I am trying to improve my understanding of the Green's Function and its use to solve second-order linear ODEs.
On
It would not print my just prior answer and i cannot see it nor can I delete it and seems the interface here is just 'hung up' on that 'muffed' answer so I try the same again.
With respect to (1) would say 'No'; Greens function method is not a general universal choice for solving the most general second order diff eq. It is usually for inhomogenous eq's and more often for those such as wave equation, Laplacian and the like as opposed to your particular example which I guess you deleted because I don't see it now. For now will give a couple of rather elementary examples and afterward example a little less trivial so you may be able to have more of an idea of under what circumstances it may be applied and depending upon what you call interesting etc. it may qualify for your note (2) and (3). Green's fct is also quite often used in partial diff eq's and more than one dimension.
Briefly or simply put in it's most basic and familiar form the greens fct is often put in terms of primed and unprimed coordinate(s) in which it is generally symmetric and such that applying differential operator the result is zero everywhere except where $x\!=\!x'\;$ or it gives $\delta(x\!-\!x')$ and in short it's formal definition is the solution{written in a rather sloppy and non-rigorous form} to $Dop\; G(x\!-\!x')\!=\!\delta(x\!-\!x')$ usually also satisfying some boundary conditions depending upon the case considered where have written $Dop$ to mean a general(or partial) differential operator usually linear. Though it can be and often is considered in finite regions of various boundary conditions, for now just consider open infinite space in one dimension. and the examples to follow should make it more clear.
For example to solve for $g(x)$ in $\;\left(1-\frac{d^2}{d\:x^2}\right)g(x)=f(x)\;$ we form $G(x\!-\!x')$ such that $\left(1-\frac{d^2}{d\:x^2}\right)G(x\!-\!x')=\delta(x\!-\!x')$ so then $\left(1-\frac{d^2}{d\:x^2}\right)\int G(x\!-\!x')f(x')\:dx'=\int\delta(x\!-\!x')f(x')\:dx'=f(x)$ and $$g(x)\!=\!\int G(x\!-\!x')f(x')\:dx'$$
Depending upon the form of $f(x)$ using $\int_{-\infty}^{\infty}e^{ik(x-x')}dk\!=\!2\pi\delta(x\!-\!x')$ obtain
$$\left(1-\frac{d^2}{d\:x^2}\right)\int_{-\infty}^{\infty}\left \{\frac1{2\pi}\!\int_{-\infty}^{\infty}\frac1{(1+k^2)}e^{ik(x-x')}dk\right\}f(x')dx'\;$$
$$=\int_{-\infty}^{\infty}\left \{\frac1{2\pi}\!\int_{-\infty}^{\infty}e^{ik(x-x')}dk\right\}f(x')dx'\!=\!f(x)$$ So $$G(x)\!=\!\int_{-\infty}^{\infty}\frac1{2\pi(1+k^2)}e^{ikx}dk$$
and $$\;g(x)\!=\!\int_{-\infty}^{\infty}\left \{\frac1{2\pi}\!\int_{-\infty}^{\infty}\frac1{(1+k^2)}e^{ik(x-x')}dk\right\}f(x')dx'\!=\!\int G(x\!-\!x')f(x')\:dx'\;$$ So for this particular example we need $f(x)$ such that the double integral may be evaluated perhaps by interchanging order of integration or whatever. Also it should be clear that we could also solve $\;C\!-\!\frac{d^2}{d\:x^2}\;$ where $\;C\;$ is any positive constant replacing the prior $\;1$.
Now the above is not the only way to form a delta function and perhaps related is that often in other than rectangular coordinates, solutions to the homogenous eq. are not valid over all space but consider {such as perhaps in spherical coord's for example} one solution goes to 0 at inf but becomes infinite at 0 and another solution vice-versa so a solution is to put them together sort of back to back such that there is a discontinuity in first derivative so that taking 2nd derivative yields proportional to delta fct. In fact can demonstrate this same idea in open infinite one dim rectangular. Suppose we require a solution, $g(x)$ say, to $\left(1\!-\!\frac{d^2}{d\:x^2}\right)g(x)\!=\!f(x)$ and require the Green's fct to vanish at infinity. A solution may be written in terms of a Green's fct $$G(x)\!=\!\frac1{2}e^{-|x|}\;$$ because $\;\left\{1\!-\!\frac{d^2}{d\:x^2}\right\}G(x)\!=\!\delta(x)\;$ and $$g(x)\!=\!\int_{-\infty}^{\infty}G(x-x')f(x')\:dx'\;$$ and this green's fct is same as just prior expressed in a fourier integral type form.
