$
\text{14.}\quad y^{\prime \prime}+4 y^{\prime}+4 y=0, y(-1)=2, y^{\prime}(-1)=1
$
The characteristic equation is,
$$
\begin{aligned} r^{2}+4 r+4 &=0 \\(r+2)^{2} &=0 \\ r_{1}=r_{2} &=-2 \end{aligned}
$$
since it has repeated roots, the general solution is:
$$
\begin{array}{l}{y(t)=c_{1} e^{-2 t}+c_{2} t e^{-2 t}} \\ {y^{\prime}(t)=-2 c_{1} e^{-2 t}+c_{2} e^{-2 t}-2 c_{2} t e^{-2 t}}\end{array}
$$
$\operatorname{plug}$ in $y(-1)=2, \quad y^{\prime}(-1)=1$
we get $$\left\{\begin{array}{l}{c_{1} e^{2}-c_{2} e^{2}=2} \\ {-2 c_{1} e^{2}+c_{2} e^{2}+2 c_{2} e^{2}=1}\end{array}\right.$$
$$
\therefore\left\{\begin{array}{l}{c_{1}=7 e^{-2}} \\ {c_{2}=5 e^{-2}}\end{array}\right.
$$
substitute $C_{2}$ and $C_{2}$ into $y(t)$
$$
=7 e^{-2} e^{-2 t}+5 e^{-2} t e^{-2 t}
$$
$\therefore$ The general solution is $y(t)=7 e^{-2(t+1)}+5 t e^{-2(t+1)}$
