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?
differential-equations complex-numbers
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up vote
0
down vote
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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?
differential-equations complex-numbers
add a comment |Â
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?
differential-equations complex-numbers
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
differential-equations complex-numbers
asked Sep 7 at 1:59
sangstar
727213
727213
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