Find the solution of the given initial problem.
$$
y^{\prime \prime}+y^{\prime}-2 y=0, y(0)=1, y^{\prime}(0)=1
$$
We assume $y=e^{r t}$, then $r$ must be a root of the characteristic equation:
$$
\begin{array}{l}{r^{2}+r-2=0} \\ \therefore (r-1)(r+2)=0 \\ {r_{1}=1 \quad r_{2}=-2}\end{array}
$$
$\therefore$ the general solution is $y=c_{1} e^{t}+c_{2} e^{-2 t}$
$$
\begin{array}{l}{y^{\prime}=c_{1} e^{t}-2 c_{2} e^{-2 t}} \\ {\operatorname{plug} \text { in } y(0)=1, y^{\prime}(0)=1}\end{array}
$$
$$
\left\{\begin{array}{l}{C_{1}+C_{2}=1\qquad {{\small 1}}} \\ {C_{1}-2 C_{2}=1\quad~~ {{\small 2}}}\end{array}\right.
$$
${{\small 1}}-{{\small 2}}$ , we get
$$
\begin{array}{r}{3 C_{2}=0} \\ {C_{2}=0}\end{array}
$$
$$
\therefore C_{1}=1
$$
$\therefore$ the solution is $y=e^{t}$
