Notice that $\displaystyle \mathcal{L}(cf) = c\mathcal{L}(f)$ and $\displaystyle \mathcal{L}(f+g) = \mathcal{L}(f) + \mathcal{L}(g)$ where $\displaystyle \mathcal{L}$ is $\displaystyle \tfrac{d}{dx}$ or $\displaystyle \tfrac{d^2}{dx^2}$ if we want to abuse notation.