Why can a linear "ordinary" differential equation have non-linear coefficients of independent variable? The confusion origins from the fact that $y$ = $x^2$ + $x$ + $1$ is a non linear equation but $y\,'$ = $x^2$ + $x$ + $1$ is a linear differential equation.


*

*Why is the non-linearity in independent variable not significant in the case of differential equations?

*Does the word linear have different meanings for "normal" (not differential) equations and differential equations?

*What would be the best way to make some geometric sense of linear differential equations? (like in the case of linear equation in two variable it is a line.) 


Mentions of "differential equation(s)" in the above questions refer to only the subset of "ordinary differential equation(s)"
 A: The "linear" here means that if $u$ and $v$ are solutions of the (homogeneous) ODE, so is $\alpha u + \beta v$. That the coefficients of the derivatives aren't linear doesn't matter.
A: You're confused about what the independent variable is. A differential equation is a statement about functions, not about numbers.
In this case, the equation $y' = x^2 + x + 1$ is a statement about two functions: 


*

*$y$

*$(x^2 + x + 1)$.


Neither of them need be linear in $x$. However it is linear in $y$. To see why (hah) first let's call $g(x) = (x^2 + x + 1)$ so we don't get distracted by $x$'s:
$$ y' = g $$
This is linear in the same way that the equation $3x = 7$ is a linear equation: on the left side we have a linear function, and on the right side we have a constant.
Let's look at a more complicated example just to be sure we understand. 
$$3xy' - 2y' + 7y/x + 3x = -x^2 + 2y$$
Rearrange so all the terms with "$y$" are on the left side and everything is on the right, and factor out so it looks like a polynomial in $y$.
$$ \big(3x - 2\big)y' + \bigg(-2 + \frac{7}{x}\bigg)y = -x^2 - 3x $$
This is a differential equation, so it's a statement about functions, not numbers. Let's call $a(x) = 3x-2$, $b(x) = -2 + 7/x$, and $c(x) = -x^2 - 3x$ just so the $x$'s don't get in the way:
$$ay' + by = c$$
Again on the left side we have a linear function of $y$, and on the right side we have a constant.

Why is the left side a linear function of $y$? Let's call $Ly = ay' + by$. You should work this out: $$L(y_1 + y_2) = Ly_1 + Ly_2$$ and $$L(\lambda y) = \lambda Ly.$$
This problem lives in the realm of linear algebra: $L$ is a linear operator, and we're solving the inhomogeneous equation
$$ Ly = c $$
A: Let's rewrite your two equations to
$$
f(x,y) = x^2+x+1-y = 0
$$
and
$$
F(y)= x^2+x+1-y'=0.
$$
It is instructive to identify which kind of maps we are dealing with: $f:\mathbb{R}^2\rightarrow \mathbb{R}$ is a function of two real variables, whereas $F:C^1(\mathbb{R})\rightarrow C(\mathbb{R})$ is a function that takes as argument a continuously differentiable function. I.e. by interpreting $y$ as a function of $x$ (and thus as element $y \in C^1(\mathbb{R}))$, $F$ only takes one argument.
We can further write $F(y)=F_0(y)+c$, where $c=x^2+x+1$ is viewed as an element in $C^1(\mathbb{R})$ and $F_0(y)=-y'$. Then $F_0$ is a linear map between the vector-spaces $C^1(\mathbb{R})$ and $C(\mathbb{R})$, which means that $F$ is an affine linear map.
On the other hand, $f$ is non-linear. This is the reason to call the first equation non-linear and the second equation (affine) linear.
A: You only need to understand the definition. First, a differential equation is an equation in the differential of an unknown function, and the function itself. Thus it is linear provided it involves only a linear combination of the differential of the unknown, and the unknown, the coefficient being given functions of the independent variables. That's all.
Thus, $f(x)\mathrm dy+yg(x)\mathrm dx=0$ is linear for any given functions $f,g$ whereas $y\mathrm dy+\mathrm dx=0$ is nonlinear because the coefficient of $\mathrm dy,$ an unknown, is itself the unknown $y.$
In general, it is just an equation of the form $$\sum_{k\ge 0} f_k(x,\mathrm dx)\mathrm d^ky=F(x,\mathrm dx),$$ where the functions $f_k, \,F$ are given, and the equation is homogeneous in the differentials.
Hope this helps.
A: First consider an example from the theory of linear equations in algebra.  Let $M$ be a given matrix and let $v$ be an unknown vector.  Then we can talk about solutions to the equation
$$
Mv=0,
$$
where $0$ is the zero vector.  This equation represents a system of linear equations and, as one might expect, we say that the matrix equation is linear.  It has the important property that if $v_1$ and $v_2$ are two solutions to the equation, then $av_1+bv_ 2$ is also a solution, where $a$ and $b$ are arbitrary scalars.  We refer to this property as linearity.
Consider now the equation $Mv=c$, where $c$ is a given vector.  If $c$ is not the zero vector then we do not have the property of linearity anymore since if $Mv_1=c$ and $Mv_2=c$ we have $M(av_1+bv_2)=ac+bc$, which is not equal to $c$ for arbitrary $a$ and $b$.  What we do have is that if $u$ is a solution to $Mu=c$ and $v_1$ and $v_2$ are solutions to $Mv=0$ then $u+av_1+bv_2$ is a solution to $Mu=c$.
What is important in all this is that $M$ is a linear operator: it satisfies
$$M(av_1+bv_2)=aMv_1+bMv_2.
$$
From this, everything said above follows: linear combinations of solutions to the homogeneous equation $Mv=0$ are also solutions of the homogeneous equation; a solution to the inhomogeneous equation $Mu=c$ plus an arbitrary linear combination of solutions to the homogeneous equation is also a solution to the inhomogeneous equation $Mv=c$.
The same terminology applies to linear differential equations.  The analog of the matrix $M$ is the linear differential operator, of which
$$
L=x\frac{d^2}{dx^2}+(2x^2+1)\frac{d}{dx}+(x^3+x+1)
$$
is an example.  If $f$ is a function, then $L$ acts on $f$ as follows:
$$
Lf(x)=xf''(x)+(2x^2+1)f'(x)+(x^3+x+1)f(x).
$$
Now suppose $f_1(x)$ and $f_2(x)$ are solutions to the homogenous equation $Lf(x)=0$.  You can easily check using basic properties of $\frac{d}{dx}$ that $af_1(x)+bf_2(x)$ is also a solution, where $a$ and $b$ are arbitrary numbers.  The geometric meaning of this property is superposition: solutions can simply be added together or rescaled and they remain solutions.
If $g(x)$ is a solution to the inhomogeneous equation $Lg(x)=P(x)$, where $P$ is some non-zero function, then $g(x)+af_1(x)+bf_2(x)$ also solves the inhomogeneous equation.
Let key property of $L$ here is again that of being a linear operator: $$
L(af_1(x)+bf_2(x))=aLf_1(x)+bLf_2(x).
$$
The ability to combine solutions allows a nice theory of linear differential equations to be constructed.  Non-linear differential equations (an example of which would be $\frac{d}{dx}f(x)+[f(x)]^2=0$) do not have this combining property, which makes the theory considerably more difficult.
The example in your question, $y'=x^2+x+1$, is an inhomogeneous equation.  The solutions to the homogenous equation $y'=0$ are the constant functions.  The fact that an arbitrary integration constant can be added to any particular solution to your equation to find another solution is linearity at work.
