Answer
$y=\frac{1}{7}e^{4x-4}-\frac{1}{7}e^{3-3x}$
Work Step by Step
$y''-y'-12y=0$
Use auxiliary equation
$r^{2}-r-12=0$
$r=\frac{1±\sqrt (1^{2}-4(1)(-1))}{2(1)}$
$r=\frac{1±\sqrt 49}{2}$
$r_{1}=4$
$r_{2}=-3$
$y=c_{1}e^{r_{1}x}+c_{2}e^{r_{2}x}$
$y=c_{1}e^{4x}+c_{2}e^{-3x}$
$y(1)=0$
$y(1)=0=c_{1}e^{4x}+c_{2}e^{-3x}$
$y'(1)=1$
$y'(1)=1=4c_{1}e^{4x}-3c_{2}e^{-3x}$
$c_{1}e^{4x}+c_{2}e^{-3x}=0$
$4c_{1}e^{4x}-3c_{2}e^{-3x}=1$
Solve the system of equations:
$c_{2}=-c_{1}e^{7}$
$4c_{1}e^{4}+3c_{1}e^{4}=1$
$c_{1}=\frac{1}{7}e^{-4}$
$c_{2}=-\frac{1}{7}e^{3}$
Hence, $y=\frac{1}{7}e^{4x-4}-\frac{1}{7}e^{3-3x}$
