A bit simpler: the general solution is $y=\bigl(C_1 + C_2 (t+1)\bigr) e^{-2 (t+1)}$: it is equivalent (with the different constants to $\bigl(C_1 + C_2 t\bigr) e^{-2 t}$ but we simply moved the "time" to start from $0$ rather than from $-1$: $t_{\text{new}}=t+1$.

Then $y(-1)=C_1=2$ and $y'(-1)= -2 C_1+ C_2=1$. Then $C_1=2$, $C_2=5$ and $y=(7+5t)e^{-2t+2}$. Indeed graph is similar to one sketched by Yeming Wen: it tends to $+0$ as $t\to +\infty$, crosses $y=0$ as $t=-7/5$ and fast tends to $-\infty$ as $t\to -\infty$.