 Solution

First solve for $v=y^{\prime }$. Multiplication by $e^x$ and use of $v(1)=0$ giveIntegrating from zero to $x$, and using $y(0)=0$, yields $y(x)=-{\int }_0^x{e}^{-s}{\int }_s^1{e}^{t}f(t)dtds$. Reversing the order of integration givesThus $G(x,t)$ equals $G_1$ for $t<x$ and $G_2$ for $x<t$. It is continuous at $x=t$, vanishes at $x=0$, and has $G_x(1,t)=0$. Its derivative jumps from $-1$ to zero across $x=t$, so the jump is $+1$, matching the coefficient of $y^{\prime \prime }$. Both pieces satisfy $G_{xx}+G_x=0$ away from the jump. These facts also verify the Green representation directly. The homogeneous solutions are $A+B{e}^{-x}$, and the two boundary conditions force $A=B=0$, proving uniqueness.

For $f(t)=e^{-t}$, the formula givesHence $y(x)=-x{e}^{-x}$. Indeed $y^{\prime }=(x-1)e^{-x}$ and $y^{\prime \prime }=(2-x)e^{-x}$, which verify the equation and both boundary conditions.