Find the particular solution of the differential equation satisfying the initial condition y(0)=0

Could someone please show me how to solve this problem step by step? I will be very greatful.
How do I solve
y'+8y=4
by using a Integrating factor I(x) and using the "backward product rule" in the process?
Thanks a lot in advance!

There is no need to use an integrating factor as you have suggested.

Find the general solution by separating the variables then integrating:
y' + 8y = 4
dy / dx + 8y = 4
dy / dx = 4 - 8y
dy / dx = -(8y - 4)
dy / (8y - 4) = -dx
∫ 8 / (8y - 4) dy / 8 = - ∫ 1 dx
ln|8y - 4| / 8 = -x + C
ln|8y - 4| / 8 = C - x
ln|8y - 4| = C - 8x
8y - 4 = ℮^(C - 8x)
8y - 4 = ℮ᶜ / ℮^(8x)
8y - 4 = C / ℮^(8x)
8y = 4 + C / ℮^(8x)
y = ½ + C / ℮^(8x)

Find the particular solution by solving for the constant:
When x = 0, y= 0
½ + C = 0
C = -½
y = ½ - 1 / [2℮^(8x)]