Linearity of Partial Differential Equations

[BustedBreaks]

Is this linear homogeneous, linear inhomogeneous etc...
$$u_{t}-u_{xx}+xu=0$$

From that first one I get this
$$\frac{u_{t}-u_{xx}}{u}=-x$$
which I'm not sure is linear.

Edit:
Similar questions involve the following equations:
$$iu_{t}-u_{xx}+\frac{u}{x}=0$$

and

$$u_{x}+e^{y}u_{y}=0$$

Another Edit:

I think I see the answer. I can rewrite the first equation like this:
$$(u_{t}-u_{xx}+xu)(\frac{1}{x})=0(\frac{1}{x})$$

and get a linear equation:

$$\frac{u_{t}}{x}-\frac{u_{xx}}{x}+u=0$$
UGH... Another Edit..

I'm not sure which approach is correct, the first or the second...

 

[LCKurtz]

What makes such an equation linear is the fact that it is linear in the dependent variable u. That means u and its derivatives appear to the first degree. No u² or uu_x and the like.

More specifically, if your differential operator is denoted L(u), it has the property that:

L(u + v) = L(u) + L(v) and L(cu) = cL(u) for constant c.

For example, given the equation

c(x,y)u_xx+ f(x,y)u + g(x,y)u_yy = h(x,y)

if you call L(u) = c(x,y)u_xx+ f(x,y)u + g(x,y)u_yy the equation becomes:

L(u) = h(x,y)

If you try the above two conditions on L, you will see that it satisfies them because of the linearity of taking derivatives. It wouldn't if it had terms like u², sin(u) etc.