For x != pi/2 + kpix≠π2+kπ divide by cosxcosx:
dy/dx +y secx = tanxdydx+ysecx=tanx
Solve first the homogeneous equation:
dy/dx + ysecx= 0dydx+ysecx=0
which is separable:
dy/dx = -ysecxdydx=−ysecx
dy/y = - secxdxdyy=−secxdx
int dy/y = -int secxdx∫dyy=−∫secxdx
lnabs y = -ln abs(secx+tanx) +Cln|y|=−ln|secx+tanx|+C
y = c/(secx+tanx)y=csecx+tanx
Using the method of variable coefficient look now for a particular solution of the complete equation in the form:
bar y = (c(x))/ (secx+tanx)¯y=c(x)secx+tanx
(dbary)/dx = (c'(x) (secx+tanx) -c(x)(secxtanx +sec^2x))/(secx+tanx)^2
(dbary)/dx = (c'(x) (secx+tanx) -secx c(x)(secx +tanx))/(secx+tanx)^2
(dbary)/dx = (c'(x) -secx c(x))/(secx+tanx)
Substitute in the complete equation:
(dbary)/dx +ysecx = tanx
(c'(x) -secx c(x))/(secx+tanx)+(c(x) secx)/(secx+tanx) = tanx
(c'(x))/(secx+tanx)= tanx
c'(x)= secxtanx+tan^2x
Using the trigonometric identity: tan^2x = sec^2x-1
c'(x)= secxtanx+sec^2x-1
c(x) = int (secxtanx+sec^2x-1)dx
c(x) = secx+tanx -x +c_1
Choose the solution for c_1 =0:
bary = (secx+tanx -x)/(secx+tanx) = 1-x/(secx+tanx)
The complete solution is then:
y = c/(secx+tanx) -x/(secx+tanx) +1
y= 1-(x-c)/(secx+tanx)
In fact:
dy/dx = ((x-c)secx-1)/(secx+tanx)
cosx dy/dx +y =((x-c)-cosx)/(secx+tanx)+1-(x-c)/(secx+tanx) =
:. = 1- cosx/(secx+tanx)
:. = (secx+tanx- cosx)/(secx+tanx)
:. = (1+sinx- cos^2x)/(1+sinx)
:. = (sin^2x+sinx)/(1+sinx) = sinx
Note that we can write the general solution also as:
y= 1-((x-c)cosx)/(1+sinx)
If we let x=0 then:
y(0) = 1+c
so that from the initial condition:
y(0) = 2 we get c=1.
y= 1-((x-1)cosx)/(1+sinx)