This question is from DE book by Braun(Pg no 10, Q no 17),
Find a continuous solution of the initial-value problem $y'+y= g(t), y(0)= 0$ where $g(t)=\begin{cases}2, &0 \leq t\leq 1, \\0, &t > 1\end{cases} $
since the intial condition is given at (0,0), therefore we consider $g(t)=2$ and so solving $y'+y=2$ gives integrating factor $\mu(t)=Ce^t \Rightarrow y=2+Ce^{-t} \Rightarrow C=-2 \Rightarrow y=2(1-e^{-t})$, the answer given in the text it this
$y(t) = \begin{cases} 2(1-e^{-t}), &0\leq t\leq 1\\ 2(e-1)e^{-t}, &t > 1 \end{cases} $
even if we consider $g(t)=0$ we get $\ln|y|=-t+C$ and we cannot proceed further since there is no initial condition, $y(1)=?$ my question is how to solve for $y$ at $t>1$ and also, do we find left and right limit at $t=1$ to prove that the answer is a continuous function?