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Show that the operator $$(D-1)(D+2)$$ is the same as the operator $$D^{2}+D-2$$.
$$\begin{array}{l}{x^{\prime}+2 y=0} \\ {x^{\prime}-y^{\prime}=0}\end{array}$$
$$\begin{array}{l}{x^{\prime}=x-y} \\ {y^{\prime}=y-4 x}\end{array}$$
$$\begin{array}{l}{x^{\prime}+y^{\prime}-x=5} \\ {x^{\prime}+y^{\prime}+y=1}\end{array}$$
$$\begin{array}{l}{x^{\prime}=3 x-2 y+\sin t} \\ {y^{\prime}=4 x-y-\cos t}\end{array}$$
$$\begin{array}{l}{(D+1)[u]-(D+1)[v]=e^{t}} \\ {(D-1)[u]+(2 D+1)[v]=5}\end{array}$$
