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

 A: 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.
A: 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.
A: *

*Assume there are constants $A$ and $B$ such that the function $A\exp(-x)+B\exp(-4x)$, is identically zero. The constants $A$ and $B$ are (over)determined by filling in (for example) $x=\ln1=0,\ln2,\ln3$ so that the only possibility becomes $A=B=0.$

*The solutions of a homogeneous equation or system of equations form a vector space. This can be verified directly with abstract solutions in the original equation, or it could even be a definition of homogeneity. What remains to be shown is that, in the case of a second-order ODE, the dimension of the solution space is 2. This can be concluded from the decomposed form (2) of the differential operator: there are only 2 places where you can supply arbitrary constants.
More explicitly, if $y$ is a solution then $(D+1)(D+4)y=0$ so $(D+4)y$ is determined up to a constant. But for any specific $C$ the equation $(D+4)y$ determines $y$ up to another constant.
A: The equation can be written in the following form:
$$L[y]=0 \tag{1}$$
where $L=\frac{d^2}{dx^2} + 5\frac{d}{dx} + 4$ is a second-order linear operator on the space of twice differentiable functions $C^2$. Let $y_1,y_2\in C^2$ be two solutions, that is $L[y_1] = L[y_2] =0$ (assuming we know they exist). By linearity $$L[Ay_1+By_2] = AL[y_1] + BL[y_2] = 0$$ where $A$, $B$ are constants. Hence $y_g = Ay_1 + By_2$ is also a solution.
Now $(1)$ is of second order, so the general solution must involve 2 arbitrary constants. $y_1$ and $y_2$ must be linearly independent to form the general solution, otherwise, if $y_2\equiv Cy_1$ where $C$ is some constant, $y_g$ collapses into $(A+BC)y_1=Dy_1$ so there is really only one arbitrary constant.
How do we know if $y_1$ and $y_2$ are linearly independent? Take arbitrary $x=x_0$ and write:
$$Ay_1(x_0) + By_2(x_0)=0 \tag{2}$$
$\color{red}{\text{By definition of linear independence the only solution to}}$ $\color{red}{(2)}$  $\color{red}{\text{must be}}$ $\color{red}{A=B=0}$. Now differentiating $Ay_1(x) + By_2(x)$ with respect to $x$ and evaluating again at $x=x_0$ we get another equation:
$$Ay'_1(x_0) + By'_2(x_0)=0 \tag{3}$$
Again the only solution has to be $A=B=0$. $(2)$ and $(3)$ together give an criterion for linear independence, that is, taken as a system their determinant (known as the Wronskian):
$$W(x_0) = \left|\begin{matrix}y_1(x_0) & y_2(x_0) \\ y'_1(x_0) & y'_2(x_0)\end{matrix}\right|$$
must not vanish for any permitted $x_0$
A: Given the general form of a second order ODE is 
$$a\frac{\mathrm{d}^2 y}{\mathrm{d}x^2}+b\frac{\mathrm{d}y}{\mathrm{d}x}+cy=0\tag{1}$$
where $a,b,c$ are known real coefficients. The most important property of $(1)$ is linearity. This means that if we have two solutions $y_1$ and $y_2$, then any linear combination of $y_1$ and $y_2$ is also a solution of $(1)$ i.e. 
$$y(x)=Ay_1(x) + By_2(x)$$ is a solution of $(1)$ for any choice of constants $A$ and $B$. This is a direct consequence of the linearity of derivatives, which allows us to write
$$\begin{align}\\ & a\frac{\mathrm{d}^2 y}{\mathrm{d}x^2}+b\frac{\mathrm{d}y}{\mathrm{d}x}+cy  = a\frac{\mathrm{d}^2 }{\mathrm{d}x^2}\left(Ay_1 + By_2\right)+b\frac{\mathrm{d} }{\mathrm{d}x}\left(Ay_1 + By_2\right)+c\left(Ay_1 + By_2\right) \\ & = A\left(a\frac{\mathrm{d}^2 y_1}{\mathrm{d}x^2}+b\frac{\mathrm{d}y_1}{\mathrm{d}x}+cy_1\right)+B\left(a\frac{\mathrm{d}^2 y_2}{\mathrm{d}x^2}+b\frac{\mathrm{d}y_2}{\mathrm{d}x}+cy_2\right) \\ & =0 \end{align}$$
So the coefficients of $A$ and $B$ are zero.

Linearity is satisfied if for some function $f$:
   $$f(x+y)=f(x)+f(y)$$ and $$f(ax)=af(x)$$ where $a$ is a constant.

