Is ψ(t,0,0) = sin(2πt) a Solution to This Differential Equation?

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SUMMARY

The differential equation dx/dt = -x^3 + 2*x + sin^3(2πt) - 2*sin(2πt) + 2π*sin(2πt) does not have the solution ψ(t,0,0) = sin(2πt). To prove this, one must show that ψ' + ψ^3 - 2ψ equals sin^3(2πt) - 2*sin(2πt) + 2π*sin(2πt). However, the correct term required on the right side is 2π*cos(2πt), not 2π*sin(2πt), indicating that ψ(t,0,0) is not a valid solution.

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DaVikes84
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I need to prove that the solution of this differential equation:

dx/dt = -x3 + 2*x + sin3(2*pi*t) - 2*sin(2*pi*t) + 2*pi*sin(2*pi*t)

has the solution:

ψ(t,0,0) = sin(2*pi*t)

I know that I need to get all of the x's on one side and the t's on the other then integrate, but I can't figure out how to get the x's and t's together. Is there a little trick or something to solving this?

Thanks a lot.
 
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DaVikes84 said:
I need to prove that the solution of this differential equation:

dx/dt = -x3 + 2*x + sin3(2*pi*t) - 2*sin(2*pi*t) + 2*pi*sin(2*pi*t)

has the solution:

ψ(t,0,0) = sin(2*pi*t)

I know that I need to get all of the x's on one side and the t's on the other then integrate, but I can't figure out how to get the x's and t's together. Is there a little trick or something to solving this?

Thanks a lot.

No integration is necessary. All you would need to do is show that \psi' + \psi^3 - 2\psi = \sin^3(2\pi t) - 2\sin(2\pi t) + 2\pi\sin(2\pi t)
which, unfortunately, is not the case; there needs to be 2\pi\cos(2\pi t) on the right instead of 2\pi \sin(2\pi t) for that to work.
 

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