Answer
There is no solution to this boundary value problem.
Work Step by Step
$y''-8y'+17y=0$
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.
