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The Method of Elimination: x" = 6x + 2y, y" = 3x + 7y
- Thread starter bearej50
- Start date
Re: The Method of Elimination
What have you tried?
I'd suggest trying trial solutions x=exp(lambda t), y = exp(mu t), then get some equations for lambda and mu.
bearej50 said:FIND THE GENERAL SOLUTION OF THE LINEAR SYSTEM
x'' = 6x + 2y, y'' = 3x + 7y
What have you tried?
I'd suggest trying trial solutions x=exp(lambda t), y = exp(mu t), then get some equations for lambda and mu.
Re: The Method of Elimination
Alternatively, noting the title of your post...
The method of elimination involves rearranging the equations to eliminate one of the variables.
For example, the first equation can be used to find y in terms of x and x''. Then you can substitute this into the second equation to get an equation just involving x, x'' and x''''.
bearej50 said:FIND THE GENERAL SOLUTION OF THE LINEAR SYSTEM
x'' = 6x + 2y, y'' = 3x + 7y
Alternatively, noting the title of your post...
The method of elimination involves rearranging the equations to eliminate one of the variables.
For example, the first equation can be used to find y in terms of x and x''. Then you can substitute this into the second equation to get an equation just involving x, x'' and x''''.
One way to solve a system like this is to, first, find the characteristic polynomial.
\(\displaystyle det(A-{\lambda}I)=\begin{vmatrix}6-{\lambda}&2\\3&7-{\lambda}\end{vmatrix}\)
\(\displaystyle {\lambda}^{2}-13{\lambda}+36=0, \;\ {\lambda}=4, \;\ {\lambda}=9\)
4 and 9 are the eigenvalues.
Now, use these in the original system to find the related eigenvectors.
Have you seen this method before?.
\(\displaystyle det(A-{\lambda}I)=\begin{vmatrix}6-{\lambda}&2\\3&7-{\lambda}\end{vmatrix}\)
\(\displaystyle {\lambda}^{2}-13{\lambda}+36=0, \;\ {\lambda}=4, \;\ {\lambda}=9\)
4 and 9 are the eigenvalues.
Now, use these in the original system to find the related eigenvectors.
Have you seen this method before?.