y' + ysecx = secx , y(0) = 4 Find the particular solution of the differential equation that satisfies the initial condition

Given y'+y*secx=secx
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'+y*secx=secx--------(1)
y'+p(x)y=q(x)---------(2)
on comparing both we get,
p(x) = secx and q(x)=secx
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 secx dx) *(secx)) dx +c)/e^(int secx dx)
first we shall solve
e^(int secx dx)=e^(ln(secx +tanx)) = secx+tanx     
so
proceeding further, we get
y(x) =((int e^(int secx dx) *(secx)) dx +c)/e^(int secx dx)
=(int ((secx+tanx)*(secx)) dx +c)/(secx+tanx)
=(int ((sec^2x+tanx*(secx)) dx +c)/(secx+tanx)
=(int (sec^2x) dx+int (tanx*(secx)) dx +c)/(secx+tanx)
=(tanx+secx +c)/(secx+tanx)
so y(x)=(tanx+secx +c)/(secx+tanx)=1 +c/(secx+tanx)
 
Now we have to find the particular solution at y(0) =4
so y(x) =1 +c/(secx+tanx)
=> y(0) = 1+c/(sec(0)+tan(0)) =4
=> 1+c=4
c=3
so the particular solution is
y(x) = 1+ 3/(secx+tanx)