University of Florida/Egm4313/s12.team11.gooding/R2/2.8

${\displaystyle \lambda ^{2}+\lambda +3.25=0\!}$
Using the quadratic equation to find roots we get:
${\displaystyle \lambda _{1}={\frac {-1+i{\sqrt {(}}12)}{2}}\!}$
${\displaystyle \lambda _{2}={\frac {-1-i{\sqrt {(}}12)}{2}}\!}$
Therefore:

${\displaystyle y_{h}(x)=e^{-{\frac {1}{2}}x}(c_{1}\cos(x{\sqrt {3}})+c_{2}\sin(x{\sqrt {3}})\!}$

${\displaystyle y'(x)=-{\frac {1}{2}}e^{-{\frac {1}{2}}x}(c_{1}\cos(x{\sqrt {3}})+c_{2}\sin(x{\sqrt {3}})+e^{-{\frac {1}{2}}x}(-{\sqrt {3}}c_{1}\sin {{\sqrt {3}}x}+{\sqrt {3}}c_{1}\cos {{\sqrt {3}}x})\!}$
${\displaystyle y''(x)={\frac {1}{4}}e^{-{\frac {1}{2}}x}(c_{1}\cos(x{\sqrt {3}})+c_{2}\sin(x{\sqrt {3}})-{\frac {1}{2}}e^{-{\frac {1}{2}}x}(-{\sqrt {3}}c_{1}\sin {{\sqrt {3}}x}+{\sqrt {3}}c_{1}\cos {{\sqrt {3}}x})-\!}$
${\displaystyle {\frac {1}{2}}e^{-{\frac {1}{2}}x}(-{\sqrt {3}}c_{1}\sin {{\sqrt {3}}x}+{\sqrt {3}}c_{1}\cos {{\sqrt {3}}x})e^{-{\frac {1}{2}}x}(-3c_{1}\cos(x{\sqrt {3}})-3c_{2}\sin(x{\sqrt {3}})\!}$
Substituting ${\displaystyle y,y',y''\!}$ into the original equation, the result is

 ${\displaystyle y''+y'+3.25y=0\!}$

${\displaystyle \lambda ^{2}+0.54\lambda +(0.0729+\pi )=0\!}$
Using the quadratic equation to find roots we get:
${\displaystyle \lambda _{1}={\frac {-0.27+i{\sqrt {(}}\pi )}{2}}\!}$
${\displaystyle \lambda _{2}={\frac {-0.27-i{\sqrt {(}}\pi )}{2}}\!}$
Therefore:

${\displaystyle y_{h}(x)=e^{-0.27x}(c_{1}\cos(x{\sqrt {\pi }})+c_{2}\sin(x{\sqrt {\pi }})\!}$

 ${\displaystyle y''+0.54y'+(0.0729+\pi )y=0\!}$