Complex Eigenvalues: Example 1

Example: Solve the initial value problem

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Answer: First, note that the matrix coefficient is

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Next, we need to find the eigenvalues which are given as roots of the characteristic equation

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This is a quadratic equation. Its roots are given by the quadratic formulas

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Set tex2html_wrap_inline435, and find an associated eigenvector V. Set

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The vector V must satisfy the equation

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This is equivalent to the system

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From the quadratic equation we get (check it)

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then we have

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This clearly implies that the two equations of the system are the same. Therefore, we use only the first to get

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Hence, we have

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Choose

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The independent solutions which will generate the general solution to the system are the real and the imaginary parts of the complex solution

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Since

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and

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where tex2html_wrap_inline463 , we have

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where

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and

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Therefore, the general solution is

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The initial condition tex2html_wrap_inline473 implies

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which gives

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Therefore, the desired particular solution is

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where

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[Differential Equations] [First Order D.E.]
[Geometry] [Algebra] [Trigonometry ]
[Calculus] [Complex Variables] [Matrix Algebra]

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Author: Mohamed Amine Khamsi

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