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July 1, 2015

July 1, 2015

Posted by **waqas** on Thursday, January 3, 2013 at 6:20am.

(cos^3x)dy/dx +ycosx=sinx

- calculus -
**Steve**, Thursday, January 3, 2013 at 10:21amHmmm. rearrange things a bit to get

y' + sec^2 x y = tan x sec^2 x

that is

y' + P(x)y = Q(x)

integrating factor is thus e^∫sec^2 x dx = e^(tan x)

now plug and chug