This is always a confusion with my students (and quite a few professors), so it makes sense to be confused every now and then. I feel there are a few things that are being conflated in your post that need to be clarified:

(i) A linear *function* and a linear* model* are not necessarily the same thing. A linear function has this form:

\(y = mx + b\) for \((m, b)\) some constants, right? However this:

\(y = mx^2 + b\) is not a linear function. It's a quadratic function. HOWEVER, if you turn these into regression equations like:

\(Y = \beta_{0} + \beta_{1}X + \epsilon\)

\(Y = \beta_{0} + \beta_{1}X^{2} + \epsilon\)

They are still linear models. We don't really talk about the "DV being not linear" but I intuit what you mean is the *relationship *between the DV and IVs is linear. It is of course not a linear relationship in the case of \(Y = \beta_{0} + \beta_{1}X^{2} + \epsilon\) where the relationship is quadratic. But the model itself is linear. The expression I've heard before to describe it is saying it is *linear in the parameters. *As such, a regression model with first-order interaction effects like:

\(Y = \beta_{0} + \beta_{1}X_{1} + \beta_{2}X_{2} + \beta_{3}X_{1}X_{2} + \epsilon\)

is still a linear model, even if there is a product term. It's a polynomial of degree 2.

(ii) A non-linear model is one where the parameters cannot be expressed in linear terms. Most obvious example for me is:

\(Y = e^{\beta_{0} +\beta_{1}X_{1} + \epsilon}\)

However, notice that this relationship could be linearized such that:

\( log(Y )= \beta_{0} +\beta_{1}X_{1} + \epsilon\)

But that changes the dependent variable from \(Y\) to \(log(Y)\).

Another example which may be closer related to your time series query perhaps would be something like:

\( Y= sin(\beta_{0} +\beta_{1}X_{1} + \epsilon)\)

Because the sine function induces periodicity that is difficult to model with a single line (unless you do piecewise regression).

The way I remember these things is usually thinking that if it has a transcendental function involved (exponentiation, trigonometric functions, etc.) It's a non-linear model. But that does not mean said model cannot be linearized. Logistic regression is perhaps a good example of this, as well as most general**IZED** linear models where you try to find a linear solution to a very much non-linear problem.