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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?

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up vote 3 down vote accepted

You already got the solution for $0 \le t \le 1$. What is $y(1)$ in that solution? Now solve on $1 < t < \infty$ with that as initial condition.

Actually there's a technical quibble here. The solution you get is not differentiable at $t=1$: it has one-sided derivatives from left and right there, but they are different. So it's not really a solution of the differential equation at $t=1$. It's what is called a weak solution.

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