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Exercise:
A system of linear differential s is given by dot x - x - y dot y - x - y abcliste abc Show that the system describes a stable node. abc Plot the direction field and add a few orbits. abcliste

Solution:
abcliste abc The system of linear differential s can be described by the matrix bf A leftmatrix- & - - & - matrixright The trace and determinant of this matrix are tau -- - Delta ----- It follows for the eigenvalues lambda_ fractaupmsqrtdelta^-Delta frac-pmsqrt-^- frac-pmsqrt which leads to lambda_- and lambda_-. Since both eigenvalues are real negative numbers the fixed po at the origin is a stable node. abc The diagram below shows the vector field and some orbits phase portrait. In red are the straight line solutions corresponding to the eigenvectors of the system. center includegraphicswidthtextwidth#image_path:stablnode# center abcliste
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Exercise:
A system of linear differential s is given by dot x - x - y dot y - x - y abcliste abc Show that the system describes a stable node. abc Plot the direction field and add a few orbits. abcliste

Solution:
abcliste abc The system of linear differential s can be described by the matrix bf A leftmatrix- & - - & - matrixright The trace and determinant of this matrix are tau -- - Delta ----- It follows for the eigenvalues lambda_ fractaupmsqrtdelta^-Delta frac-pmsqrt-^- frac-pmsqrt which leads to lambda_- and lambda_-. Since both eigenvalues are real negative numbers the fixed po at the origin is a stable node. abc The diagram below shows the vector field and some orbits phase portrait. In red are the straight line solutions corresponding to the eigenvectors of the system. center includegraphicswidthtextwidth#image_path:stablnode# center abcliste
Contained in these collections:

Attributes & Decorations
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Differential equations
Tags
phase portrait, vector field
Content image
Difficulty
(2, default)
Points
0 (default)
Language
ENG (English)
Type
Calculative / Quantity
Creator by
Decoration