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Question
Solve the following:
`("d"y)/(""dx) + y cos x = sin x cos x`
Solution
It is of the form `("d"y)/("d"x) + "P"y` = Q
Here P = cos x
Q = sin x cos x
`int "P""d"x = int cos x "d"x = sinx`
I.F = `"e"^(int pdx) = "e"^(sinx)`
The required solution is
y(I.F) = `int "Q" ("I.F") "d"x + "c"`
y(esinx) = `int "Q" ("I" - "F") "d"x + "c"`
y(esinx) = `int "te"^"t" "dt" + "c"`
y esinx = `int ("t") ("e"^"t") - int ("e"^"t") "dt" + "c"`
y esinx = `["te"^"t" - "e"^"t"] + "c"`
y esinx = `"e"^"t" ["t" - 1] + "c"`
y esinx = `"e"^sinx [sin x - 1] + "c"`
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