Why do the concepts of linear algebra apply to differential equations? A lot of the stuff we do to solve differential equations are taken word for word from linear algebra. The concept of linear independence, determinant of the Wronskian used to determine independence, adding a particular solution to the kernel to get the general solution - all this stuff that I'm used to applying on vectors seems to work with functions. Why?
 A: Because differentiation is a linear operator, and hence can be represented by a matrix acting on a function.
Specific example for OP
$f''+3f'+2f=0 \implies f=ae^{x}+be^{2x}\equiv (a,b)\cdot(e^{x},e^{2x})\equiv \mathbf{a\cdot f}$ 
Therefore, the solutions to the differential equations can be represented as vectors in a function space with the basis $(e^{x},e^{2x})$
Second Addition
As you can see by all the answers on this page, solutions to linear ODEs can be represented by vectors in the space of functions. If you want more details, you can check out this nice paper.
A: Here's a simple application of the theory of linear algebra to differential equations. 
Consider the differential equation
$$
y^{\prime\prime}+y=0\tag{1}
$$
How do we find all of the solutions to (1)? We know that (1) is solved by both $y=\cos(t)$ and $y=\sin(t)$ but are there any other solutions?
To answer this question note that if $y_1$ and $y_2$ solve (1) and if $\lambda_1$ and $\lambda_2$ are scalars, then
$$
y=\lambda_1\cdot y_1+\lambda_2\cdot y_2
$$
is also a solution to (1). This means that the collection $V$ of all solutions to (1) is a vector space. In fact, one may show that $\dim V=2$ and that $\sin(t)$ and $\cos(t)$ are linearly independent solutions to (1). Hence all solutions to (1) are of the form
$$
\lambda_1\cdot\cos(t)+\lambda_2\cdot \sin(t)
$$
for $\lambda_1,\lambda_2\in\Bbb R$.
More generally, one may show that the collection of solutions $V$ to a differential equation of the form
$$
a_n\cdot y^{(n)}+a_{n-1}\cdot y^{(n-1)}+\dotsb+a_1\cdot y^\prime+a_0\cdot y=0\tag{2}
$$
is a vector space with $\dim V=n$. So, to find all solutions to (2) we need only find $n$ linearly independent solutions to (2).
