## Multivariable Calculus, 7th Edition

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