User:Egm4313.s12.team19.Horwitz.R2 Code test page

A Code test page 2
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R2.8
Contributed by Vladimir Horwitz.

Problem Statement:
Answer questions 8 and 15 from Kreyszing.

Solution:
Prob. 15)


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! $$ \ y'' + .54 y' + (.0729 + \pi) y = 0 $$ !! (Eq. 1)
 * $$ \ \lambda^2 + .54 \lambda + .0729 + \pi = 0 $$
 * (Eq. 2)
 * The characteristic equation has these roots.
 * $$ \lambda = -.27 + \sqrt {\pi} i, \bar \lambda = -.27 - \sqrt {\pi} i $$
 * (Eq. 3, a and b)
 * }
 * $$ \lambda = -.27 + \sqrt {\pi} i, \bar \lambda = -.27 - \sqrt {\pi} i $$
 * (Eq. 3, a and b)
 * }

Due to the fact that this roots are complex, the following equations from Kreyszig 2011, page 57 case 3, can be applied.


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! $$ \ y = e^{-.27} * (A \cos (\sqrt \pi x) + B \sin (\sqrt \pi x)) $$ (Eq. 4)
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! And following this, we can say !!
 * $$ y' = -.27 e^{-.27} \ (A \cos (\sqrt \pi x) + B \sin (\sqrt \pi x)) \ + \ e^{-.27} \ (-A \sqrt \pi \sin (\sqrt \pi x) + B \sqrt \pi \cos (\sqrt \pi x)) $$
 *  (Eq. 5)
 * and
 * $$ y'' = .0729 e^{-.27} \ (A \cos \sqrt \pi x \ + B \sin \sqrt \pi x) \ + \ e^{-.27} \ (- \pi A \cos \sqrt \pi x \ - \pi B \sin \sqrt \pi x) + \downarrow $$
 * $$ \uparrow -.27 e^{-.27} \ (-A \sqrt \pi \sin \sqrt \pi x \ + B \sqrt \pi \cos \sqrt \pi x) \ - \ .27 e^{-.27} \ (-A \sqrt \pi \sin \sqrt \pi x \ + B \sqrt \pi \cos \sqrt \pi x) $$
 * (Eq. 6)
 * Then, after major calculator work, we find that
 * $$ 0 = (0) * e^{-.27} * (A \cos \sqrt \pi x \ + B \sin \sqrt \pi x) $$
 * (Eq. 7)
 * }
 * Then, after major calculator work, we find that
 * $$ 0 = (0) * e^{-.27} * (A \cos \sqrt \pi x \ + B \sin \sqrt \pi x) $$
 * (Eq. 7)
 * }
 * (Eq. 7)
 * }