y' + ytanx = secx + cosx , y(0) = 1 Find the particular solution of the differential equation that satisfies the initial condition

Given y'+ytanx=secx+cosx
when the first order linear ordinary Differentian equation has the form of
y'+p(x)y=q(x)
then the general solution is ,
y(x)=((int e^(int p(x) dx) *q(x)) dx +c)/ e^(int p(x) dx)
so,
y'+ytanx=secx+cosx--------(1)
y'+p(x)y=q(x)---------(2)
on comparing both we get,
p(x) = tanx and q(x)=secx +cosx
so on solving with the above general solution we get:
y(x)=((int e^(int p(x) dx) *q(x)) dx +c)/e^(int p(x) dx)
=((int e^(int (tanx) dx) *(secx+cosx)) dx +c)/e^(int tanx dx)
first we shall solve
e^(int (tanx) dx)=e^(ln(secx))= secx     
so proceeding further, we get
y(x) =((int e^(int (tanx) dx) *(secx+cosx)) dx +c)/e^(int tanx dx)
=((int secx *(secx+cosx)) dx +c)/(secx )
=((int (sec^2 x+cosxsecx)) dx +c)/(secx )
= ((int (sec^2 x)dx +int 1 dx) +c)/secx
= (tanx+x +c)/secx
= sinx +(x+c)/secx
y(x) = sinx +(x+c)/secx
 
 
to find the particular solution of the differential equation we have y(0)=1
 
on substituting x=0 we get y=1 and so we can find the value of the c
y(0)= sin0+(0+c)/sec0 =0+0+c/1 = c
but y(0)=1
=> 1=c
=> c=1
so y=sinx+(x+1)/secx