Find the particular solution of the following equation 
my task is to Find the particular solution of the following equation at y(0) =1. the equation is
$$dy/dx + 4y = 7$$

I have made this much progress as far as separating the Ys and the Xs and taking the anti derivatives (I hope it's correct) but don't know how to reach an answer
dy/dx + 4y = 7    || .dx
$dy + 4y = 7 dx$
$4 y^2/2 = 7x +c$
.....
correct answer is:
$y = 7/4 - 3/4e^{-4x}$
update:
I managed to get 7/4 and Ce^-4x) but still don't get where -3/4 comes from
thank you!

We have 
$$\frac{dy}{dx}=7-4y,$$ and therefore 
$$\frac{dy}{7-4y}=dx.$$
Continue.  
Added: Integrate. We get $-\frac{1}{4}\ln(|7-4y|)=x+C$. Put $x=0$. We get $C=-\frac{1}{4}\ln 3$. Thus
$$\ln(7-4y)=-4x+\ln 3.$$
Note this is valid only when $7-4y\gt 0$. Exponentiate. We get $7-4y=3e^{-4x}$.
Solve for $y$. We get $y=\frac{1}{4}\left(7-3e^{-4x}\right)$. 
