A question about Green's function

[karlzr]

There is a second order nonhomogeneous equation of motion with nonzero initial condition given at $t=-\infty$:
$D^2 y(x)=f(x)$ with $y(-\infty)=e^{-i x}$
where I have used the shorthand notation $D^2$ for the full differential operator. Also I have the two solutions $y_1(x)$ and $y_2(x)$ to the homogeneous equation with $y_1(x \to -\infty) \approx e^{-ix}$ and $y_2(x \to -\infty) \approx e^{ix}$. So how do I construct the particular solution using $y_1(x)$ and $y_2(x)$?

I know Green's function can be constructed using the two homogeneous solutions. So the naive solution I got is
$y(x)=y_1(x)+\int^x_{-\infty} dx' G(x,x')f(x')$ with
$G(x,x')=\frac{y_1(x)y_2(x')-y_2(x)y_1(x')}{W(y_2(x'),y_1(x'))}$.

Is it wrong? since I get some ridiculous result (divergence) when I use this solution to do calculation, because the lower bound is $-\infty$.

 

[MisterX]

I know Green's function can be constructed using the two homogeneous solutions. So the naive solution I got is
$y(x)=y_1(x)+\int^x_{-\infty} dx' G(x,x')f(x')$ with
$G(x,x')=\frac{y_1(x)y_2(x')-y_2(x)y_1(x')}{W(y_2(x'),y_1(x'))}$.

Is it wrong? since I get some ridiculous result (divergence) when I use this solution to do calculation, because the lower bound is $-\infty$.

Where are you getting this out of curiosity? There are other ways to obtain the Green's function, at least.

 

[karlzr]

Where are you getting this out of curiosity? There are other ways to obtain the Green's function, at least.

Actually I obtained this expression from the approach of variation of parameters. It plays the role of Green's function in this case.