1. ## Green's Function

$y'' - y = f(x)$
Subject to $y (\pm \infty) = 0$

I started by looking for a Green's Function of the form

$G_{xx} (x, \xi) - G(x, \xi) = \delta(x - \xi)$

For all $x \ne \xi, \delta(x - \xi) = 0$

And thus we have two cases.

Case 1: $x < \xi$

$G_{xx} (x, \xi) - G( x, \xi) = 0$
Implying that $G(x, \xi) = c_1 e^x + c_2 e^{-x}$
Applying our left-hand BC,
$G( -\infty, \xi) = 0 = c_2 e^{\infty}$
So to ensure that our solution is bounded at either extreme, we insist that $c_2$ is 0.

Case 2: $x > \xi$

$G_{xx} (x, \xi) - G(x, \xi) = 0$
Which also reduces to the form $G(x, \xi) = c_3 e^x + c_4 e^{-x}$
Applying our right-hand BC, $G( \infty, \xi) = 0 = c_3 e^{\infty}$
So similarly, we insist that $c_3 = 0$.

Combining our results, we have then that

$G(x, \xi) =$
$c_1 e^x; x < \xi$
$c_4 e^{-x}; x > \xi$

However, we want to make sure that there's continuity at $x = \xi$ and a finite jump of 1 in the first derivative, so we get that

$c_1 e^{\xi} = c_4 e^{-\xi}$
And
$- c_4 e^{-\xi} - c_1e^{\xi} = 1$

Which can be combined to give
$c_4 = \frac{e^{\xi}}{-2}$ and $c_1 = -\frac{1}{2 e^{\xi}}$.

All I'm really looking to do is make sure that I've correctly found the Green's function up to this point.

2. Let $u := \dot{y}-y$. Then your equation is $\dot{u}+u=f(x)$, which has solution $u(x) = e^{-x}(\int^x e^s f(s) \,\mathrm{d}s + A)$. Plugging back in you get $y(x) = e^x(\int^x e^{-s}u(s) \,\mathrm{d}s + B)$.

3. I'm not entirely sure I understand. I'm specifically trying to solve the ODE using this method, and I was wondering if, up until this point, I have worked it through correctly.

4. Your Green function is correct, I am pretty sure.

5. Alright, so, with an application of Green's Theorem and a little reduction, I should end up with

$u(x) = \frac{-1}{2}(\int_{ -\infty}^{x} \frac{f(\xi) e^{\xi}}{e^x} d \xi + \int_{x}^{\infty} \frac{f(\xi) e^x}{e^{\xi}} d \xi)$

Where $u(x)$ is a solution to the ODE?

6. I think maybe it should be $u(x) = \frac{-1}{2}(\int_{ -\infty}^{x} \frac{f(\xi) e^{\xi}}{e^x} d \xi {\color{red}-} \int_{x}^{\infty} \frac{f(\xi) e^x}{e^{\xi}} d \xi)$; I'll double-check.

edit: okies.

Let $u(x) = \frac{-1}{2}(e^{-x}\int_{-\infty}^x e^{\xi}f(\xi)\,\mathrm{d}\xi - e^x \int_{x}^{\infty} e^{-\xi}f(\xi) \,\mathrm{d}\xi)$

then $u'(x) = \frac{-1}{2}(-e^{-x}\int_{-\infty}^x e^{\xi}f(\xi)\,\mathrm{d}\xi + e^{-x}e^xf(x) - e^x \int_{x}^{\infty} e^{-\xi}f(\xi) \,\mathrm{d}\xi - e^xe^{-x}f(x))$ $= \frac{-1}{2}(-e^{-x}\int_{-\infty}^x e^{\xi}f(\xi)\,\mathrm{d}\xi - e^x \int_{x}^{\infty} e^{-\xi}f(\xi) \,\mathrm{d}\xi )$

and $u''(x) = \frac{-1}{2}(e^{-x}\int_{-\infty}^x e^{\xi}f(\xi)\,\mathrm{d}\xi - e^{-x}e^xf(x) - e^x \int_{x}^{\infty} e^{-\xi}f(\xi) \,\mathrm{d}\xi - e^xe^{-x}f(x))$. This should satisfy the DE.

7. I see why that negative sign should be there, but I don't see how it comes about when working through to the solution. Did I solve for the coefficients incorrectly?