## Calculus 8th Edition

Since we cannot determine the value of $c_{2}$, we can say that the given problem has no solution.
$y''+4y'+20y=0$ Use auxiliary equation $r^{2}+4r+20=0$ $r=\frac{-4±\sqrt ((4^{2}-4(1)(20))}{2(1)}$ $r=\frac{-4±\sqrt (-64)}{2}$ $r_{1}=-2+4i$ $r_{2}=-2-4i$ $r_{1}=α+βi$ $r_{2}=α-βi$ $α=-2$ $β=4$ $y=e^{αx}(c_{1}cosβx+c_{2}sinβx$ $y=e^{-2x}(c_{1}cos4x+c_{2}sin4x$ $y(0)=1$ $1=e^{-2(0)}(c_{1}cos4(0)+c_{2}sin4(0)$ $1=c_{1}(1)+c_{2}(0)$ $c_{1}=1$ $y(\pi)=2$ $2=e^{-2(\pi)}(c_{1}cos4(\pi)+c_{2}sin4(\pi)$ $2=e^{-2\pi}[c_{1}(1)+c_{2}(0)]$ $c_{1}=\frac{2}{e^{-2\pi}}$ Since we cannot determine the value of $c_{2}$, we can say that the given problem has no solution.