y' + 3y = e^(3x) Solve the first-order differential equation

Given y'+3y=e^(3x)
when the first order linear ordinary differential 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'+3y=e^(3x)--------(1)
y'+p(x)y=q(x)---------(2)
on comparing both we get,
p(x) = 3 and q(x)=e^(3x)
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 3 dx) *(e^(3x))) dx +c)/e^(int 3 dx)
first we shall solve
e^(int 3 dx)=e^(3x)     
so
proceeding further, we get
y(x) =((int e^(int 3 dx) *(e^(3x))) dx +c)/e^(int 3 dx)
=((int e^(3x) *(e^(3x))) dx +c)/e^(3x)
=((int e^(6x) ) dx +c)/e^(3x)
= (e^(6x)/6 +c)/e^(3x)
=(e^(6x)/6 +c)*e^(-3x)
so y(x)=(e^(6x)/6 +c)*e^(-3x)