Extra negative in in Laplace transform solution

[abalmos]

I have the correct solution of this problem ($$cos(t)+u_{3\pi}[1-cos(t-3\pi)]$$). As you can tell from my solution below I am off by a extra negative sign on the cos(t) coefficient. I have had this same issue on several other problems, so I believe I must be missing something on the procedure. Any guidance would be much appreciated!

Solve IVP $$y'' + y = u_{3\pi}(t) \left\{y(0)=1, y'(0)=0$$

My solution:
$$s^{2}Y(s) + sY(0) + Y'(0) + Y(s) = \frac{e^{-3\pi s}}{s}$$
$$Y(s)(s^{2} +1) + s = \frac{e^{-3\pi s}}{s}$$
$$Y(s) = \frac{e^{-3\pi s}}{s(s^{2}+1)} - \frac{s}{(s^{2}+1)}$$

$$H(t) = \frac{1}{s(s^{2}+1)}$$
$$h(t) = Y(H(t))$$

$$y = u_{3\pi}h(t-3\pi}) - cos(t)$$

$$h(t) = \frac{1}{s} - \frac{s}{s^{2}+1}$$
$$h(t) = 1 - cos(t)$$

so ...

$$-cos(t) + u_{3\pi}[1-cos(t-3\pi)]$$


I have seen in other examples (http://tutorial.math.lamar.edu/Classes/DE/IVPWithStepFunction.aspx , example 2) which have similar setups, and I noticed that when the little h is introduce (the inverse transformation to the $$u_{3\pi}$$ term) the sign of the $$cos(t)$$ term changes but I do not understand why.

Thank you so much for help and guidance you can provide!

- Andrew Balmos

 

[mighty2000]

Dear Andrew,

Thank you for sharing your solution and your question. It seems like you have a good understanding of the procedure for solving this type of problem, but you may be missing a key concept when it comes to the step function.

When we use the Laplace transform to solve differential equations with step functions, we are essentially splitting the solution into two parts: one for the case where the step function is "on" (equal to 1) and one for when it is "off" (equal to 0). In your solution, you correctly identified the function h(t) that corresponds to the step function H(t), but you may have missed that h(t) is only valid for t > 0. This means that for t < 0, the solution should be 0, since the step function is "off" and does not affect the solution.

So, when you introduce the inverse transformation h(t) and plug it into the solution for y, you need to take into account the fact that for t < 0, h(t) = 0. This is why the sign of the cos(t) term changes - it is being multiplied by 0 for t < 0. In the solution you referenced from Lamar University, they explicitly mention this and include a piecewise function to handle this case.

In summary, the extra negative sign on the cos(t) coefficient is not incorrect, it is just accounting for the case where t < 0. I hope this helps clarify things for you. Keep up the good work and don't be afraid to ask for guidance when needed. Science is a collaborative field and we all learn from each other. Best of luck in your studies!