User:Egm4313.s12.team3.pliskow/R2work/Problem8PartAVerify

Verification of General Solution: Report 2, Problem 8, Part A

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$$\displaystyle y''+y'+3.25y=0 $$     (8.1)
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Let's verify that the solution to the above ODE is indeed what was determined:


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$$\displaystyle y=e^{-\frac{x}{2}}[Acos(\sqrt{3}x)+Bsin(\sqrt{3}x)] $$    (1)
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Before verifying that (1) satisfies the original ODE (8.1), we must calculate the first derivative:


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$$\displaystyle y'=e^{-\frac{x}{2}}[-\sqrt{3}Asin(\sqrt{3}x)+\sqrt{3}Bcos(\sqrt{3}x)]-\frac{1}{2}e^{-\frac{x}{2}}[Acos(\sqrt{3}x)+Bsin(\sqrt{3}x)] $$    (2)
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$$\displaystyle y'=e^{-\frac{x}{2}}cos(\sqrt{3}x)[\sqrt{3}B-\frac{1}{2}A]+e^{-\frac{x}{2}}sin(\sqrt{3}x)[-\sqrt{3}A-\frac{1}{2}B] $$    (3) And the second derivative:
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$$\displaystyle y''=e^{-\frac{x}{2}}[-3Acos(\sqrt{3}x)-3Bsin(\sqrt{3}x)]-\frac{1}{2}e^{-\frac{x}{2}}[-\sqrt{3}Asin(\sqrt{3}x)+\sqrt{3}Bcos(\sqrt{3}x)]-\frac{1}{2}e^{-\frac{x}{2}}[-\sqrt{3}Asin(\sqrt{3}x)+\sqrt{3}Bcos(\sqrt{3}x)]+\frac{1}{4}e^{-\frac{x}{2}}[Acos(\sqrt{3}x)+Bsin(\sqrt{3}x)] $$    (4)
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$$\displaystyle y''=e^{-\frac{x}{2}}cos(\sqrt{3}x)[-3A-\frac{\sqrt{3}}{2}B-\frac{\sqrt{3}}{2}B+\frac{1}{4}A]+e^{-\frac{x}{2}}sin(\sqrt{3}x)[-3B+\frac{\sqrt{3}}{2}A+\frac{\sqrt{3}}{2}A+\frac{1}{4}B] $$    (5) With the derivatives calculated, we can check to see if it satisfies the original ODE.
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$$\displaystyle y''+y'+3.25y=e^{-\frac{x}{2}}cos(\sqrt{3}x)[-3A-\frac{\sqrt{3}}{2}B-\frac{\sqrt{3}}{2}B+\frac{1}{4}A]+e^{-\frac{x}{2}}sin(\sqrt{3}x)[-3B+\frac{\sqrt{3}}{2}A+\frac{\sqrt{3}}{2}A+\frac{1}{4}B]+ e^{-\frac{x}{2}}cos(\sqrt{3}x)[\sqrt{3}B-\frac{1}{2}A]+e^{-\frac{x}{2}}sin(\sqrt{3}x)[-\sqrt{3}A-\frac{1}{2}B]+3.25e^{-\frac{x}{2}}[Acos(\sqrt{3}x)+Bsin(\sqrt{3}x)] $$    (6)
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$$\displaystyle y''+y'+3.25y=e^{-\frac{x}{2}}cos(\sqrt{3}x)[-3A-\sqrt{3}B+\frac{1}{4}A+\sqrt{3}B-\frac{1}{2}A+3.25A]+e^{-\frac{x}{2}}sin(\sqrt{3}x)[-3B+\sqrt{3}A+\frac{1}{4}B-\sqrt{3}A-\frac{1}{2}B+3.25B]
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$$    (7)
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$$\displaystyle y''+y'+3.25y=e^{-\frac{x}{2}}cos(\sqrt{3}x)[-3A+\frac{1}{4}A-\frac{1}{2}A+3.25A-\sqrt{3}B+\sqrt{3}B]+e^{-\frac{x}{2}}sin(\sqrt{3}x)[\sqrt{3}A-\sqrt{3}A+\frac{1}{4}B-\frac{1}{2}B+3.25B-3B] $$    (8)
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$$\displaystyle y''+y'+3.25y=e^{-\frac{x}{2}}cos(\sqrt{3}x)[0]+e^{-\frac{x}{2}}sin(\sqrt{3}x)[0] $$    (8)
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$$\displaystyle y''+y'+3.25y=0 $$    (8)
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Since (8) matches the form of the original ODE, we can conclude that the general solution does indeed satisfy the ODE.
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