Solution to differential equation $\left( 1-\lambda\frac{\partial}{\partial z}\right)w(x,y,z)-g(x,y,z+h)=0$ I'm trying to solve the following differential equation:
$\left( 1-\lambda\frac{\partial}{\partial z}\right)w(x,y,z)-g(x,y,z+h)=0$
here $g(x,y,z+h)$ is a known function that however i will leave unspecified moreover we are dealing with Real variables and constants.
The paper that i'm following reports the following solution:
$w(x,y,z)= \frac{1}{2\lambda}\int_{0}^{\infty}e^{-\frac{s}{2\lambda}}g(x,y,z+h+s)ds$
I'm sure this comes from an exponential integrating factor, in fact arranging the original differential equation in the following way:
$\frac{-1}{2 \lambda}w(x,y,z)e^{-\frac{z}{2 \lambda}}+e^{-\frac{z}{2 \lambda}}\frac{\partial}{\partial z}w(x,y,z)=-\frac{1}{2 \lambda}e^{-\frac{z}{2 \lambda}}g(x,y,z+h)$
we obtain the following differential equation:
$\frac{\partial}{\partial z} \left( e^{-\frac{z}{2 \lambda}}w(x,y,z)\right)=-\frac{1}{2 \lambda}e^{-\frac{z}{2 \lambda}}g(x,y,z+h)$
now i'm quite clueless on how to proceed, any ideas?
thanks in advance
 A: First, we may note that the $x,y$ dependence is irrelevant, along with the shift in the argument of $g(x,y,z)$. To that end I'll introduce $W(z)=w(x,y,z)$ and $G(z)=g(x,y,z+h)\to g(z)$. Then we just have a simplified ODE given as $(1-\lambda D_z)W(z)=G(z)$.
This can indeed be handled by an integrating factor by observing
$$\dfrac{d}{dz}\left(e^{-z/\lambda}W\right)=-\frac{1}{\lambda}e^{-z/\lambda}\left(W-\lambda \frac{dW}{dz}\right)=-\frac{1}{\lambda}e^{-z/\lambda}G(z),$$ which if integrated on the interval $[z,\infty)$ implies
$$e^{-z/\lambda}W(z)-(e^{-z/\lambda}W)_{z=\infty}=\frac{1}{\lambda}\int_{z}^\infty e^{-z'/\lambda}G(z')\,dz'$$
We can drop the boundary term at $z=\infty$ on the LHS if we assume that $W(z)$ does not grow exponentially fast as $z\to\infty$. Shifting the exponential prefactor to the RHS then gives
$$W(z)=\frac{1}{\lambda}\int_{z}^\infty e^{-(z'-z)/\lambda}G(z')\,dz'
=\frac{1}{\lambda}\int_{0}^\infty e^{-s/\lambda}G(z+s)\,ds$$
where $s:=z'-z$. Returning to the original $(x,y,z)$ dependence yields the integral representation
$$w(x,y,z)=\frac{1}{\lambda}\int_{0}^\infty e^{-s/\lambda}g(x,y,z+h+s)\,ds.$$ This is very close to the desired form; what remains mysterious are the factors of two. This may explained by going back to the paper cited in the comments to the OP: Those equations are defined with the operator $1-2\lambda \dfrac{\partial}{\partial z}$. Consequently we should make the replacement $\lambda\mapsto 2\lambda$ in the results above when comparing with the formulae found in the paper, yielding the result in the OP.
