Asked by waqas

integrate it by integrating factor

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

Answers

Answered by Steve
Hmmm. 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
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