Question

Solve the following:

`("d"y)/("d"x) + y tan x = cos^3x`

Answer 1

It is of the form `("d"y)/("d"x) + "P"y` = Q

Here P = tan x

Q = cos³x

`int "Pd"x = int tan x  "d"x`

= `int sinx/cosx  "d"x`

= `- int (- sinx)/cosx  "d"x`

= – log cos x

= log sec x

I.F = `"e"^(int Pdx)`

= `"e"^(log  sec x)`

= sec x

The required solution is

y(I.F) = `int "Q" ("I.F")  "d"x + "c"`

y(sec x) = `int cos^3x (sec x)  "d"x + "c"`

y(sec x) = `int cos^3x 1/cosx  "d"x + "c"`

y(sec x) = `int cos^2x  "d"x + "c"`

y(sec x) = `int ((1 + cos 2x)/2)  "d"x + "c"`

y(sec x) = `1/2 int (1 + cos2x)  "d"x + "c"`

y(sec x) = `1/2 [x + (sin2x)/2] + "c"`