The general solution to the equation

$y^{\text{'}\text{'}}4y^{\text{'}}3y=0$

has the form $c_1{e}^{at}{c}_2{e}^{bt}$ where $a=b=y(0)=1,y^{\text{'}}(0)=0c_1=c_2=a.$

$a=$

$b=$

Now suppose that the initial conditions are $y(0)=1,y^{\text{'}}(0)=0.$ Then solve for:

$c_1=$

$c_2=$