Find the steadystate solution having the form xCcost for the

Solution

Let x = Ccos(wt-)
=>x\' = -Cwsin(wt-)
=>x\'\' = -Cw²cos(wt-)

x\"+x\'+x = 2cos(3t)

=>-Cw²cos(wt-)-Cwsin(wt-)+Ccos(wt-) = 2cos(3t)

=>-Cw²cos(wt)cos()-Cw²sin(wt)sin()-Cwsin(wt)cos()+Cwcos(wt)sin()+Ccos(wt)cos()+Csin(wt)sin() = 2cos(3t)

=>((-Cw²cos())+(Cwsin())+(Ccos()))cos(wt) + ((-Cw²sin())+(-Cwcos())+(Csin())sin(wt) = 2cos(3t)

Comparing L.H.S. and R.H.S. we get,

(-Cw²cos())+(Cwsin())+(Ccos()) = 2, (-Cw²sin())+(-Cwcos())+(Csin() = 0, w = 3

Substituting w = 3 in other two equations we get,

(-9C+C)cos() + 3Csin() = 2, (-3C)cos() + (-9C+C)sin() = 0

=>-8Ccos() + 3Csin() = 2.........(1), -3Ccos()-8Csin() = 0.......(2)

Multiplying equation (1) with 8 on both sides we get, -64Ccos()+24Csin() = 16............(3)

Multiplying equation (2) with 3 on both sides we get, -9Ccos()-24Csin() = 0............(4)

Equations (3) + (4) => -73Ccos() = 16............(5)

From (4) we get, -9Ccos() = 24Csin() =>tan() = -9/24 = -3/8 => = -20.56^o =>cos() = 0.936

From (5) we get, C = -0.234

Therefore, x = -0.234cos(3t+20.56^o)