For a less trivial example I had occasion to analyze the kinetic energy or Laplacian plus a constant, which took as unity , on a group of particles in quantum theory of total $3n$ rectangular dim's. in hyperspherical coordinates. Without going into detail, after certain angular solutions we have at last the solution to the hyperradius coord. with parameter(s) depending upon particular discrete quantum numbers according angular eigensolutions, such as $\mathit{lk}_2$ which may be considered as total generalized angular momentum of which detail we may not concern here $$\left\{-{\frac{d^2}{{\mathit{dx}_1}^2}}-\frac{(3n\!-\!1)\dfrac{d}{\mathit{dx}_1}}{x_{1}}+\mathit{lk}_2 \frac{(\mathit{lk}_2+3n-2)}{{x_{1}}^2}+1\right\}\mathit{fr}(x_1,y_1)=0$$ except at $x_1\!=\!y_1$ and the variables $x_1,y_1$ in which it is symmetric are analogous to the prime and unprimed coord's prior. A delta fct singularity is effected by a discontinuity in the first derivative at $x_1\!=\!y_1\;.$ We set the magnitude of the Wronskian between two solutions of which one is regular at zero and the other at infinity equal to unity after multiplied by a volume normalization $x_1^{3n\!-\!1}$ integrating factor. More specifically the correct solution for n even is of form $$\mathit{fr}(x_1,y_1)=\frac{\pi\,i\,J_{(3n\!/2+\mathit{lk}_2\!-\!1)}(i\,\mathit{rl})\left(J_{(3n\!/2+\mathit{lk} _2\!-\!1)}(i\,\mathit{rg})+iY_{(3n\!/2+\mathit{lk}_2\!-\!1)}(i\,\mathit{rg})\right)}{2x_1^{(3n\!/2\!-\!1)}y_1^{(3n\!/2\!-\!1)}}$$ where $\mathit{rl},\mathit{rg}$ is the lesser,greater of $x_1,y_1.\,$ For n odd $\;\mathit{fr}\;$ can be written in terms of elementary functions of finite power series times an exponential which is actually simpler but we don't write that detail here because it is somewhat involved and we can demonstrate this property for odd n more directly on a similar but simpler case.{ Note in the following $r$ means the magnitude of a hyperradius of $3n$ rectangular dim's and not just one dim $r$ nor in general only 3 dim's.} Instead to demonstrate we attempt a solution $G_1$ to $\{1-\nabla^2\}G_1=0,r\ne 0$, for a similar but different green's fct which has a simple terminating solution just as the prior that did not write only for odd n but not even and easier to show than the prior, and has singularity at $r=0$ in form ${\rm{G1}}\!=\!{\rm{G}}\,e^{-r}(1+\mathit{Ps})$ where $\mathit{Ps}$ is a power series in $r$ to be determined with the constant term removed,\begin{gather*}-\left (\frac{d^{2}{\rm{G1}}}{dr^{2}}+\frac{(3n-1)}{r}\frac{d{\rm{\,G1}}}{dr} \right)+{\rm{G1}}=0,r\ne 0.\end{gather*} For n odd the series terminates after (3n-1)$/$2 terms. For example for n$=$3\begin{gather*} {\rm{G1}}=\frac{(15+15r+6r^{2}+r^{3})e^{-r}} {32\pi ^{4}r^{7}}\text{ and for n$=$5 }\\[8pt] {\rm{G1}}=\frac{(10395+10395r+4725{r}^{2}+1260{r}^{3}+210{r}^{4}+21{r}^{5}+r^{6})e^{-r}}{256\pi^{7}r^{13}}\end{gather*} and in general for n odd $>$ 2 \begin{gather*}{\rm{G1}}=\Gamma{\left (\frac{3n}{2}\right)}\frac{\left(\sum _{i=1}^{\frac{3(n-1)}{2}}\prod_{j=1}^{i}{\frac{(3n-2j-1)}{j(3n-j-2)}}r^{i}+1\right)e^{-r}}{2\pi^{3n/2}(3n-2)r^{3n-2}}.\end{gather*}
In general, Greens functions do not tend to be well-behaved at all points in space, which can lead to problems.
For example consider finding a solution to electromagnetic radiation problem, using the Green function, but ask for solution at the point where the source, i.e. some oscillating charge distribution is located. See, for example, W. C. Chew "Waves and Fields in Inhomogeneous Media". You then need to think of ways of isolating your observer from the source etc.
Another interesting area is the null-space of Green functions, i.e. non-trivial sources that do not lead to excitation. Again, in electromagnetism, there is a lot of info on this in A. J. Devany, "Mathematical foundations of Imaging, Tomography and Wavefield Inversion". There such things are called the non-radiating configurations.