Real analysis question involving a linear ODE 
Where do I start with this one? This question is really quite difficult..
 A: Let $I=(u,v)$.
(i): A solution on $I$ is given by
$$f(x)=\exp\left(-\int_{x_0}^x a(t) dt\right)\tag{2}$$
where $x_0\in I$ is arbitrary.
Indeed, $f^\prime(x)=-a(x) \exp\left(-\int_{x_0}^x a(t) dt\right) = -a(x) f(x)$. Also, $f$ has no zeroes.
(ii):
Let $\tilde{f}$ be any other solution. Let us define $g(x)=\frac{\tilde{f}(x)}{f(x)}$. Also $g$ is differentiable on $I$ with derivative
$$g^\prime(x) = \frac{1}{f(x)^2} (\tilde{f}^\prime(x) f(x) - \tilde{f}(x) f^\prime(x))=\frac{1}{f(x)^2}(-a(x)\tilde{f}(x)f(x)+a(x)\tilde{f}(x)f(x))=0$$
Therefore $g$ is a constant function. That is, there exists $c\in\mathbb{C}$ such that
$$\tilde{f}=c\cdot f$$
(iii): Define 
$$f(x)=y_0 \exp\left(-\int_{x_0}^x a(t) dt\right)$$
$f$ solves the equation $(1)$ and $f(x_0)=y_0$.
(iv): By (ii), the set of solutions of $(1)$ is given by
$$U=\left\{c\cdot f \,\,|\,\,c\in\mathbb{C}\right\}$$
where $f$ is as in $(2)$. Since $f\in C^1(I,\mathbb{C})$, $U\subset C^1(I,\mathbb{C})$ and if $g=c_1\cdot f,h=c_2\cdot f\in U$, $\lambda,\mu\in\mathbb{C}$, then also
$$\lambda g+\mu h=(\lambda c_1+\mu c_2) f\in U$$
So $U$ is a subvectorspace. It has dimension one, because $\{f\}$ is a basis.
Additional information:
The "schematic" way of arriving at the solution in $(2)$ goes as follows: rewrite the equation as
$$\frac{y^\prime}{y} = -a$$
Now integrate
$$\int\frac{y^\prime}{y} = -\int a$$
This gives
$$\log y = -\int a + \mathrm{const}$$
So
$$y(x)=c\cdot\exp\left(-\int_{x_0}^x a(t) dt\right)$$
This method is called separation of variables and can be applied whenever your ODE is of the form
$$y^\prime = F(x)G(y)$$
A: The equation of interest is
$f'(x) + a(x)f(x) = 0;  \tag{0}$
we have:
(i.)  set
$f(x) = e^{-\int_{x_0}^x a(s) ds}; \tag{1}$
then, since $\int_{x_0}^x a(s) ds$ is continuously differentiable by virtue of the hypothesis that $a(x)$ is continuous,
$f'(x) = -a(x)e^{-\int_{x_0}^x a(s) ds} = -a(x)f(x), \tag{3}$
or
$f'(x) + a(x)f(x) = 0, \tag{4}$
showing that $f(x)$ satisfies (0).  $f(x)$ is continously differentiable on $I$ since $\int_{x_0}^x a(s) ds$ is, and $f(x)$ vanishes nowhere since $0$ is not in the range of $e^z$, $z \in \Bbb C$;
(ii.)  if $\tilde f(x)$ also solves (4) on $I$, consider $f^{-1} \tilde f(x)$; we have
$(f^{-1} \tilde f)' = -f^{-2}f' \tilde f + f^{-1} \tilde f' = -f^{-2}(-af) \tilde f + f^{-1} (-a \tilde f) = af^{-1} \tilde f - a f^{-1} \tilde f = 0, \tag{5}$
which shows that $f^{-1} \tilde f$ is a constant $c \in \Bbb C$ on $I$.  Thus we have
$\tilde f(x) = cf(x); \tag{6}$
(iii.)  for $y_0 \in \Bbb C$, consider $f(x) = y_0 e^{-\int_{x_0}^x a(s) ds}$; it is easy to see, by direct differentiation,  that this $f(x)$ satisfies (0) and we also have
$f(x_0) = y_0e^{-\int_{x_0}^{x_0} a(s) ds} = y_0 e^0 = y_0 (1) = y_0; \tag{7}$
the uniqueness of $f(x) = y_0e^{-\int_{x_0}^x a(s) ds}$ follows from (ii.), since any solution $\tilde f(x)$ is of the form $ce^{-\int_{x_0}^x a(s) ds}$; equating $f(x_0)$ with $\tilde f(x_0)$ yields $c = y_0$, showing the uniqueness of $f(x)$.
(iv.)  since every solution is of the form $c e^{-\int_{x_0}^x a(s) ds}$ for some $c \in \Bbb C$, we see that the solutions form a subspace of $C^1(I, \Bbb C)$, since, denoting  $e^{\int_{x_0}^x a(s)ds}$ by $E(x)$, $c_1 E(x) + c_2 E(x) = (c_1 + c_2) E(x)$ and $c_1(c_2 E(x)) = (c_1 c_2) E(x)$ for all $c_1, c_2 \in \Bbb C$.  We see the vector space axioms are satisfied by the set $\{ ce^{\int_{x_0}^x a(s) ds}, c \in \Bbb C \} \subset C^1(I, \Bbb C)$.  Furthermore, $\{ E(x) \}$ is clearly a basis of this subspace, implying that is of dimension one.  QED.
Call it a wrap, it's in the can!
Hope this helps.  Cheerio,
and as always,
Fiat Lux!!!
