Originally Posted by
algorithm Hi,
y''(t) + 3y'(t) + y(t) = 2x(t)
Where x(t) = u(t), and the system is initially at rest. We are interested only when t>=0, so x(t) = 1 always.
I can solve it by Laplace transforms:
Original eqn:
y''(t) + 3y'(t) + y(t) = 2
Taking Laplace of both sides (and assuming system was at rest):
Y(s)[s^2 + 3s + 2] = 2/s
Y(s) = [2/s]/[s^2 + 3s + 2]
Y(s) = A/s + B/(s + 1) + C/(s + 2)
Finally: y(t) = 1 - 2exp[-t] + ext[-2t]
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The problem is when I try solving it by equating coefficients. I get a different answer:
Original eqn:
y''(t) + 3y'(t) + y(t) = 2
To get complementary function, y_comp(t):
y''(t) + 3y'(t) + y(t) = 0
Auxiliary eqn: s^2 + 3s + 1 = 0; s = -1, -2
y_comp(t) = Aexp[-t] + Bexp[-2t]
To get particular integral, y_pt(t):
Forcing function: x(t) = 1
Gives y_pt(t) = 1
General solution:
y(t) = y_comp(t) + y_pt(t) = Aexp[-t] + Bexp[-2t] + 1
Solving for the constants A and B (problem is here):
y'(t) = -Aexp[-t] -2Bexp[-2t]
y''(t) = Aexp[-t] + 4Bexp[-2t]
Given the system was at rest, I get:
0 = -A - 2B (1)
0 = A + 4B (2)
Solving these gives A=B=0. But this means that y(t) = 1, which is different to the Laplace solution of y(t) = 1 - 2exp[-t] + ext[-2t].
Thanks for any help.