## Calculus 8th Edition

$y''-8y'+17y=0$ Use auxiliary equation $r^{2}-8r+17=0$ $r^{2}-8r+16=-1$ $(r-4)^{2}=-1$ $r-4=±i$ $r_{1}=4-i$ $r_{2}=4+i$ $α=4$ $β=1$ Formula 11 $y=e^{αx}(c_{1}cosβx+c_{2}sinβx)$ $y=e^{4x}(c_{1}cosx+c_{2}sinx)$ $y(0)=3$ $y=e^{4(0)}(c_{1}cos(0)+c_{2}sin(0))$ $c_{1}=3$ →(1) $y(\pi)=2$ $2=e^{4(\pi)}(c_{1}cos(\pi)+c_{2}sin(\pi))$ $c_{1}=-\frac{2}{e^{4\pi}}$ →(2) Equation 1 and Equation 2 both contradict each other. Therefore, there is no solution to this boundary value problem.