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**Belowzero78** Question: Use Laplace transforms to solve the initial value problem $\displaystyle \frac{dy}{dt} + y = f(t) $, where $\displaystyle f(t)=\left\{\begin{array}{cc}3&\mbox{ for }

0\leq t <7\\-3 & \mbox{ for } t\geq7\end{array}\right. $, and $\displaystyle y(0) = -5 $.

My Solution is: $\displaystyle f(t)=\left\{\begin{array}{cc}-5e^{-t}&\mbox{ for }

0\leq t <7\\-5e^{-t} + 6e^{7-t} - 6 & \mbox{ for } t\geq7\end{array}\right. $

I am confident for the first interval $\displaystyle 0\leq t < 7 $ that the answer is correct, but since i got the incorrect answer for the 2nd part, i was wondering since at $\displaystyle t\geq7 $, when $\displaystyle t = 7 $, the function $\displaystyle y(t) $ is still continuous, but not differentiable. What should be the correct answer and whoever contributes, could you please explain why?

Hope to hear from you all soon!