Solving Homogenous Linear Equation Initial Value Problem $Y''-4Y'-5Y=0 $at $Y(0)=2, Y'(0)=1$

Question

I've gone through this problem a few times and think I might be making a mistake somewhere along the way, because my answer isn't matching that of the back of the textbook. I was given the equation: $$Y'' - 4Y' - 5Y = 0 ; Y(0) = 2, Y'(0) = 1$$ I then wrote the auxiliary equation of: $$r^2 - 4r - 5 = 0$$ From there I factored to solve for values of r $$(r+1)(r-5) $$ so my values of r came to be: r = -1 and r = 5, so I wrote my two Y equations $$Y_1 = e^{5t} \ \ Y'_1 = 5e^{5t}$$ $$Y_2 = e^{-t} \ \ Y'_2 = -e^{-t} $$ Then I set up a system of equations using the two initial values of Y(0) = 2 and Y'(0) = 1 to get $$2 = C_1 + C_2$$ $$1 = 5C_1 - C_2$$ I Found C1 = 1/2 and C2 = 3/2, so I rewrote the general solution equation with values for C1 and C2 $$Y(t) = \frac{1}{2}e^{5t} + \frac{3}{2}e^{-t}$$ However, the answer in the back of my textbook says: $$2e^{5(t+1)} + e^{-(t+1)}$$ I'm really not seeing any clear place I messed up or how I would get to the answer in the back from my work.