# If $\frac{dy}{dx}=A(x)y+B(x)$ and if $A(x)$, $B(x)$ are bounded and integrable, show the fixed point theorem solves the initial value problem

Given the linear differential equation $\frac{dy}{dx}=A(x)y+B(x)$, show that if $A(x)$ and $B(x)$ are bounded and integrable on $I=\{x|a \leq x \leq b\}$, then the fixed point theorem yields a solution to the initial value problem on $I$.

The following is the fixed point theorem:

If f is a mapping on a complete metric space S into S such that $d(f(x),f(y)) \leq kd(x,y)$ for all $x,y \in S$ with $0<k<1$, then the mapping has a unique fixed point.

Let $M=\|A(t)\|_\infty$ and choose $\delta>0$ such that $k\equiv M\delta<1$. Define $S=C[a,a+\delta]$ with norm $\|\cdot\|=\|\cdot\|_\infty$. Define $T: S\to S$ by $$Ty=\int_a^x(A(t)y(t)+B(t))dt.$$ Then, for $y_1,y_2\in S$, $$\|Ty_1-Ty_2\|=\bigg|\int_0^xA(t)(y_1(t)-y_2(t))dt\big|\le M\|y_1-y_2\|(x-a)\le k \|y_1-y_2\|.$$ Now you can use the fixed point theorem for $T$ in $S$. If $\delta\ge b$, done. If $\delta<b$, choose $n$ such that $b\in(a+(n-1)\delta, a+n\delta)$ and you use the same method in $[a+\delta,a+2\delta]$, $\cdots$, $[a+(n-1)\delta, a+n\delta]$.