Answer
$y=e^{-\frac{1}{2}x}(\sqrt 2sin\frac{\sqrt 2}{2}x)$
Work Step by Step
$4y''+4y'+3y=0$
Use auxiliary equation
$4r^{2}+4r+3=0$
$r=\frac{-4±\sqrt (4^{2}-4(4)(3))}{2(4)}$
$r=\frac{-4±\sqrt (-32)}{8}$
$r=-\frac{1}{2}±\frac{\sqrt 2}{2}i$ $r_{1}=α+βi$
$r_{2}=α-βi$
$α=-\frac{1}{2}$
$β=\frac{\sqrt 2}{2}$
$y=e^{αx}(c_{1}cosβx+c_{2}sinβx)$
$y=e^{-\frac{1}{2}x}(c_{1}cos\frac{\sqrt 2}{2}x+c_{2}sin\frac{\sqrt 2}{2}x)$
$y(0)=0$
$0=e^{-\frac{1}{2}(0)}(c_{1}cos\frac{\sqrt 2}{2}(0)+c_{2}sin\frac{\sqrt 2}{2}(0))$
$0=c_{1}+0$
$c_{1}=0$
$y'=-\frac{1}{2}e^{-\frac{1}{2}x}(c_{1}cos\frac{\sqrt 2}{2}x+c_{2}sin\frac{\sqrt 2}{2}x)+e^{-\frac{1}{2}x}(-\frac{c_{1}\sqrt 2}{2}sin\frac{\sqrt 2}{2}x+\frac{c_{2}\sqrt 2}{2}cos\frac{\sqrt 2}{2}x)$
$y'=\frac{e^{-\frac{1}{2}x}}{2}[-(c_{1}cos\frac{\sqrt 2}{2}x+c_{2}sin\frac{\sqrt 2}{2}x)+(-c_{1}\sqrt 2sin\frac{\sqrt 2}{2}x+c_{2}\sqrt 2cos\frac{\sqrt 2}{2}x)]$
$y'=\frac{e^{-\frac{1}{2}x}}{2}[-c_{1}cos\frac{\sqrt 2}{2}x-c_{2}sin\frac{\sqrt 2}{2}x)-c_{1}\sqrt 2sin\frac{\sqrt 2}{2}x+c_{2}\sqrt 2cos\frac{\sqrt 2}{2}x]$
$y'(0)=1$ $1=\frac{e^{-\frac{1}{2}(0)}}{2}[-c_{1}cos\frac{\sqrt 2}{2}(0)-c_{2}sin\frac{\sqrt 2}{2}(0))-c_{1}\sqrt 2sin\frac{\sqrt 2}{2}(0)+c_{2}\sqrt 2cos\frac{\sqrt 2}{2}(0)]$
$1=\frac{1}{2}[-c_{1}(1)-c_{2}(0)-c_{1}\sqrt 2(0)+c_{2}\sqrt 2(1)]$
$-c_{1}+c_{2}\sqrt 2=2$
$-0+c_{2}\sqrt 2=2$
$c_{2}=\frac{2}{\sqrt 2}$
$c_{2}=\sqrt 2$
$y=e^{-\frac{1}{2}x}(c_{1}cos\frac{\sqrt 2}{2}x+c_{2}sin\frac{\sqrt 2}{2}x)$
$y=e^{-\frac{1}{2}x}(\sqrt 2sin\frac{\sqrt 2}{2}x)$
