Q25
For the differential equation d^2 x d t^2 + 6 dx dt + 8 x = 0 with initial conditions x(0) = 1 and . dx dy |_ t = 0 = 0, the solution is
For the differential equation with initial conditions x(0) = 1 and the solution is
A.
x(t) = 2e-6t - e-2t
B.
x(t) = 2e-2t - e-4t
AnswerC.
x(t) = -e-6t + 2e-4t
D.
x(t) = e-2t + 2e-4t
Answer: Option B
Solution
Answer: Option B
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