Nonhomogeneous: Undetermined coefficients

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SUMMARY

The discussion centers on solving the nonhomogeneous ordinary differential equation (ODE) given by (d²x/dt²) + (ω²)x = Fsin(ωt) using the method of undetermined coefficients. Participants emphasize the importance of first solving the corresponding homogeneous equation, which is x'' + ω²x = 0, to find the fundamental solution. The particular solution is proposed in the form yₚ = Axsin(ωt) + Bxcos(ωt), noting that if the fundamental solution includes sine and cosine, the particular solution must include terms multiplied by x. The confusion regarding the roots of the homogeneous equation is clarified, leading to the conclusion that r = ±ωi.

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andrewdavid
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(d^2x/dt^2)+(w^2)x=Fsin(wt), x(0)=0,x'(0)=0

Hope that's readable. First part is second derivative of x with respect to t. w is a constant and F is a constant. I need to find a solution to this using method of undetermined coeffecients and I'm confused with all the different variables. Anyone get me started at least?
 
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Well, first off start by solving the homogenous equation to find the fundamental solution.

\ddot{x} + \omega^{2}x = 0

After that try a Particular solution of the type

y_{p} = A x \sin(\omega t) + B x\cos(\omega t)

Remember that if the fundamental solution has already sin and cos, you will need to try a xsin and xcos, like this case.
 
Last edited:
I got my homogenous equation x''+(w^2)x=0 but I can't find my roots with that w^2 in there.
 
What seems to be the problem? Show me your work.
 
Here, i will start you off

\ddot{x} + \omega^{2}x = 0

we assume a as a solution

x(t) = e^{rt}

So we substitute in our ODE

r^{2}e^{rt} + \omega^{2}e^{rt} = 0

so

e^{rt}(r^{2} + \omega^{2}) = 0

because e^{rt} cannot be equal to 0

r^{2} + \omega^{2} = 0

which ends up as

r = \pm \omega i
 
Last edited:
I figured it out, thanks a lot for your help, I was just being dumb.
 

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