Answer
$${y}=\exp^{x}$$
Work Step by Step
1) $$y''+y'-2y=0,\quad{y}(0)=1,\quad{y'}(0)=1$$Let $y-e^{\lambda{x}}$ so that $(\ln{y})'=\lambda$.
$${\lambda}^2+{\lambda}-2=0$$
$$(\lambda-1)(\lambda+2)=0$$
$$\lambda_{1,2}=1,-2$$
The general solution is ${y}=c_{1}\exp^{x}+c_{2}\exp^{-2x}$. Substituting in the constraints, we obtain
$c_{1}+c_{2}=1$ and $c_{1}-2c_{2}=1\rightarrow{c}_{1}+c_{2}=c_{1}-2c_{2}\rightarrow{c}_{2}=0\rightarrow{c}_{1}=1$.
$$\therefore{y}=\exp^{x}$$
