Find the general solution to the ODE 6p′(q) + 24q(p) = 16 with instructions please?

6p'+24qp=16=>
p'+4qp=8/3
This is linear & its integrating factor
is a=e^(2q^2)
e^(2q^2)dp+4pqe^(2q^2)dq=8e^(2q^2)dq/3
=>
d[pe^(2q^2)]=(8/3)e^(2q^2)dq
=>
p=(8/3)e^(-2q^2)[Se^(2q^2)dq+C]

I rewrote q(p) as q p (q times p): Not sure whether this is what you meant.

6 p' + 24 q p = 16

divide both sides by 6
p' + 4 q p = 8/3 ---------(1)

Linear equation of first order

Integrating factor e^∫4q dq = e^(2q^2)

Multiply equation (1) by the integrating factor e^(2q^2)
e^(2q^2) p' + 4 q p e^(2q^2) = (8/3) e^(2q^2) ---------(2)

The left side of (2) is d/dq ( p times the integraing factor) or
(p e^(2q^2) )'

(p e^(2q^2))' = (8/3) e^(2q^2)
∫(p e^(2q^2))' dq = (8/3)∫ e^(2q^2) dq + C

p e^(2q^2) = (8/3)∫ e^(2q^2) dq + C
p = (8/3) ∫e^(2q^2) dq / e^(2q^2) + C / e^(2q^2)---- answer

∫e^(2q^2) dq = sqrt (2 pi) erfi (sqrt(2) x)
http://www.wolframalpha.com/input/?i=erfi

I'm not sure of this solution as I cannot distinguish between q(p) and qp