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December 20, 2014

December 20, 2014

Posted by **shasha** on Monday, March 14, 2011 at 7:36am.

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 equations-missing one initial condition -
**MathMate**, Monday, March 14, 2011 at 8:57amNormalize 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.

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