The eqn is : y" +y' +y = (sin(x))^2 and i'm in need of a particular solution, y(p). I tried y=Asin(x)^2, but am not getting anywhere. y = A*sin(x)^2 y' = Asin(2x) y'' = 2Acos(2x) best, jblorien
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$\displaystyle \sin^2 x = \frac{1}{2} - \frac{1}{2}\cos 2x$
solved it, no worries.
Last edited by jblorien; Feb 11th 2008 at 12:56 PM.
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