The given differential equation is:
y' = 2xy + x
Rewriting it in differential form, we get:
(2xy + x)dx - dy = 0
This is a first-order linear differential equation, which can be written in the form:
y' + P(x)y = Q(x)
where P(x) = -2x and Q(x) = x.
Therefore, the differential equation is a first-order linear differential equation.