Definition linear ODE An ordinary differential equation is said to be linear if 
$$F(t,y(t),...,y^{(n)}(t))=0$$
is linear in every derivative. I run into a little problem when using this definition for the equation
$$y'y=0$$
because we have that $$F(t,\alpha y_1(t)+\beta y_2(t),y'(t))=(\alpha y_1(t)+\beta y_2(t))y'(t)=\alpha y_1(t)y't(t) + \beta y_2(t)y'(t)\\
=\alpha F(t,y_1(t),y'(t)) + \beta F(t,y_2(t),y'(t))$$
and vice versa for $y'(t)$. This shows that the ODE is linear. But an equivalent definition of linearity states that it has to have the form
$F(t,y(t),...,y^{(n)}(t))=\sum_{k=0}^n a_i(t)y^{(i)}(t) - g(t)=0$
With this definition the ODE is not linear anymore. Not sure what my mistake is there.

The problem is $y^{(n)}$ are dependent on $y$, so your definition is not sound. Derivative operators are linear by default, so the equation is linear if it is linear in $y$
Let 
$$ F(y) = y'y $$ 
Then
$$ F(ay_1 + by_2) = (ay_1' + by_2')(ay_1+by_2) $$
while
$$ aF(y_1) + bF(y_2) = ay_1'y_1 + by_2'y_2 $$
which means
$$ F(ay_1+by_2) \ne aF(y_1) + bF(y_2) $$
Therefore $F(y)$ is not a linear operator.
