Answer
Any function in the form
$y=e^{-2x}(cos4x+csin4x)$
satisfies the boundary condition, where $c∈ℝ$.
Work Step by Step
$y''+4y'+20y=0$
Use auxiliary equation
$r^{2}+4r+20=0$
$r^{2}+4r+4=-16$
$(r+2)^{2}=-16$
$r+2=±4i$
$r_{1}=-2+4i$
$r_{2}=-2-4i$
$α=-2$
$β=4$
Formula 11
$y=e^{αx}(c_{1}cosβx+c_{2}sinβx)$
$y=e^{-2x}(c_{1}cos4x+c_{2}sin4x)$
$y(0)=1$
$1=e^{-2(0)}(c_{1}cos4(0)+c_{2}sin4(0))$
$c_{1}=1$
$y(\pi)=e^{-2\pi}$
$e^{-2\pi}=e^{-2\pi}(c_{1}cos(4\pi)+c_{2}sin(4\pi))$
$c_{1}=1$
This implies that $c_{1}=1$ and that $c_{2}$ can take any value.
Let's call $c_{2}=c$
Therefore, any function in the form
$y=e^{-2x}(cos4x+csin4x)$
satisfies the boundary condition, where $c∈ℝ$.
