Question

Solve this initial value problem using Laplace transformations. y''-y'-2y=-8cos(t)-2sin(t)
y(pi/2)=1 ; y'(pi/2)=0

Answer 1

how much of this can you do?

otherwise asked, which step is bothering you?

Answer 2

I have this much done, but I think I did something wrong.

Answer 3

looks good except you have to take the laplace of both sides.
so you have to take the laplace of -8cost - 2sint

your either manipulating a function entirely of s or entirely of t. they are never mixed

Answer 4

do you have to replace t with t+\[\pi\]\[\div\]2?

Answer 5

ive actually never encountered the case where the boundary conditions aren't y(0) and y'(0) but instead y(pi/2)

@satellite73 might know about this. i'll look it up for myself too

Answer 6

ok cool thanks.

Answer 7

i looked it up. glad i did.

yes you do need to replace it by it but in this problem, i'm pretty convinced that it is inconsequential besides the final answer being replaced by y(t - pi/2) instead of y(t)

it would have real consequences throughout the process only if f(t) [on the right side of the otherwise homogeneous equation] had any t^n terms. in that case you would need to expand (t-pi/2)^n

{last example on this page: http://tutorial.math.lamar.edu/Classes/DE/IVPWithLaplace.aspx }