Q:

Problem on a function that satisfies $f''(x) -2f'(x) +f(x) ≥ e^{x}$ for all $x \in [0,1]$.

If $f(x) :[0,1] \to R$ is a twice differentiable function such that $f(0)=f(1)=0$ and satisfies $f''(x) -2f'(x) +f(x) ≥ e^{x}$ for all $x \in [0,1]$. Then for $0<x<1$ find the range of $f(x)$.
I think this question involves the use of $O.D.E$, but I have no idea how to start.
I thought of forming a function $h(x)$ whose derivative is $ f''(x) - 2f'(x) + f(x) - e^{x}$. So $h(x)$ can be $f'(x) - 2f(x) + \int_{0}^x f(t)dt - e^{x}$
Then $h(x)$ is strictly incresing.
But that doesn't help.
I just need a hint how can I approach$?$

A:

Let $g(x)=f(x)e^{-x}$, so that $g'(x)=[f'(x)-f(x)]e^{-x}$ and therefore
$$
g''(x)=\bigl([f''(x)-f'(x)]-[f'(x)-f(x)]\bigr)e^{-x}\geq 1.
$$
Thus $h(x)=g(x)+x(1-x)/2$ satisfies $h(0)=h(1)=0$ and $h''(x)\geq 0$, hence by convexity of $h$ we have that $h(x)\leq 0$ for all $x\in [0,1]$. There is no lower bound for $h(x)$ since the functions $h(x)=nx(x-1)$ satisfy the conditions, for all positive $n$, and in particular $h(\tfrac12)=-n/4$ can be arbitrarily negative.
Thus $g(x)$ has a quadratic upper bound given by $-x(1-x)/2$ (and no lower bound), consequently
$$
f(x)\leq -\frac{x(1-x)e^x}{2}.
$$
Thus the range of $f(x)$ is the interval $[\min f(x),0]$ where $\min f(x)$ can be arbitrarily negative.

