$(1$ point$)$ The differential equation

$y^{\text{'}\text{'}}-y=0$

has one of the following two parameter families as its general solution

$y=c_1{e}^x+c_2{e}^{-x}$

$y=c_{1}cos(x)+c_{2}sin(x)$

$y=c_{1}tan(x)+c_{2}sec(x)$

$y=c_1+c_{2}x$

Find the solution such that $y(0)=8$ and $y^{\text{'}}(0)=4$ :