Answer
$y=-e^{-x}+4e^{0.5x}$
Work Step by Step
$2y''+y'-y=0$
Use auxiliary equation
$2r^{2}+r-1=0$
$(2r-1)(r+1)=0$
$r_{1}=-1$
$r_{2}=\frac{1}{2}$
$y=c_{1}e^{r_{1}x}+c_{2}e^{r_{2}x}$
$y=c_{1}e^{-x}+c_{2}e^{0.5x}$
$y'=-c_{1}e^{-x}+\frac{1}{2}c_{2}e^{0.5x}$
Given:$y(0)=3$
$y(0)=3=c_{1}e^{0}+c_{2}e^{0}$
$c_{1}+c_{2}=3$
$y'(0)=3=-c_{1}e^{0}+\frac{1}{2}c_{2}e^{0}$
$-c_{1}+\frac{1}{2}c_{2}=3$
$c_{1}+c_{2}=3$
$-c_{1}+\frac{1}{2}c_{2}=3$
$c_{1}=3-c_{2}$
$-(3-c_{2})+\frac{1}{2}c_{2}=3$
$-3+c_{2}+\frac{1}{2}c_{2}=3$
$\frac{3}{2}c_{2}=6$
$c_{2}=4$
$c_{1}=-1$
$y=-e^{-x}+4e^{0.5x}$
