I want to find the Green function for $$u'' +2u'-3u=f(x)$$ for which $0<x< \infty, u(0) = 0, \lim_{x\to \infty} u(x)=0$.
However, first I have to solve this bvp and I do not know how to solve it because of the condition that $\lim_{x\to \infty} u(x)=0$.
To add, in the question it is said that we can take $f(x)=e^{-6x}$. During my research, I saw that to use the boundary condition we should know have that $\lim_{x\to \infty} f(x)=c$ for some constant $c$. We can see that the function $f$ satisfies this condition.
So you need to solve $$ u''(t)+2u'(t)-3u(t)=\delta(t-s) $$ or $$ u''(t)+2u'(t)-3u(t)=0\text{ for }t\ne s $$ with the conditions $u(0)=0$, $u(s^-)=u(s^+)$, $u'(s^-)+1=u'(s^+)$ and $u(\infty)=0$. On the first part the solution has thus to be $u(t)=A(e^{t}-e^{-3t})$ and on the second segment $u(t)=Be^{-3t}$ The connecting condition is $$ A(e^{s}-e^{-3s})=Be^{-3s}\implies B=A(e^{4s}-1)\\~\\ A(e^s+3e^{-3s})+1=-3Be^{-3s}=-3A(e^{s}-e^{-3s})\implies 1=-4Ae^s $$ so that in combination of all the different solutions the Green function is \begin{align} G(t,s)&=\begin{cases} -\frac14 e^{-s} (e^{t}-e^{-3t})&t\le s,\\ -\frac14(e^{3s}-e^{-s})e^{-3t}&t>s. \end{cases}\\ &=-\frac{e^{2s}}4(e^{\min(s,t)}-e^{-3\min(s,t)})\,e^{-3\max(s,t)}\\ &=-\frac{e^{-s}}4(e^{t-4\max(0,t-s)}-e^{-3t}) \end{align}