Differential Equations with Discontinuous Forcing Functions

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SUMMARY

The forum discussion focuses on solving the initial-value problem for the differential equation y'' = 1 - u(t-1) with initial conditions y(0) = 0 and y'(0) = 0. The user applied the Laplace transform, resulting in the equation Y(s) = (1 - e^{s})/s^{3}. The correct solution, as provided in the textbook, is y(t) = (1/2)t^{2} - (1/2)u(t-1)(t-1)^{2}. The user initially struggled with the solution but later identified a mistake related to factoring their answer.

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Homework Statement


Solve the given initial-value problem.
[itex]y'' = 1 - u(t-1)[/itex]
[itex]y(0) = 0[/itex]
[itex]y'(0) = 0[/itex]

Homework Equations

The Attempt at a Solution


First I took the Laplace transform of both sides:
[itex]\mathcal{L}(y'') = \mathcal{L}(1 - u(t-1))[/itex]
[itex]s^{2}Y(s) - sy(0) - y'(0) = \mathcal{L}(1) - \mathcal{L}(u(t-1))[/itex]
[itex]s^{2}Y(s) = \frac{1-e^{s}}{s}[/itex]
[itex]s^{2}Y(s) = (1-e^{s})\frac{1}{s}[/itex]
[itex]Y(s) = (1-e^{s})\frac{1}{s^{3}}[/itex]
At this point I am sort of stuck, the solution given in the back of the book is : [itex]\frac{1}{2}t^{2} - \frac{1}{2}u(t-1)(t-1)^{2}[/itex]
I'm having a hard time seeing how my work is going to end up as the solution given, so I am thinking maybe I didn't do something right here..
 
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I think I may have figured out what I was doing wrong, I forgot to factor my answer...I'll post a solution momentarily...
 
Ok, so did figure out what I was doing wrong...I'm sorry if I've wasted anyone's time
 

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