Answer
Since we cannot determine the value of $c_{2}$, we can say that the given problem has no solution.
Work Step by Step
$y''+4y'+20y=0$
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.
