Q:

Linear ODE question

We have an first order ODE :    
Equation1 :   $y' + y = x$  ? 
We can view the left-hand side as an operator acting on $y$.    
In that case $L=(d/dx + 1)$ 
$L(y_1) = x$
$L(y_2)=x$
$L(y_1+y_2)=x$
So, clearly $L(y_1+y_2) = x \neq L(y_1)+L(y_2) = 2x$    
So why is $y'+y=x$ is a linear ODE ?

A:

As we discussed earlier in another thread, the following
$$
L(y):=y'+y
$$
is a linear operator. Note it is not $y'+1$. And note that linear means
$$
L(\alpha y+ \beta z)=\alpha L(y)+ \beta L(z)
$$
for every scalars $\alpha,\beta$, and every differentiable functions $y,z$. The fact that $L$ is linear is merely the fact that differentiation is linear. What you wrote $L(y_1+y_2)=L(y_1)+L(y_2)$ does not suffice to claim linearity of $L$.
Now your ODE can be written
$$
L(y)=f
$$
with $f(x)=x$.
The solution set is either empty, or an affine subspace $$y_p+\ker L$$ where $y_p$ can be any particular solution. This is exactly like linear systems 
$$
AX=B
$$
where $A$ is a rectangular matrix. The solution set is either empty (the system is not compatible), or an affine subspace
$$
X_p+\ker A
$$
where $X_p$ is any particular solution.
People say linear to stress out the fact that there is a linear part in the equation which yields the affine structure of the solution set. One could also say that the equation is affine, by writing it $L(y)-f=0$ and observing that $y\longmapsto L(y)-f$ is affine. But nobody says that, as far as I know.

