Matrix Exponential: Solve $u''+2u'+u=0$ for $u(0)=u_0, u'(0)=u_1$ I am trying to solve the problem $u''+2u'+u=0$ for $u(0)=u_0, u'(0)=u_1$ using the matrix exponential. I first wrote the linear system as $x'_1=x_2$ and $x'_2=-x_1-2x_2$. I then found $e^{At}= $$ \left[ \begin{array}{ c c } (t+1)e^{-t} & te^{-t} \\\ -te^{-t} & (1-t)e^{-t} \end{array} \right] $$ $ My book then says that $\vec{x}=e^{A(t-t_0)}x_0$. My question is how do I use this exponential matrix and the initial conditions to find the final answer? Thanks.

The solution $x(t)=e^{A(t-t_0)}x_0$ of $\dot{x}(t)=Ax(t)$ has the initial value $x_0$ for $t=t_0=0$ and $x$ is defined as $x=(x_1,x_2)=(u,u')$. Thus $x_0$ should be defined as $x_0=(u(0),u'(0))=(u_0,u_1)$.