Problem R2.8
Find a general solution. Check your answer by substitution.
![{\displaystyle y''+y'+3.25y=0\!}](https://wikimedia.org/api/rest_v1/media/math/render/svg/b8c9d923da1aaf51f512e5a748be3ab1334302da)
Let:
![{\displaystyle \lambda =d/dx\!}](https://wikimedia.org/api/rest_v1/media/math/render/svg/d593269a908d556490dcbb9f7325b021671b7b3a)
![{\displaystyle \lambda ^{2}+\lambda +3.25=0\!}](https://wikimedia.org/api/rest_v1/media/math/render/svg/4f94421263cb6a7288bb55470d6ae297350bfe36)
Using the quadratic equation to find roots we get:
![{\displaystyle \lambda _{1}={\frac {-1+i{\sqrt {(}}12)}{2}}\!}](https://wikimedia.org/api/rest_v1/media/math/render/svg/1186ab707155414819afd55c385a97db271dee61)
![{\displaystyle \lambda _{2}={\frac {-1-i{\sqrt {(}}12)}{2}}\!}](https://wikimedia.org/api/rest_v1/media/math/render/svg/2c24752201325784decdc72f88175788283b2194)
Therefore:
![{\displaystyle y_{h}(x)=e^{-{\frac {1}{2}}x}(c_{1}\cos(x{\sqrt {3}})+c_{2}\sin(x{\sqrt {3}})\!}](https://wikimedia.org/api/rest_v1/media/math/render/svg/1e6b884d91bb36baed3ad789cf7ede3fc2b5fe04)
![{\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})\!}](https://wikimedia.org/api/rest_v1/media/math/render/svg/01469d07393d98b4bd2136537fc5428de013fbf8)
![{\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})-\!}](https://wikimedia.org/api/rest_v1/media/math/render/svg/8c904f820750563254e68a35f804552e37f3da27)
![{\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}})\!}](https://wikimedia.org/api/rest_v1/media/math/render/svg/8ad47734f11e7edd8807720131b00dd2c841e477)
Substituting
into the original equation, the result is
![{\displaystyle y''+y'+3.25y=0\!}](https://wikimedia.org/api/rest_v1/media/math/render/svg/b8c9d923da1aaf51f512e5a748be3ab1334302da)
![{\displaystyle y''+0.54y'+(0.0729+\pi )y=0\!}](https://wikimedia.org/api/rest_v1/media/math/render/svg/6065a541cd08d27c8c5a852ba563ac280655c9fd)
Let: ![{\displaystyle {\frac {d}{dx}}=\lambda \!}](https://wikimedia.org/api/rest_v1/media/math/render/svg/beced72adb12a7e92607091e65c182bd79d1fbd4)
![{\displaystyle \lambda ^{2}+0.54\lambda +(0.0729+\pi )=0\!}](https://wikimedia.org/api/rest_v1/media/math/render/svg/b856ed309454c3914708a2a11317c652c94c8529)
Using the quadratic equation to find roots we get:
![{\displaystyle \lambda _{1}={\frac {-0.27+i{\sqrt {(}}\pi )}{2}}\!}](https://wikimedia.org/api/rest_v1/media/math/render/svg/bd63d9853cde8d9ea8306c497f42e0afe441a442)
![{\displaystyle \lambda _{2}={\frac {-0.27-i{\sqrt {(}}\pi )}{2}}\!}](https://wikimedia.org/api/rest_v1/media/math/render/svg/7bda3babe6472032c84cefa10b570ee77537a482)
Therefore:
![{\displaystyle y_{h}(x)=e^{-0.27x}(c_{1}\cos(x{\sqrt {\pi }})+c_{2}\sin(x{\sqrt {\pi }})\!}](https://wikimedia.org/api/rest_v1/media/math/render/svg/372f6d34926267e93d5f518369751be2ca5778ac)
![{\displaystyle y'(x)=-0.27e^{-0.27x}(c_{1}\cos(x{\sqrt {\pi }})+c_{2}\sin(x{\sqrt {\pi }})+e^{-0.27x}(-{\sqrt {\pi }}c_{1}\sin {{\sqrt {\pi }}x}+{\sqrt {\pi }}c_{1}\cos {{\sqrt {\pi }}x})\!}](https://wikimedia.org/api/rest_v1/media/math/render/svg/7b9e8fb9295bf3adfec3628fe5be1fed2950289f)
![{\displaystyle y''(x)=0.0729e^{-0.27x}(c_{1}\cos(x{\sqrt {\pi }})+c_{2}\sin(x{\sqrt {\pi }})-0.27e^{-0.27x}(-{\sqrt {\pi }}c_{1}\sin {{\sqrt {\pi }}x}+{\sqrt {\pi }}c_{1}\cos {{\sqrt {\pi }}x})-\!}](https://wikimedia.org/api/rest_v1/media/math/render/svg/33e8969c5b82b591ea65706ee3a3aac71d5fee82)
![{\displaystyle 0.27e^{-0.27x}({\sqrt {\pi }}c_{1}\sin {{\sqrt {\pi }}x}+{\sqrt {\pi }}c_{1}\cos {{\sqrt {\pi }}x})+e^{-0.27x}(-\pi (c_{1}\cos(x{\sqrt {\pi }}))-\pi (c_{2}\sin(x{\sqrt {\pi }})))\!}](https://wikimedia.org/api/rest_v1/media/math/render/svg/9a0a7e73eb726080318b04dc2079d2eba917beac)
Substituting
into the original equation, the result is
![{\displaystyle y''+0.54y'+(0.0729+\pi )y=0\!}](https://wikimedia.org/api/rest_v1/media/math/render/svg/6065a541cd08d27c8c5a852ba563ac280655c9fd)
Egm4313.s12.team11.gooding 03:41, 7 February 2012 (UTC)