# A question about Green's function

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##.

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.

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.