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 $\color{red}{\mathrm{red}}$.


*

*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: $y=Ae^{-x}+Be^{-4x}$?   


If you recall from linear algebra, abstract functional spaces can be considered as vector spaces. We define the zero function to serve as the zero vector, and pointwise addition/multiplication as the vector space operations.
For a collection of normal vectors, to show linear independence, we want to show none of the chosen vectors can be written as a linear combination of any of the others; each vector describes a 'different' part of the space. For a vector space of functions, e.g. the space of differentiable functions, to show linear independence, we must show there are no non-zero scalars $a,b$ such that
\begin{align*}
af(t)+bg(t)=0
\end{align*}
for $\textit{all}$ values of $t$ in the domain. It isn't enough that we can find  one or two values of $t$ where the sum equals zero, but instead for all values of the domain.
Now, to show the two functions in your problem are linearly independent. Suppose we could find two non zero numbers $a$ and $b$ such that
\begin{align*}
ae^{-t}+be^{-4t}=0
\end{align*}
For all $t$. Differentiate this expression.
\begin{align*}
-ae^{-t}-4be^{-4t}=0
\end{align*}
Adding the two equations together gives
\begin{align*}
-3be^{-4t}&=0
\end{align*}
The exponential function is strictly positive, so we must have $b=0$. This implies $ae^{-t}=0$, and similarly $a=0$. 
Now, why must a linear combination also be a solution? Well the differential equation described is $\textbf{linear}$, so any element of the span of linearly independent solutions will always yield another solution. They all get mapped to zero.
