How can I classify whether a differential equation is linear or not? The definition on out textbook is:

The Ordinary Differential Equation $$F(t; y, y^{(1)},\dots, y^{(n)})=0$$ is said to be
  linear if $F$ is a linear function of the variables $y, y^{(1)}, \dots, y^{(n)}$,
  otherwise, the equation is non-linear. A similar definition applies to
  PDEs.

I cannot understand that the function is linear if it is a linear function. Additionally and directly my questions are:


*

*What's the meaning of some function is a linear function of some variable?

*Can anybody give me a clear explanation or some examples that can help me understand it?
 A: A differential equation is linear if is of the form
$$a_n(x)y^{(n)} + a_{n-1}y^{(n-1)} + ... + a_1(x)y^{(1)}(x) + a_0(x)y(x) = a(x)$$
Where all the $a$'s are differentiable functions and $y^{(n)} = \mathrm d^ny/\mathrm dx^n$. This is other way to see it. Here just change $x$ to $t$ to get the same variable as you used in your question. From now on I'll also use $t$ as variable.
We say that some operator is linear if it has the properties that, say, $L$ is a linear operator in some $\mathbb{R}$-vector space $U$ iff

$$L(\alpha u + \beta v) = \alpha L(u) + \beta L(v)$$

when $\alpha,\beta \in \mathbb{R}$ and $u,v \in U$. So what he meant by the definition is that you cannot have things like
$$(dy/dx)^2, y^2(d^2y/dx^2)^3/2, \text{ etc...}$$
If that happens then $F$ wouldn't have the properties above. Define the operator $F$ as acting on a space of functions such that 
$$F(y) := F(t,y(t),\dots,y^{(n)}(t)) = a_n(t)y^{(n)}(t) + \dots + a_0(t)y(t) + a(t)$$
Then you can show that $F$ satisfies that, for $\alpha,\beta \in \mathbb{R}$ and for $y,z$ functions $n$-differentiable with respect to $t$ that

$$F(\alpha y + \beta z) = \alpha F(y)  + \beta F(z)$$

Now I give you an example of a non-linear operator (i.e. a non-linear differential equation). Suppose now that you define the operator as 
$$\tilde{F}(y):= y^2 + \frac{dy}{dx}$$
You can see that 
$$\tilde{F}(\alpha y + \beta z) = (\alpha y(t) + \beta z(t))^2 + \alpha\frac{dy}{dt} + \beta \frac{dz}{dt} =\\ \alpha^2 y^2(t) + \beta^2z^2(t) + 2\alpha\beta y(t)z(t) + \alpha\frac{dy}{dt} + \beta \frac{dz}{dt}$$ 
$$\alpha \tilde{F}(y) = \alpha y^2(t) + \alpha \frac{dy}{dt}$$
$$\beta \tilde{F}(z) = \beta z^2(t) + \beta \frac{dz}{dt}$$
And you can see that
$$\tilde{F}(\alpha y + \beta z) \neq \alpha \tilde{F}(y) + \beta \tilde{F}(z)$$
This is what he meant as he was saying that $ F$ must be linear.
