Deriving the second order differential equation general solution with complex roots

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This is actually essentially a question on what to do with the constants. I'm trying to derive the underdamped solution for a spring.



For my trial solution $x(t) = Aexpleft(atright)$, I have



$$a = -Gamma pm omega$$



$$therefore x(t) = Aexp(left(-Gamma + omega)tright) + Bexp(left(-Gamma - omega)tright)$$



Anyway, I eventually get to (and I'm pretty sure my math is right):



$$x(t) = exp(-Gamma t) left ((A+B)cos(omega t) + i(A-B)sin(omega t)right)$$



So, if I let $A+B = C$, that looks nicer.



However, I feel tentative about just saying $$i(A-B) = D$$ without explicitly saying that $D$ is complex, and even then I don't think that's right. I don't think there's supposed to be a complex part to this solution - so what do I do? How do I turn this into a real constant?










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    This is actually essentially a question on what to do with the constants. I'm trying to derive the underdamped solution for a spring.



    For my trial solution $x(t) = Aexpleft(atright)$, I have



    $$a = -Gamma pm omega$$



    $$therefore x(t) = Aexp(left(-Gamma + omega)tright) + Bexp(left(-Gamma - omega)tright)$$



    Anyway, I eventually get to (and I'm pretty sure my math is right):



    $$x(t) = exp(-Gamma t) left ((A+B)cos(omega t) + i(A-B)sin(omega t)right)$$



    So, if I let $A+B = C$, that looks nicer.



    However, I feel tentative about just saying $$i(A-B) = D$$ without explicitly saying that $D$ is complex, and even then I don't think that's right. I don't think there's supposed to be a complex part to this solution - so what do I do? How do I turn this into a real constant?










    share|cite|improve this question























      up vote
      0
      down vote

      favorite









      up vote
      0
      down vote

      favorite











      This is actually essentially a question on what to do with the constants. I'm trying to derive the underdamped solution for a spring.



      For my trial solution $x(t) = Aexpleft(atright)$, I have



      $$a = -Gamma pm omega$$



      $$therefore x(t) = Aexp(left(-Gamma + omega)tright) + Bexp(left(-Gamma - omega)tright)$$



      Anyway, I eventually get to (and I'm pretty sure my math is right):



      $$x(t) = exp(-Gamma t) left ((A+B)cos(omega t) + i(A-B)sin(omega t)right)$$



      So, if I let $A+B = C$, that looks nicer.



      However, I feel tentative about just saying $$i(A-B) = D$$ without explicitly saying that $D$ is complex, and even then I don't think that's right. I don't think there's supposed to be a complex part to this solution - so what do I do? How do I turn this into a real constant?










      share|cite|improve this question













      This is actually essentially a question on what to do with the constants. I'm trying to derive the underdamped solution for a spring.



      For my trial solution $x(t) = Aexpleft(atright)$, I have



      $$a = -Gamma pm omega$$



      $$therefore x(t) = Aexp(left(-Gamma + omega)tright) + Bexp(left(-Gamma - omega)tright)$$



      Anyway, I eventually get to (and I'm pretty sure my math is right):



      $$x(t) = exp(-Gamma t) left ((A+B)cos(omega t) + i(A-B)sin(omega t)right)$$



      So, if I let $A+B = C$, that looks nicer.



      However, I feel tentative about just saying $$i(A-B) = D$$ without explicitly saying that $D$ is complex, and even then I don't think that's right. I don't think there's supposed to be a complex part to this solution - so what do I do? How do I turn this into a real constant?







      differential-equations complex-numbers






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      asked Sep 7 at 1:59









      sangstar

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