Let $f: (a,b) \times \mathbb{R} \rightarrow \mathbb{R}$ be of class $C^1$ in $D:=(a,b) \times \mathbb{R}$ and satisfies condition $| f(t,x)| \leq A+B|x| \textrm{ for } (t,x) \in D,$ where $A,B$ are fixed real constants and let $t_0 \in (a,b)$.
How to prove using the fixed point method that for arbitrary $x_0\in \mathbb{R}$ there exist exactly one solution $x: (a,b)\rightarrow \mathbb{R}$ of differential equation $\frac{dx}{dt}=f(t,x) $ with condition $x(t_0)=x_0$ ?
Thanks.
Added.
Maybe it would be. Let $X=\{x:(a,b) \rightarrow \mathbb{R}: \sup_{t\in (a,b)} e^{-B\gamma|t-t_0|} |x(t)| <\infty, x(t_0)=x_0 \}$, $d(x,y)=\sup_{t\in (a,b)} e^{-B\gamma|t-t_0|} |x(t)-y(t)|$ for $x,y \in X$, where $\gamma$ is a suitable positive constant. Then $(X,d)$ is a complete metric space and $Tx(t):=x_0+\int_{t_0}^t f(s,x(s))ds$, for $x \in X$ and $t\in (a,b)$, maps X into itself (because $|f(s,x(s))|\leq A+Be^{B\gamma|t-t_0|}\cdot sup_{t\in (a,b)} |x(s)|e^{-B\gamma|t-t_0|} |x(t)|$ and $| \int_{t_0}^t e^{B \gamma |s-t_0|} ds| \leq \frac{1}{B \gamma} e^{B\gamma|t-t_0|}$). However I don't know is it $T$ a contraction with some $\gamma>0$ and whether or not each solution of the differential equation belongs to $X$.