In a video I was watching regarding how to solve these, the lecturer said that
the form of a Bernoulli Differential Equation is y'+P(x)y=q(x)y^n
where n>1

This means that if n = 1, it wouldn't be a Bernoulli differential equation and would be a first order linear differential equation, but if n=1, y'+P(x)y=q(x)y doesn't take the form of y'+P(x)y=q(x), so how is this a first order linear differential equation?

, here is the video in case anyone wants to check it out.

standard form for a linear first order ode
y' + f(x) y = g(x)

a n=1 example of a Bernoulli like eqn
is y' + P(x) y = q(x) y
rewrite it as y' + (P(x) - q(x))y = 0.
this is in standard form with f(x) = P(x)-q(x) and g(x) = 0.

Ah, but my original question was whether this was a Bernoulli equation or a First Order Linear DFQ(if it is, why), b/c in the video, they stated n=1 as a First Order LinDFQ