The Green function can be obtained by solving the equation $$ G''(x)-K^2G(x)=\delta(x) $$ being $\delta(x)$ the Dirac distribution. This is the proper definition. The general solution of this can be obtained by writing $$ G(x)=a\cdot\theta(-x)e^{Kx}+b\cdot\theta(x)e^{-Kx)} $$ being $a$ and $b$ two constant to be fixed and $\theta(x)$ the Heaviside step function. Putting this into the equation yields $$ G''(x)-K^2G(x)=-K b \delta(x) - K a \delta(x) + b \delta'(x) - a \delta'(x)=\delta(x). $$ So, it is $b=c$ and $b=-1/(2K)$. So, finally $$ G(x) = -\frac{1}{2K}\left[\theta(-x)e^{Kx}+\theta(x)e^{-Kx)}\right]. $$ I think you can start from here.

The Green function can be obtained by solving the equation $$ G''(x)-K^2G(x)=\delta(x) $$ being $\delta(x)$ the Dirac distribution. This is the proper definition. The general solution of this can be obtained by writing $$ G(x)=a\cdot\theta(-x)e^{Kx}+b\cdot\theta(x)e^{-Kx)} $$ being $a$ and $b$ two constant to be fixed and $\theta(x)$ the Heaviside step function. Putting this into the equation yields $$ G''(x)-K^2G(x)=-K b \delta(x) - K a \delta(x) + b \delta'(x) - a \delta'(x)=\delta(x). $$ So, it is $b=c$ and $b=-1/(2K)$. So, finally $$ G(x) = -\frac{1}{2K}\left[\theta(-x)e^{Kx}+\theta(x)e^{-Kx}\right]. $$ I think you can start from here.