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