User:Egm4313.s12.team3.katz/Brian Katz Report 1 Questions

=Problem 5: Solving Ordinary Differential Equations by find the general solution=

Problem Statement
Given ODE's of this kind find the general solution and check the results using substitution. Similar ODE's and classification are shown in [[media:Iea.s12.sec2d.djvu|Sec 2 p. 2-5]].
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$$  \displaystyle y''+4y'+(\pi^2 +4)y=0 $$ $$ \displaystyle y''+2\pi y'+\pi^2 y=0 $$
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Part A

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$$  \displaystyle y''+4y'+(\pi^2 +4)y=0 $$ (5.1) Using the general form
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$$  \displaystyle \lambda^2+a\lambda+b=0 $$
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Equation (5.1) becomes
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$$  \displaystyle \lambda^2+4\lambda+(\pi^2+4)=0 $$ $$   \displaystyle \lambda_1=\frac{-4+(\sqrt{16-4(\pi^2+4)})}2 \; \; \; \; \; \; \; \; \; \; \; \lambda_2=\frac{-4-(\sqrt{16-4(\pi^2+4)})}2 $$ $$   \displaystyle \lambda_1=\frac{-4+(\sqrt{-4\pi^2)}}2 \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \lambda_2=\frac{-4-(\sqrt{-4\pi^2)}}2 $$ $$   \displaystyle \lambda_1=\frac{-4+2\pi i}{2} \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \lambda_2=\frac{-4-2\pi i}{2} $$ $$   \displaystyle \lambda_1=-2+\pi i \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \lambda_2=-2-\pi i $$
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$$  \displaystyle y_1=e^{-2x} \cos{\pi x} \;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\; y_2=e^{-2x} \sin{\pi x} $$ Calculating the derivatives:
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$$  \displaystyle y=Ae^{-2x} \cos{\pi x}+Be^{-2x} \sin{\pi x} $$ $$ \displaystyle y'=[-2Ae^{-2x} \cos{\pi x}-A\pi e^{-2x} \sin{\pi x}]+[\pi Be^{-2x} \cos{\pi x}-2Be^{-2x} \sin{\pi x}] $$ $$ \displaystyle y''=[4Ae^{-2x} \cos{\pi x}\;+2\;\pi Ae^{-2x} \sin{\pi x}\;+\;2\pi Ae^{-2x} \sin\pi x\;-\;\pi^2 Ae^{-2x} \cos{\pi x}]\;+\;[-2B\pi e^{-2x} \cos{\pi x}\;-\;\pi^2 Be^{-2x} \sin{\pi x}\;+\;4Be^{-2x} \sin{\pi x}\;-\;2\pi Be^{-2x} \cos{\pi x}] $$ Simplifying some of the terms:
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$$  \displaystyle 4y'=[-8Ae^{-2x} \cos{\pi x}-4A\pi e^{-2x} \sin{\pi x}]+[4\pi Be^{-2x} \cos{\pi x}-8Be^{-2x} \sin{\pi x}] $$ $$ \displaystyle (\pi^2 +4)y=(\pi^2 +4)Ae^{-2x} \cos{\pi x}+(\pi^2 +4)Be^{-2x} \sin{\pi x} $$
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Finally subbing the terms into the original equation:
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$$  \displaystyle y''\;+\;4y'\;+\;(\pi^2\;+\;4)y=[e^{-2x} \cos{\pi x}]\cdot[4A\;-\;\pi^2 A\;-\;2\pi B\;-\;2\pi B\;-\;8A\;+\;4\pi B\;+\;\pi^2 A\;+\;4A]\;+\;[e^{-2x} \sin{\pi x}]\cdot[2\pi A\;+\;2\pi A\;-\;\pi^2 B\;+\;4B\;-\;4\pi A\;-\;8B\;+\;\pi^2 B\;+\;4B] $$ $$ \displaystyle [e^{-2x} \cos{\pi x}]\cdot[0]+[e^{-2x} \sin{\pi x}]\cdot[0]=0 $$
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Part B

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$$  \displaystyle y''+2\pi y'+\pi^2 y=0 $$ (5.2) Using the general form
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$$  \displaystyle \lambda^2+a\lambda+b=0 $$ Equation (5.2) becomes
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$$  \displaystyle \lambda^2+2\pi \lambda+\pi^2=0 $$ $$   \displaystyle \lambda_{1,2}=\frac{-2\pi \pm \sqrt{4\pi^2 - 4\pi^2}}{2} $$ $$   \displaystyle \lambda_{1,2}=-\pi $$
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$$  \displaystyle y_{1,2}=(c_1+c_2x)e^{-\pi x} $$ Calculating the derivatives:
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$$  \displaystyle y=(c_1+c_2x)e^{-\pi x} $$ $$ \displaystyle y'=c_2e^{-\pi x}-\pi (c_1+c_2x)e^{-\pi x} $$ $$ \displaystyle y''=-\pi c_2e^{-\pi x}\;+\;[(-\pi (c_1\;+\;c_2x)\;\cdot\; -\pi e^{-\pi x})(-\pi c_2e^{-\pi x})] =-2\pi c_2e^{-\pi x}\;+\;\pi^2 (c_1\;+\;c_2x)e^{-\pi x} $$ Finally subbing the terms into the original equation:
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$$  \displaystyle y''+2\pi y'+\pi^2 y=\pi^2 (c_1+c_2x)e^{-\pi x}-2\pi c_2e^{-\pi x}+2\pi c_2e^{-\pi x}-2\pi^2 (c_1+c_2x)e^{-\pi x}+\pi^2(c_1+c_2x)e^{-\pi x} = 0 $$
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