Show that a solution of the IVP is bounded

This would be a fairly straightforward application of Gronwall Inequality:

$$x'\cdot x = Ax\cdot x \leq M|x|^2 \implies \left(\frac{1}{2}|x|^2\right)' \leq 2M\left(\frac{1}{2}|x|^2\right)$$

$$\frac{1}{2}|x|^2 \leq \frac{1}{2}|x_0|^2 e^{2Mt} \implies |x| \leq |x_0|e^{Mt}$$

We may assume

$\vert x(t) \vert \ne 0, \; \forall t \in [0, T], \tag 0$

for if $\vert x(t) \vert = 0$ for some $t$, then uniqueness of solutions implies $x(t) = 0$ for all $t \in [0, T]$, in which case the desired result trivially holds. We may then argue as follows:

$\dfrac{d}{dt} \vert x(t) \vert^2 = \dfrac{d}{dt} \langle x(t), x(t) \rangle = 2\langle x(t), \dot x(t) \rangle$ $= 2 \langle x(t), A(t)x(t) \rangle \le 2M \langle x(t), x(t) \rangle = 2M \vert x(t) \vert^2; \tag 1$

with

$\vert x(t) \vert \ne 0, \tag 2$

(1) may be written

$\dfrac{1}{\vert x(t) \vert^2}\dfrac{d}{dt} \vert x(t) \vert^2 \le 2M; \tag 3$

that is,

$\dfrac{d}{dt} \ln \vert x(t) \vert^2 \le 2M, \tag 4$

or

$2 \dfrac{d}{dt} \ln \vert x(t) \vert \le 2M,\tag 5$

or

$\dfrac{d}{dt} \ln \vert x(t) \vert \le M, \tag 6$

which we integrate 'twixt $0$ and $t$:

$\ln \left ( \dfrac{\vert x(t) \vert}{\vert x(0)\vert} \right ) = \ln \vert x(t) \vert - \ln \vert x(0) \vert = \displaystyle \int_0^t \dfrac{d}{ds} \ln \vert x(s) \vert \; ds \le \int_0^t M \; ds = Mt; \tag 7$

thus

$ \dfrac{\vert x(t) \vert}{\vert x(0)\vert} \le e^{Mt}, \tag 8$

or

$\vert x(t) \vert \le \vert x(0) \vert e^{Mt}, \tag 9$

$OE\Delta$.