Solution
= Solution
For $\lambda=2$, the Green representation of the <boundary value problem> is
$$
y(x)=\int_0^\infty G(x;\xi)\bigl[-(\xi+3)e^{-\xi}\bigr],d\xi.
$$
Splitting the integral at $\xi=x$ gives
$$
y(x)=e^{-x}\int_0^x(\xi+2),d\xi
+(x+2)\int_x^\infty e^{-\xi},d\xi.
$$
Therefore
$$
\boxed{y(x)=e^{-x}\left(\frac{x^2}{2}+3x+2\right)}.
$$
It satisfies $y(0)=2y'(0)$ and tends to zero as $x\to\infty$.