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Finding steady-state solution to 2nd-order DE

  • #1

Homework Statement


I want to get the particular solution for:

Lq''+Rq'+(1/c)q = Esin(xt)

Homework Equations


Lq''+Rq'+(1/c)q = Esin(xt)

The Attempt at a Solution


I got the particular solution and its derivatives:

q = A*Sin(xt) + B*cos(xt)
q' = x*A*cos(xt) - x*B*sin(xt)
q'' = -x^2*A*sin(xt) - x^2*B*cos(xt)

I plug these into the 2nd order equation giving me:

A[-x^2*L*sin(xt) + R*x*cos(xt) + (1/c)*Sin(xt)] + B[-x^2*L*cos(xt)-x*R*sin(xt)+(1/c)*cos(xt)] = Esin(xt)

I want to get the form of A and B in the attached image

Also, alpha = x in my typing.
 

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Answers and Replies

  • #2
Orodruin
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Identifythe factors of sin(xt) on both sides of the wquation and then do the same for cos(xt). That gives you two equations for two unknowns.
 
  • #3
Identifythe factors of sin(xt) on both sides of the wquation and then do the same for cos(xt). That gives you two equations for two unknowns.
I'm not sure if I understand. So I factor out sin(xt) on both sides and factoring cos(xt) giving me:

sin(xt)*[-L*x^2*A - R*x*B + (1/c)*A] + cos(xt)*[-L*x^2*B + R*x*A + (1/c)*B]
 
  • #4
Orodruin
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That's just one side of the equation. You need to do it on both sides and then identify the factors on either side.
 
  • #5
That's just one side of the equation. You need to do it on both sides and then identify the factors on either side.
But the other side is just Esin(xt).

So I have:

(-L*x^2*B) + (R*x*A)+([1/c]*B) = 0
(-L*x^2*A) - (R*x*B) + ([1/c]*A) = E
 
  • #6
Orodruin
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Right. Solve for A and B.
 
  • #7
Right. Solve for A and B.
I got the answer, thank you.
 

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