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)+(-24A-6B)sin(4t)
Compare coefficients of cos(4t) and sin(4t):
-24A-6B=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.
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
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