I had made a post in the past about the same problem and unfortunately I wasn't clear enough(adsbygoogle = window.adsbygoogle || []).push({});

so I am trying again.

I am studying an article and there I found without any proof that the solution of:

Let ##g \in \mathbb{C}## and let ##u:(0,\infty)\to \mathbb{C}##

$$ -u''+\lambda^2u=f\,\, on \,(0,\infty);\quad u(0)=g,\quad \lim_{x\to\infty}u(x)=0,$$

is given by using a Green's function and it is the following:

For $$z\in (0,\infty),$$

##u(z)=\frac{1}{2\lambda^2}\Bigl(e^{-\lambda z}\int_{0}^{\lambda z}e^yf(y/\lambda)dy+e^{\lambda z}\int_{\lambda z}^{\infty}e^{-y}f(y/\lambda)dy-

e^{-\lambda z}\int_{0}^{\infty}e^{-y}f(y/\lambda)dy\Bigr)+ge^{-\lambda z}.##

I would be very grateful if someone explain me how can green's function be find for this problem, because I am interested to find a respectively solution for the problem,

$$ -u''+\lambda^2u'=f\,\, on \,(0,\infty);\quad u(0)=g,\quad \lim_{x\to\infty}u(x)=0.$$

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# Solve second order ode with Green function

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