Answer
See below
Work Step by Step
Given $V=C^2(I)$
We can write set $S$ as $S=\{y \in C^2(I):y''+2y'-y=1\}$.
Assume that $y_0 \in C^2(I)$ such as $y_0(t) =0 \forall t \in I$.
By then we have $y''_0(t)=0,y_0'(t)=0 \forall t \in I$
We can see that $y_0''(t)+2y_0'(t)-y_0(t)=0+2(0)-0=0 \forall t\in I \rightarrow y_0 \notin S$. Hence, $S$ is a subspace of $C^2(I)$
