The second question is still not answered ('how to show that this sum is a general solution'). I was looking for this myself, then came across this post.
I think I understand it now. Thought I might as well post it here even though this is a very old question. If anyone ever reads this : let me know if there's something wrong with the explanation below...
Assume we have two different solutions $x(t)=\alpha(t)$ and $x(t)=\beta(t)$ to an ODE of the form $\frac{dx}{dt}+a(t)x(t)=0$.
For the two curves to cross each other we have to have :
$\alpha(t_0)=\beta(t_0)$ and $\frac{d}{dt}\alpha(t)|_{t=t_0} \neq \frac{d}{dt}\beta(t)|_{t=t_0}$ for some $t_0$ .
With $\frac{d \alpha}{dt}+a(t) \alpha(t)=0$ and $\frac{d \beta}{dt}+a(t) \beta(t)=0$ we see that this is impossible. Because the curves cannot cross they must be uniquely determined by their value for some $t=t_0$.
Now in a second order ODE like $\frac{d^2x}{dt^2}+a(t)\frac{dx}{dt} + b(t)x=0$ , two solutions can cross each other.
But when we assume two solutions with : $\alpha(t_0)=\beta(t_0)$ and $\frac{d}{dt}\alpha(t)|_{t=t_0} = \frac{d}{dt}\beta(t)|_{t=t_0}$ and $\frac{d^2}{dt^2}\alpha(t)|_{t=t_0} \neq \frac{d^2}{dt^2}\beta(t)|_{t=t_0}$ for some $t_0$ we again see that this cannot be done.
So for two solutions $\alpha(t)$ and $\beta(t)$ with : $\alpha(t_0)=\beta(t_0)$ we see that the curves of $\frac{d\alpha}{dt}$ and $\frac{d\beta}{dt}$ can never cross each other because when $\frac{d\alpha}{dt}$ and $\frac{d\beta}{dt}$ are equal in $t_0$ then their derivatives $\frac{d}{dt}(\frac{d\alpha}{dt})$ and $\frac{d}{dt}(\frac{d\beta}{dt})$ must also be equal.
Thus curves of $\frac{dx}{dt}$ are uniquely determined by the choice of two boundary conditions : $x(t_0)$ and $\frac{dx}{dt}|_{t=t_0}$ . By integration of $\frac{dx}{dt}$ keeping the boundary conditions fixed, this in turn means that $x(t)$ must be completely determined by this choice of boundary conditions.
Same goes for higher order linear homogeneous ODE's of course.
The only thing we have to check now for the solution to be the complete solution is that we must be able to reach all possible function values with the constant coefficients we use in our solution.
This can be done using the Wronskian determinant.
UPDATE:
It is also possible for curves to cross each other while having the same derivative e.g.: $x = \sin t $ and $ x=t $ when $ t=0 $.
But that would mean that the second derivative of at least one of the curves must be zero.
They can also touch each other without crossing, but in both cases their derivatives must be equal in the point where they coincide ( at $t_0$ ). This means if we take the derivatives in the first order example above:
$\frac{d^2 \alpha}{dt^2}+\frac{d }{dt}a(t) \alpha(t) + a(t)\frac{d }{dt} \alpha(t)=0$ and $\frac{d^2 \beta}{dt^2}+\frac{d }{dt}a(t) \beta(t) +a(t) \frac{d }{dt}\beta(t)=0$
So at $t=t_0$ we must have : $\frac{d^2 \alpha}{dt^2}|_{t=t_0}=\frac{d^2 \beta}{dt^2}|_{t=t_0} $ . We can now continue this process for all higher derivatives and conclude that all higher derivatives of $\alpha$ and $\beta$ must be equal at $t=t_0$ .
This of course means that $\alpha$ and $\beta$ must be the same curves.
This last argument must be used to replace the first of course ! Whenever two curves have equal boundary conditions at some point, the two curves must be entirely the same because all their derivatives are the same at $t_0$ !
That implies they must be uniquely determined by their boundary conditions.
Wronskian tells us whether all possible boundary conditions can be satisfied. If that is the case we know we have the general solution.
