MHB Find Ansatz for this ODE (3.5.15)

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Hi - I'm given: $ y'' + y' - 2y = \frac{e^{x}}{x} $

What is a good Ansatz to find the particular solution? I've tried a few that haven't worked...Thanks
 
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ognik said:
Hi - I'm given: $ y'' + y' - 2y = \frac{e^{x}}{x} $

What is a good Ansatz to find the particular solution? I've tried a few that haven't worked...Thanks

According to W|A, the particular solution is not expressible in terms of elementary functions. Are you certain you have the problem copied correctly?
 
Copied OK, probably typo in book - if the driving function was just $e^x$ it would be more sensible
 
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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