First-order linear differential inequality 
Let $x$ be a function of  $C^1(I,R)$ where $I\subset \mathbb{R}$ , such that  $$x'(t)\leq a(t) x(t)+b(t),$$ where $a$ and $b$ are continuous functions on  $I$ in $R$ then 
  $$ x(t)\leq x(t_0) \exp\left(\int_{t_0}^{t}a(s)ds\right)+\int_{t_0}^{t}\exp\left(\int_{s}^t a(\sigma)d\sigma\right)b(s)ds$$

How to prove this proposition please ?
Thank you
 A: I, too, first guessed this would be a problem suggesting the application of Grownall's inequality, but it seems an even more elementary solution avails itself:
Given that
$x'(t) \le a(t) x(t) + b(t), \tag 1$
we have the equivalent form
$x'(t) - a(t) x(t) \le b(t); \tag 2$
and since
$\displaystyle \exp \left (-\int_{t_0}^t a(\sigma)\; d\sigma \right )  > 0 \tag 3$
we further have
$\displaystyle \exp \left (-\int_{t_0}^t a(\sigma)\; d\sigma \right )(x'(t) - a(t) x(t)) \le  \exp \left (-\int_{t_0}^t a(\sigma)\; d\sigma \right ) b(t); \tag 4$
we observe that
$\displaystyle \left ( \exp \left (-\int_{t_0}^t a(\sigma)\; d\sigma \right ) x(t) \right )' = \displaystyle -a(t) \exp \left (-\int_{t_0}^t a(\sigma)\; d\sigma \right ) x(t) + \exp \left (-\int_{t_0}^t a(\sigma)\; d\sigma \right ) x'(t)$
$= \displaystyle \exp \left (-\int_{t_0}^t a(\sigma)\; d\sigma \right )(x'(t) - a(t)x(t)); \tag 5$
thus (4) becomes
$\displaystyle \left ( \exp \left (-\int_{t_0}^t a(\sigma)\; d\sigma \right ) x(t) \right )' \le  \exp \left (-\int_{t_0}^t a(\sigma)\; d\sigma \right ) b(t); \tag 6$
we integrate (6) 'twixt $t_0$ and $t$:
$\displaystyle  \exp \left (-\int_{t_0}^t a(\sigma)\; d\sigma \right ) x(t)  - x(t_0) = \int_{t_0}^t \left ( \exp \left (-\int_{t_0}^s a(\sigma)\; d\sigma \right ) x(s) \right )'\; ds$
$\le \displaystyle \int_{t_0}^t  \exp \left (-\int_{t_0}^s a(\sigma)\; d\sigma \right ) b(s) \; ds, \tag 7$
whence
$\displaystyle  \exp \left (-\int_{t_0}^t a(\sigma)\; d\sigma \right ) x(t) \le x(t_0) + \displaystyle \int_{t_0}^t  \exp \left (-\int_{t_0}^s a(\sigma)\; d\sigma \right ) b(s) \; ds, \tag 8$
which we multiply through by
$\displaystyle  \exp \left (\int_{t_0}^t a(\sigma)\; d\sigma \right ) > 0 \tag 9$
to obtain
$x(t) \le$
$\displaystyle x(t_0) \exp \left (\int_{t_0}^t a(\sigma)\; d\sigma \right )$
$+ \exp \left (\displaystyle \int_{t_0}^t a(\sigma)\; d\sigma \right ) \displaystyle \int_{t_0}^t  \exp \left (-\int_{t_0}^s a(\sigma)\; d\sigma \right ) b(s) \; ds; \tag{10}$
finally,
$\displaystyle \exp \left (\int_{t_0}^t a(\sigma)\; d\sigma \right ) \int_{t_0}^t  \exp \left (-\int_{t_0}^s a(\sigma)\; d\sigma \right ) b(s) \; ds$
$= \displaystyle \int_{t_0}^t  \exp \left (\int_{t_0}^t a(\sigma)\; d\sigma \right ) \exp \left (-\int_{t_0}^s a(\sigma)\; d\sigma \right ) b(s) \; ds, \tag{11}$
and since
$\displaystyle  \int_{t_0}^t  \exp \left (\int_{t_0}^t a(\sigma)\; d\sigma \right ) \exp \left (-\int_{t_0}^s a(\sigma)\; d\sigma \right ) b(s) \; ds$
$= \displaystyle \int_{t_0}^t  \exp \left (\int_{t_0}^t a(\sigma)\; d\sigma -\int_{t_0}^s a(\sigma)\; d\sigma \right ) b(s) \; ds$
$= \displaystyle \int_{t_0}^t  \exp \left (\int_s^t a(\sigma)\; d\sigma \right ) b(s) \; ds, \tag{12}$
(10) becomes
$x(t) \le \displaystyle x(t_0) \exp \left (\int_{t_0}^t a(\sigma)\; d\sigma \right ) + \int_{t_0}^t  \exp \left (\int_s^t a(\sigma)\; d\sigma \right ) b(s) \; ds, \tag{14}$
which, since $s$ and $\sigma$ are in fact merely so-called "dummy" variables of integration, is indeed the desired result.
A: You can solve the differential equation 
$x′(t)+d(t)= a(t)x(t)+b(t)$
for some $d\geq 0$. When you get the explicit solution, just recall the sign $d$ had.
