Why does the sum of two linearly independent solutions of a second order homogeneous ODE give a general solution? The following is a short extract from the book I am reading: 

If given a Homogeneous ODE: 
  $$\frac{\mathrm{d}^2 y}{\mathrm{d}x^2}+5\frac{\mathrm{d} y}{\mathrm{d}x}+4y=0\tag{1}$$
  Letting 
  $$D=\frac{\mathrm{d}}{\mathrm{d}x}$$ then $(1)$ becomes 
  $$D^2 y + 5Dy + 4y=(D^2+5D+4)y$$
  $$\implies\color{blue}{(D+1)(D+4)y=0}\tag{2}$$
  $$\implies (D+1)y=0 \space\space\text{or}\space\space (D+4)y=0$$ which has solutions $$y=Ae^{-x}\space\space\text{or}\space\space y=Be^{-4x}\tag{3}$$ respectively, where $A$ and $B$ are both constants. 
Now if $(D+4)y=0$, then $$(D+1)(D+4)y=(D+1)\cdot 0=0$$
  so any solution of $(D + 4)y = 0$ is a solution of the differential equation $(1)$ or $(2)$. Similarly, any solution of $(D + 1)y = 0$ is a solution of $(1)$ or $(2)$. $\color{red}{\text{Since the two solutions (3) are linearly independent, a linear combination}}$ $\color{red}{\text{of them contains two arbitrary constants and so is the general solution.}}$ Thus $$y=Ae^{-x}+Be^{-4x}$$ is the general solution of $(1)$ or $(2)$.

The part I don't understand in this extract is marked in $\color{red}{\mathrm{red}}$.


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*Firstly; How do we know that the two solutions: $y=Ae^{-x}\space\text{and}\space y=Be^{-4x}$ are linearly independent?

*Secondly; Why does a linear combination of linearly independent solutions give the general solution. Or, put in another way, I know that $y=Ae^{-x}\space\text{or}\space y=Be^{-4x}$ are both solutions. But why is their sum a solution: $y=Ae^{-x}+Be^{-4x}$?   


Note that, in general, for each $a$ we have $D-a$ is a linear transformation in the space of differentiable functions. Then, if you want to solve $(D-a)y=0$, it is the same that $Dy=ay$. That is, $a$ is an eigenvalue.
Now, if $a\neq b$ and $Dy_1=ay_1$ and $Dy_2=ay_2$, we have that $y_1$ and $y_2$ are eigenvectors asociated to different eigenvalues. Thus they are linearly independent.
On the other hand, since $(D-a)(D-b)=0$ is the kernel of the linear transformation $(D-a)(D-b)$, which has dimension at most 2, then $y_1$ and $y_2$ is a base for this kernel. Thus any solution is a linear combination of $y_1$ and $y_2$.
Finally, if you have "more products" $(D-a_1)(D-a_2)...(D-a_n)$, all the anterior considerations can be generalized.
