I Second Order ODE with Exponential Coefficients

thatboi
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Hi all,
I have another second order ODE that I need help with simplifying/solving:
##p''(x) - D\frac{e^{\gamma x}}{A-Ae^{\gamma x}}p'(x) - Fp(x) = 0##
where ##\gamma,A,F## can all be assumed to be nonzero real numbers and ##D## is a purely nonzero imaginary number.
Any help would be appreciated!
 
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Set t = e^{\gamma x}. Then <br /> \gamma^2(1-t)t^2 \frac{d^2p}{dt^2} + \left(\gamma^2(1-t)t - \frac{\gamma D}{A} t^2\right) \frac{dp}{dt} - F(1-t)p = 0. This looks like it should be solvable by Frobenius' method.
 
I have the equation ##F^x=m\frac {d}{dt}(\gamma v^x)##, where ##\gamma## is the Lorentz factor, and ##x## is a superscript, not an exponent. In my textbook the solution is given as ##\frac {F^x}{m}t=\frac {v^x}{\sqrt {1-v^{x^2}/c^2}}##. What bothers me is, when I separate the variables I get ##\frac {F^x}{m}dt=d(\gamma v^x)##. Can I simply consider ##d(\gamma v^x)## the variable of integration without any further considerations? Can I simply make the substitution ##\gamma v^x = u## and then...

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