User:Egm6321.f10.team3.cook/HW4

=Problem 3: Find $$ \chi $$ using trial solution=

Given
Method of trial solution, see King p 513, A5.3. Where $$ r$$ are the roots given by page 25-2 of the lecture notes
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$$ {r}_{1,2} = \pm  \frac{1}{k} $$ $$
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 * $$\displaystyle (1)
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$$ {r}_{3,4} = \pm  \frac{i}{k} $$ $$
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 * $$\displaystyle (2)
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where $$ i $$ is the imaginary number and $$ k =  \sqrt[4]{ \frac{EI}{m { \omega}^{2}}} $$ given in pg 25-1 of the lecture notes

Find
Find
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\chi (x) = {e}^{rx} $$ $$
 * $$\displaystyle (3)
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Solution
First we solve $$\displaystyle (3) $$ for the root in $$\displaystyle (2) $$ using the following relationship


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{e}^{i \theta} = \cos( \theta) + i\sin( \theta) $$ $$
 * $$\displaystyle
 * $$\displaystyle
 * $$\displaystyle (4)
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The solution to $$\displaystyle (3) $$ becomes


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{\chi (x)}_{1,2}= {e}^{\frac{x}{k}}+{e}^{-\frac{x}{k}} = \cos( \frac{x}{k}) + i\sin( \frac{x}{k})+ \frac{1}{\cos( \frac{x}{k}) + i\sin( \frac{x}{k}} $$ $$
 * $$\displaystyle
 * $$\displaystyle
 * $$\displaystyle (5)
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Next we solve $$\displaystyle (3) $$ for the root in $$\displaystyle (1) $$ using the following relationship


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{e}^{\theta} +{e}^{-\theta} = 2\cosh \theta $$ $$
 * $$\displaystyle
 * $$\displaystyle
 * $$\displaystyle (6)
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The solution to $$\displaystyle (3) $$ becomes


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{\chi (x)}_{3,4}= {e}^{\frac{x}{k}}+{e}^{-\frac{x}{k}} = 2\cosh(\frac{x}{k}) $$ $$
 * $$\displaystyle
 * $$\displaystyle
 * $$\displaystyle (6)
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Using superposition, the solution to $$\displaystyle (3) $$ is


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\chi (x)= {\chi (x)}_{1,2} + {\chi (x)}_{3,4} = \cos( \frac{x}{k}) + i\sin( \frac{x}{k})+ \frac{1}{\cos( \frac{x}{k}) + i\sin( \frac{x}{k}} + 2\cosh(\frac{x}{k}) $$ $$
 * $$\displaystyle
 * $$\displaystyle
 * $$\displaystyle (6)
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