**Unformatted text preview: **1.2
+ 2x = 5 cos 4t ,
2
5 dt
dt 1
x (0) = , x (0) = 0
2 Example 6 - Continued
Given the solution
x (t ) = e −3t 25
86
50
38
cos t −
sin t + −
cos 4t +
sin 4t
51
51
102
51
transient steady −state transient term : xc (t ) s.t. limt →∞ xc (t ) = 0
steady-state term: xp (t ) Example 7 - Transient/Steady-State Solutions
The solution of
d 2x
dx
+2
+ 2x = 4 cos t + 2 sin t ,
2
dt
dt x (0) = 0, x (0) = x1 is given by
x (t ) = (x1 − 2)e −t sin t + 2 sin t Undamped Forced Motion: β = 0
Solve the IVP
d 2x
+ ω 2 x = F0 sin γ t ,
dt 2 x (0) = 0, x (0) = 0 Example
A mass weighing 16 pounds stre...

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