First-order linear differential inequality

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.