Let $y(x)$ be the general solution of $y'(x) = A_{n x n}(x) y(x)+f (x)$, $x \in I$ where $A(x)$ and $f (x)$ have continuous entries over I. Let be $\phi_{x_o}$ the fundamental matrix in $x_o \in I$. Show that for every $x_o \in I$, we have $$|y(x)|_n \leq n \|\phi_{x_o}\|\left(|y(x_o)|_n + \int_{x_0}^x\|\phi^{-1}_{x_o}\|\cdot|f(t)|_n\,dt \right).$$
We denote $|v|_n = \sqrt{\sum_n |v_k|^2}$, where $v=(v_1,v_2,...,v_n)^t$ and the matrix norm $\displaystyle\|A\|=\max_{1\leq i \leq n}\left\{\sum_{j=1}^{n} |a_{ij}|\right\}.$
My attempt:
I know that that the general solution of the ODE is given by $$y(x)=W(x)\left(c+ \int_{x_0}^x W^{-1}(s) f(s)\,ds\right)$$ $$|y|_n \leq |W(x)c|_n + \left|W(x)\int_{x_0}^x W^{-1}(s) f(s) \,ds\right|_n$$ then $$|y|_n \leq n\|W(x)\|\,|c|_n + \int_{x_0}^x\|\phi^{-1}_{x_o}\|\cdot|f(t)|_n\, dt $$
But after that step I do not know what to do, someone could help me please?
I know the next results:
$$|Av|_n \leq n\cdot \|A\|\, |v|_n$$ if $\phi(t):[a,b] \rightarrow M_n(\mathbb{R}) $ and $f:[a,b] \rightarrow \mathbb{R}^n $ have continuous entries then $$\left|\int_a^b \phi(t) f(t) \,dt\right|_n \leq \int_a^b \|\phi(t)\|\, |f(t)|_n\, dt. $$