So the given problem is $$y''-y=e^{-x},~y(0)=y'(0)=0$$ and $$y:\mathbb{R}\to\mathbb{R}$$ is a solution. They ask whether $y(x)$ is bounded and does it attain minimum in $\mathbb{R}$?
So I solved the problem to arrive at $$y(x)=-\dfrac{1}{4}e^{-x}+\dfrac{1}{4}e^x-\dfrac{1}{2}xe^{-x}$$.
Since $e^x$ is unbounded for $x>0$ so I concluded that $y(x)$ is unbounded. Is my analysis correct? I couldn't arrive at any conclusion about the minimum of $y(x)$ in $\mathbb{R}$. How can I do this?
On
To determine the minimum value of $y(x)$, find $y(x)$ for all $x$ such that $\frac{\mathrm{d}y}{\mathrm{d}x} = 0$, $\lim_{x \to -\infty}y(x)$, and $\lim_{x \to \infty}y(x)$. Then determine which of these is least.
Assuming that your solution is correct, $\lim_{x \to \infty} y(x)$ is unbounded because the $e^x$ term dominates the sum as $x$ becomes large.
Since they're asking about $\mathbb{R}$, let's look at both ends of the solution.
$$y(x)=-\dfrac{1}{4}e^{-x}+\dfrac{1}{4}e^x-\dfrac{1}{2}xe^{-x}$$ So at $x\rightarrow\infty$, $y\rightarrow\infty$
At $x\rightarrow-\infty$, $y\rightarrow\infty$
Since the function is continuous, and tends to infinity on both ends of $\mathbb{R}$, we can conclude it has a minimum.
(More precisely, there exists some $M>0$ s.t. on $(\infty,-M]\cup[M,\infty)$ the function is larger, than, say, $10^6$.
And then we may use the Weierstrass Extreme Value theorem in $[-M,M]$)