Given two solutions to a second-order homogeneous linear DE, how to show they form a basis for the solution set? Define functions $f_1,f_2 : \mathbb{R} \rightarrow \mathbb{R}$ by asserting that for all $x \in \mathbb{R}$ it holds that:
$$f_1(x) = e^{2x}\cos x, f_2(x) = e^{2x}\sin x.$$
Also, define a set $$F = \{f : \mathbb{R} \rightarrow \mathbb{R}\,|\,f''-4f'+5f=0\}.$$
It is easy to see that for all $i \in \{1,2\}$ it holds that $f_i \in F$. How can I show that $\{f_1,f_2\}$ is a basis for $F$?
Edit: Here's an incomplete attempt at a proof. The starred lines (namely, 2 and 5) are the ones I need help with.


*

*Let $F' = \{c_1 f_1 + c_2 f_2 \,|\,c_1,c_2 \in \mathbb{R}\}$.

*Prove that $\{f_1,f_2\}$ are linearly independent.*

*Conclude that $\{f_1,f_2\}$ is a basis for $F'$, and therefore that that $\mathrm{dim}(F')=2$.

*Prove that $F' \subseteq F$. (This is easy.)

*Use a theorem to show that $\mathrm{dim}(F)=2$.*

*Conclude that $F'=F$, and therefore that $\{f_1,f_2\}$ is a basis for $F$.


Line 5 is the really interesting one. What's the name of the theorem that tells us that $\mathrm{dim}(F) = 2$?
Edit 2: Line 5 follows from Theorem 3.4 in this document. (Thank you muzzlater.) So all that remains to show is Line 2.
 A: Below is a proof of the uniqueness theorem, using Wronskians and variation of parameters. 
Theorem $\ $  If  $\rm\:f,g,h\: $ are solutions on an  interval I of 
$$\rm     y'' =\ p\ y' + q\ y,\ \ \ \ p,q\ \ continuous\ on\ I $$
and the Wronskian $\rm\ \  W = g\:h'-g'h \ne 0\:$ for all $\rm\:x\in I$    
then $\,\exists\,$ constants $\rm\: c,d\:$ such that    $\rm\: f  = c\: g  + d\: h\:$ on $\rm\,I.$ 
Proof $\ $ The equations $[0],[1]$ below have unique solution $\rm\:(c,d)\:$ via det $\rm = W \ne 0\:$ on $\rm\,I.$ 
$\rm[0]\qquad           f\  =\ c\: g \: + d\: h $ 
$\rm[1]\qquad           f' =\ c\: g' + d\: h'$
Now  $\rm\:q\:[0] + p\:[1]\ $  yields, $ $ on $ $ LHS: $\rm\,\ q\:f+p\:f'\: =\ f'',\ $ similar on RHS below
$\rm[2]\qquad  f'' =\ c\: g'' + d\: h''\ $  via RHS:  $\rm\ \, q\:g+p\:g'\: =\ g'',\,\ \ q\:h+p\:h'\: =\ h''$ 
$\rm[3]\qquad           0\  =\ c'\:g \:+ d'\:h\:\ \ $  via  $\ \ [0]'-[1]$ 
$\rm[4]\qquad           0\  =\ c'\:g' + d'\:h'\ \ $  via  $\ \ [1]'-[2]$ 
$[3],[4]\:$ have solution $\rm\:(c',d') = (0,0),\:$ 
which is unique by  $\rm\ det = W = g\:h'-g'\:h \ne 0\:$ on $\rm\,I.\:$ Therefore $\rm\:c,d\:$ are constants. $\ \ $ QED
