An autonomous (strange) Cauchy problem 
Let $f \colon \mathbb R \to \mathbb R$ be defined by  $$
f(y)=\begin{cases} y & \text{ if } y \le 1 \\ 1 & \text{ if } y>1
\end{cases} $$
I have to determine explicity the solution $y_a(\cdot)$ of the
  Cauchy problem  $$\tag{CP} \begin{cases} y'=f(y) \\  y(0)=a 
 \end{cases} $$ where $a>0$.

Well, the function
$$
y_1(t) =
\begin{cases}
e^t & \text{ if } t \le 0 \\
t+1 & \text{ if } t > 0
\end{cases}
$$
works for $a=1$ (it is continuous and differentiable and satisfies (CP) for $a=1$). 
Have you got any ideas to treat the case $a\ne 1$? What can we do?
 A: If $a > 1$, then we have the following initial value problem (IVP)
$$\dot{y} (t) = 1, \qquad{} y (0) = a$$
whose solution is $y (t) = a + t$ for all $t \geq 0$. If $a \leq 1$, the we have the IVP
$$\dot{y} (t) = y, \qquad{} y (0) = a$$
whose solution is $y (t) = a \, e^t$ for all $t \in \ [0, t_*]$, where the time instant $t_*$ is when $y$ reaches $1$, i.e., $y (t_*) =1$. We obtain $t_* = \ln(1 / a)$. For $t \geq t_*$ we have another IVP
$$\dot{y} (t) = 1, \qquad{} y (t_*) = 1$$
whose solution is $y (t) = 1 + (t- t_*)$ for all $t \geq t_*$. Finally, we have that for $a \leq 1$ the solution is
$$y (t) = \left\{\begin{array}{cl} a \, e^t & \textrm{if} \quad{} t \in [0, \ln(1/a)]\\ 1 + (t - t_*) & \textrm{if} \quad{} t \geq \ln(1/a)\end{array}\right.$$ 
A: Let  $G_a:(0,\infty) \to \mathbb{R}$ be such that
$$
G_a(a)=0,\ G_a'(s)=\frac{1}{f(s)} \ \forall s>0 
$$
i.e. 
\begin{eqnarray}
G_a(s)&=&\begin{cases} 
\ln(s/a)   & \text{ if }  0<s\le 1\\
s-1-\ln a & \text{ if } s>1
\end{cases} \ \text{ for } 0<a \le 1\\
G_a(s)&=&\begin{cases} 
\ln s -a+1   & \text{ if }  0<s\le 1\\
s-a & \text{ if } s>1
\end{cases} \ \text{ for } a>1
\end{eqnarray}
Then $G_a$ is bijective with $G_a^{-1}: \mathbb{R} \to (0,\infty)$ defined by
\begin{eqnarray}\tag{1}
G_a^{-1}(\tau)&=&\begin{cases} 
ae^\tau      & \text{ if } \tau \ge -\ln a\\
\tau+1+\ln a & \text{ if } \tau <   -\ln a
\end{cases} \ \text{ for } 0 < a \le 1\\
G_a^{-1}(\tau)&=&\begin{cases} 
e^{\tau+a-1}      & \text{ if } \tau \ge 1-a\\
\tau+a & \text{ if } \tau <   1-a
\end{cases} \ \text{ for } a > 1
\end{eqnarray}
Hence
$$
y_a'=f(y_a),\ y_a(0)=a \iff t=\int_a^{y_a(t)}G_a'(s)ds=G(y_a(t)),
$$
i.e. 
$$
y_a(t)=G_a^{-1}(t),
$$
where $G_a^{-1}$ is given by (1).
