Answer
$y=c_{1}e^{(2-\sqrt 3)x}+c_{2}e^{(2+\sqrt 3)x}$
Work Step by Step
$y''-4y'+y=0$
Use auxiliary equation,
$r^{2}-4r+1=0$
$r=\frac{-(-4)±\sqrt ((-4)^{2}-4(1)(1))}{2(1)}$
$r=\frac{4±\sqrt (16-4)}{2}$
$r_{1}=\frac{4-2\sqrt 3}{2}$ = $2-\sqrt 3$
$r_{1}=\frac{4+2\sqrt 3}{2}$ = $2+\sqrt 3$
Formula 8
$y=c_{1}e^{r_{1}x}+c_{2}e^{r_{2}x}$
$y=c_{1}e^{(2-\sqrt 3)x}+c_{2}e^{(2+\sqrt 3)x}$
