User:Egm4313.s12.team1.wyattling/r5

Statement
Show that $$ y_p(x) = \sum_{i=0}^{n}y_{p,i}(x)\!$$ is indeed the overall particular solution of the L2-ODE-VC $$ y''_{p,i} + p(x)y'_{p,i} + q(x)y_{p,i} = r_i(x)\!$$ with the excitation $$ r(x) = r_1(x) + r_2(x) + ... + r_n(x) = \sum_{i=0}^{n}r_i(x)\!$$. Discuss the choice of $$ y_p(x)\!$$, in example for $$ r(x) = Kcos(wx)\!$$. Why would you need to have both $$ cos(wx)\!$$ and $$ sin(wx)\!$$ in $$ y_p(x)\!$$?

Solution
The following represent particular solutions and their derivatives, equated into a summation:
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y_p(x) = y_{p,1}(x) + y_{p,2}(x) + ... + y_{p,n}(x) \rightarrow y_p(x) = \sum_{i=0}^{n}y_{p,i}(x) $$     (4.0)
 * $$\displaystyle
 * $$\displaystyle
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y_p'(x) = y_{p,1}'(x) + y_{p,2}'(x) + ... + y_{p,n}'(x) \rightarrow y_p'(x) = \sum_{i=0}^{n}y_{p,i}'(x) $$     (4.1)
 * $$\displaystyle
 * $$\displaystyle
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y_p(x) = y_{p,1}(x) + y_{p,2}''(x) + ... + y_{p,n}(x) \rightarrow y_p(x) = \sum_{i=0}^{n}y_{p,i}''(x) $$     (4.2)
 * $$\displaystyle
 * $$\displaystyle
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Because the given ODE is in the form of L2-ODE-VC, it is linearly independent. Each $$ y_{p,i}\! $$i is a solution for each $$ r_i\! $$, and when there are multiple excitations, the solution to a sum of these excitations is the sum of the particular solutions. Using these particular solutions, we can show that they are the solutions to the following L2-ODE-VC (4.3) with the given excitation:


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y_{p,i}'' + p(x)y_{p,i}' + q(x)y_{p,i} = r_i(x) $$     (4.3)
 * $$\displaystyle
 * $$\displaystyle
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\sum_{i=0}^{n}y_{p,i}''(x) + p(x)\sum_{i=0}^{n}y_{p,i}'(x) + q(x)\sum_{i=0}^{n}y_{p,i}(x) = \sum_{i=0}^{n}r_i(x) $$     (4.4)
 * $$\displaystyle
 * $$\displaystyle
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For $$ r(x) = Kcos(wx)\!$$. You need to have both $$ cos(wx)\!$$ and $$ sin(wx)\!$$ in $$ y_p(x)\!$$ because when solving for the particular solution, it is necessary to take derivatives of the particular solution, and in the case of $$ r(x) = Kcos(wx)\!$$ the derivatives will produce extra terms since the derivative of $$ cos(wx)\!$$ will produce both $$ sin(wx)\!$$ and $$ cos(wx)\! $$ terms. Having both $$ cos(wx)\!$$ and $$ sin(wx)\!$$ are there to eliminate the extra term.

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