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Suppose we have the following second-order differential equation:

$\cos^2(x)y'' -\sin(x)y' + y = 0$

How do we determine its general solution? I couldn't even guess a particular solution; all my efforts led nowhere. I started off with something like $y = Ae^{B\cos(x)}$ but needless to say, that also looked like a dead-end.

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    I wish I knew how to substitute for x in an equation like this. My guess for a solution for an equation like this would be more along the lines of $y=k(\sec x+\tan x)^n$2012-10-09
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    Please read http://tw.knowledge.yahoo.com/question/question?qid=1011101609747 which is in another Q&A site. Most often we can only transform them to the 2nd order linear ODEs with polynomial function coefficients.2012-10-09

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