Repeated Eigenvalues: Example1

Example. Consider the system

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1.
Find the general solution.
2.
Find the solution which satisfies the initial condition

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3.
Draw some solutions in the phase-plane including the solution found in 2.

Answer. The matrix coefficient of the system is

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In order to find the eigenvalues consider the characteristic polynomial

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Since tex2html_wrap_inline95 , we have a repeated eigenvalue equal to 3. Let us find the associated eigenvector tex2html_wrap_inline97 . Set

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Then we must have tex2html_wrap_inline101 which translates into

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This reduces to y=x. Hence we may take

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Next we look for the second vector tex2html_wrap_inline109 . The equation giving this vector is tex2html_wrap_inline111 which translates into the algebraic system

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where

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Clearly the two equations reduce to the equation y - x=1 or y = 1 + x, where x may be chosen to be any number. So if we take x=0 for example, we get

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Therefore the two independent solutions are

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The general solution will then be

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In order to find the solution which satisfies the initial condition

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we must have

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This implies tex2html_wrap_inline135 and tex2html_wrap_inline137 . Hence the solution is

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The phase plane with some solutions is given in the picture below:

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Author: Mohamed Amine Khamsi

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