frac{C}{B}$$ $$\dot{V}(0)\leq c_{1}r_{1}+c_{2}r_{2}.$$
This is just algebra from here. So my question is if I addressed this problem correctly, if not, what was my error?