Real analysis question involving inhomogenous linear ODE 
So I had another problem like this but the ODE was homogenous, now there is a non zero right side.
I completed part (i), $\large c(x) = \int \frac{b(x)}{g(x)} dx$.
I am stuck on (v).
(1) is the homogenous bit, which is just
$$f'(x) +a(x)f(x) = 0$$
 A: You need to show that there is linear subspace $U\subset C^1(I,\mathbb{C})$ and a vector $v_0\in C^1(I,\mathbb{C})$ such that every solution of $(2)$ lies in $v+U$.
We claim that $U$ is a set of solutions of $(1)$, $v_0=f_0$. Indeed, by paragraph $(ii)$ we already know that every solution of $(2)$ is a sum of $f_0$ and some solution of $(1)$. Thus we completed the proof? No, we need to show that $U$ is a $1$-dimensional linear subspace of $C^1(I,\mathbb{C})$. Every solution of $(1)$ is given by the formula
$$
f(x)=\operatorname{const}\cdot \exp\left(-\int_{x_0}^x a(t)dt\right)\in C^1(I,\mathbb{C})
$$
This result tells us that every solution of $(1)$ is a function $\exp\left(-\int_{x_0}^x a(t)dt\right)$ times some constant coefficient. In other words $U$ is $1$-dimensional subspace of $C^1(I,\mathbb{C})$ with the basis consisting of single function: $\exp\left(-\int_{x_0}^x a(t)dt\right)$. And now we are done.
A: Parts (ii), (iii), and (v) are  pure linear algebra. We are given a linear map
$$L:\quad C^1(I)\to C^0(I),\qquad f\mapsto f'+a\> f\ .$$
Solving the ODE $(2)$ means finding the $f$'s in $C^1(I)$ with $Lf=b$, where  $b\in C^0(I)$ is given in advance. Denote the set of these $f$'s by ${\cal S}$.
(ii) When $f$, $f_0\in{\cal S}$ then $L(f-f_0)=Lf -Lf_0=b-b=0$; whence $g:=f-f_0$ solves the corresponding homogeneous problem $Ly=0$.
(iii) From (ii) we can conclude the following: When $f_0\in{\cal S}$ is a "particular solution" found by whichever means then any solution $f\in{\cal S}$ can be written in the form $f=f_0+ g$, where $g$ solves the corresponding homogeneous problem $Ly=0$.
(v) The solution set ${\cal K}$ of the homogeneous ODE $Ly=0$ is the kernel of the linear map $L$; therefore ${\cal K}$ is a vector subspace of $C^1(I)$, by general principles. According to (iii) we have
$${\cal S}=\bigl\{f_0+g\>\bigm|\>g\in{\cal K}\bigr\}\ ,$$
whence ${\cal S}$ is an affine subspace of $C^1(I)$.
(i) So far, ${\cal S}$ could still be empty. Now the   variation of the constant serves as a "trick" (there are others, like the Laplace transform) to provide us with a single particular solution $f_0$ of $(2)$.
Let $x\mapsto A(x)$ $\>(x\in I)$ be a primitive of $a(\cdot)$. Then ${\cal K}$  consists of the functions $Ce^{-A(x)}$ with $C\in{\mathbb C}$. As $e^{-A(x)}$ is never zero we loose nothing by writing the prospective $f_0\in{\mathbb S}$ in the form $$f_0(x):=c(x)e^{-A(x)}\ .\tag{3}$$ Introducing this "Ansatz" into $(2)$ leads to the condition
$$c'(x)=b(x)e^{A(x)}\qquad(x\in I)$$
for $c(\cdot)$, which is satisfied, e.g., by setting
$$c(x):=\int_{x_0}^x b(t)e^{A(t)}\ dt\qquad(x\in I)\ ,$$
where $x_0\in I$ can be chosen arbitrarily. This $c(\cdot)$ is in $C^1(I)$, and one has $c(x_0)=0$. I leave it to the reader to verify that the function $(3)$ so constructed is indeed in ${\cal S}$.
(iv) According to (iii) the general solution of $(2)$ is now of the form
$$f(x)=c(x) e^{-A(x)}+Ce^{-A(x)},\qquad C\in{\mathbb C}\ .$$
The initial condition $f(x_0)=y_0$ enforces $C=y_0e^{A(x_0)}$, so that we finally obtain
$$f(x)=\left(y_0+\int_{x_0}^x b(t)e^{A(t)-A(x_0)}\ dt\right)e^{A(x_0)-A(x)}\ .$$
Note that differences $A(t)-A(x_0)$ can be written as $\int_{x_0}^t a(\tau)\ d\tau$.
