Posted by shasha on Monday, March 14, 2011 at 7:36am.
The differential equation that governs the forced oscillation is shown below:
0.2 d²y/dt² + 1.2 dy/dt +2y = r(t) where r(t) is the external force.
Given that r(t) = 5 cos 4t with y(0) = 0 . find the equation of motion of the forced oscillations

Differential equationsmissing one initial condition  MathMate, Monday, March 14, 2011 at 8:57am
Normalize the equation by multiplying by 5:
0.2 d²y/dt² + 1.2 dy/dt +2y = 5 cos(4t) = r(t)
to:
d²y/dt² + 6 dy/dt + 10y = 25cos(4t)
Find the complementary solution:
m²+6m+10=0
m=3±i
So the solution to the homogeneous equation is:
yc=e^(3t)(C1*cos(t)+C2*sin(t))
Now find the particular solution by undetermined coefficients:
Assume the particular solution to be:
yp=Acos(4t)+Bsin(4t)
and substitute in y of the the original equation:
d²yp/dt² + 6 dyp/dt + 10yp = 25cos(4t)
16Acos(4t)16Bsin(4t)
+6(4Bcos(4t)4Asin(4t))
+10Acos(4t)+10Bsin(4t)
=(6A+24B)cos(4t)+(24A6B)sin(4t)
Compare coefficients of cos(4t) and sin(4t):
24A6B=0 => B=4A
6A+24B=25 => 102A=25 => A=25/102
Therefore
yp(t)=(25/102)cos(4t)+(100/102)sin(4t)
(substitute in homogeneous equation to verify that you get 25cos(4t) )
The general solution is therefore:
y=yc+yp=e^(3t)(C1*cos(t)+C2*sin(t))(25/102)cos(4t)+(100/102)sin(4t)
Initial conditions:
To solve the second order problem completely, you'll need two initial conditions. We are givn y(0)=0 at t=0.
We need another one (such as y'(0)=5 at t=0).
Substitute the initial conditions into the general solution above and solve for C1 and C2 to give the final solution of the initial value problem.