Derive from scratch without any formulas the specific integr

Derive (from scratch without any formulas) the specific integrating factor for the linear first-order equation (cosx)y\' = cos^2 x - (sin x)y, and carefully describe each step of the process. Then solve the equation.

Solution

Solution: Given first order linear differential equation is

(cos x ) y\' = cos² x - (sinx)y

or y\' + (sinx)/(cos x ) y =(cos² x)/ (cos x) [ deviding both sides by cos x ]

or dy/dx + (tanx)y = cosx ......................(1)

Multiplying both sides of equation (1) by e^{\\int(tan x)dx}, we have

e^{\\int(tan x)dx}[dy/dx + (tanx)y] = e^{\\int(tan x)dx}[cosx ]                                     (\\int = integration symbol)

implies that d/dx[ye^{\\int(tan x)dx}] = cosxe^{\\int(tan x)dx}

Now integrating above equation, we get

ye^{\\int(tan x)dx} = \\int [cosxe^{\\int(tan x)dx}] dx+ c ...............................(2)

where c is arbitrary constant of integration.

Again \\int (tanx)dx = log sec x

So, e^{\\int(tan x)dx} = e^{log sec x} = sec x.

Then equation (2) becomes

y (sec x) = \\int [(cos x). (sec x)]dx + c = \ t[1]dx + c = x +c

So y (sec x) = x +c is the required solution of the given equation.

( Here e^{\\int(tan x)dx} = e^{log sec x} = sec x is Integrating Factor)