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How do I solve this second order PDE?

$\frac{\partial^2 X}{\partial t^2}+X= \cos(\omega t)+\frac{1}{4}(1-exp(-\tau +F(\sigma)))^\frac{-3}{2} [\sin(6t-6\psi)+3\sin(4t-4\psi)+3\sin(2t-2\psi)]$

$\omega$ is a constant. $F(\sigma)$ is just a general function of $\sigma$. $\tau$ and $\sigma$ are slow time scales, i.e. $\epsilon t$ and $\epsilon^2 t$ respectively where $\epsilon$<<1. And $\psi(\sigma)$ is just a function of $\sigma$.

Is it ok if I turn this into second order ordinary differential equation? And instead of constants, use general functions of $\sigma$ and $\tau$.

$\frac{d^2 X}{dt^2}+X= \cos(\omega t)+\frac{1}{4}(1-exp(-\tau +F(\sigma)))^\frac{-3}{2} [\sin(6t-6\psi(\sigma))+3\sin(4t-4\psi(\sigma))+3\sin(2t-2\psi(\sigma))]$

Then solving the homogeneous part gives me X = $F_1 (\tau,\sigma)\cos(t)+F_2 (\tau,\sigma)\sin(t)$.

What would be the particular solution for the non-homogeneous part?

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  • $\begingroup$ you're missing a parenthesis somewhere. Do you mean $(1-e^{-\tau+F(\sigma)})^{-3/2}$? $\endgroup$ – Dylan Nov 15 '15 at 6:58
  • $\begingroup$ You can try using variation of parameters if you know the solutions to the homogeneous equation $\endgroup$ – Dylan Nov 15 '15 at 7:07
  • $\begingroup$ @Dylan yes. Thank you. I have just corrected it. $\endgroup$ – p.gurung Nov 15 '15 at 7:42

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