If then, .
Regards
Fernando Revilla
Question: Use Laplace transforms to solve the initial value problem , where , and .
My Solution is:
I am confident for the first interval that the answer is correct, but since i got the incorrect answer for the 2nd part, i was wondering since at , when , the function 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!
If then, .
Regards
Fernando Revilla
The function does not satisfy the differential equation, so there must be some mistake.
Regards.
Fernando Revilla
Take into account that your solution does not satisfy the initial condition . You should obtain .
Regards.
Fernando Revilla