Easier Ways to Find General Solutions of Higher Dimensional ODE's
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We have often used in the 2x2 case the Jordan form of the matrix A (2x2) in the system X' = AX, which yields a nice solution with standard basis vectors, which we transpose by the transformation matrix T to get back the original ODE's general solution (assuming invertible T).
Now with 3x3 and above, calculating the inverse matrix becomes prohibitively tedious, and I wanted to see if there is a faster way when there are repeated eigenvalues.
Example: A = ((0,0,1),(0,1,0),(1,0,0)), then the eigenvalues are -1, 1, 1, and clearly because x_2' = x_2 (the y component), x_2(t) = c_2 * e^t. I'm not sure what to do next; I know often we "guess" a solution for the components and check, but I'm stuck here...
ordinary-differential-equations
asked Nov 25 '18 at 5:16
MathGuyForLifeMathGuyForLife
1007
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add a comment |
$begingroup$
We have often used in the 2x2 case the Jordan form of the matrix A (2x2) in the system X' = AX, which yields a nice solution with standard basis vectors, which we transpose by the transformation matrix T to get back the original ODE's general solution (assuming invertible T).
Now with 3x3 and above, calculating the inverse matrix becomes prohibitively tedious, and I wanted to see if there is a faster way when there are repeated eigenvalues.
Example: A = ((0,0,1),(0,1,0),(1,0,0)), then the eigenvalues are -1, 1, 1, and clearly because x_2' = x_2 (the y component), x_2(t) = c_2 * e^t. I'm not sure what to do next; I know often we "guess" a solution for the components and check, but I'm stuck here...
ordinary-differential-equations
asked Nov 25 '18 at 5:16
MathGuyForLifeMathGuyForLife
1007
$endgroup$
$begingroup$
so then, is the solution just the same format as in the case of 3 distinct eigenvalues? i.e. ae^-t(-1,0,1)+be^t(1,0,1)+ce^t(0,1,0) ?
$endgroup$
– MathGuyForLife
Nov 25 '18 at 5:22
add a comment |
$begingroup$
We have often used in the 2x2 case the Jordan form of the matrix A (2x2) in the system X' = AX, which yields a nice solution with standard basis vectors, which we transpose by the transformation matrix T to get back the original ODE's general solution (assuming invertible T).
Now with 3x3 and above, calculating the inverse matrix becomes prohibitively tedious, and I wanted to see if there is a faster way when there are repeated eigenvalues.
Example: A = ((0,0,1),(0,1,0),(1,0,0)), then the eigenvalues are -1, 1, 1, and clearly because x_2' = x_2 (the y component), x_2(t) = c_2 * e^t. I'm not sure what to do next; I know often we "guess" a solution for the components and check, but I'm stuck here...
ordinary-differential-equations
asked Nov 25 '18 at 5:16
MathGuyForLifeMathGuyForLife
1007
$endgroup$
We have often used in the 2x2 case the Jordan form of the matrix A (2x2) in the system X' = AX, which yields a nice solution with standard basis vectors, which we transpose by the transformation matrix T to get back the original ODE's general solution (assuming invertible T).
Now with 3x3 and above, calculating the inverse matrix becomes prohibitively tedious, and I wanted to see if there is a faster way when there are repeated eigenvalues.
Example: A = ((0,0,1),(0,1,0),(1,0,0)), then the eigenvalues are -1, 1, 1, and clearly because x_2' = x_2 (the y component), x_2(t) = c_2 * e^t. I'm not sure what to do next; I know often we "guess" a solution for the components and check, but I'm stuck here...
ordinary-differential-equations
ordinary-differential-equations
asked Nov 25 '18 at 5:16
MathGuyForLifeMathGuyForLife
1007
asked Nov 25 '18 at 5:16
MathGuyForLifeMathGuyForLife
1007
asked Nov 25 '18 at 5:16
MathGuyForLifeMathGuyForLife
1007
asked Nov 25 '18 at 5:16
MathGuyForLifeMathGuyForLife
1007
asked Nov 25 '18 at 5:16
MathGuyForLifeMathGuyForLife
1007
1007
$begingroup$
so then, is the solution just the same format as in the case of 3 distinct eigenvalues? i.e. ae^-t(-1,0,1)+be^t(1,0,1)+ce^t(0,1,0) ?
$endgroup$
– MathGuyForLife
Nov 25 '18 at 5:22
add a comment |
$begingroup$
so then, is the solution just the same format as in the case of 3 distinct eigenvalues? i.e. ae^-t(-1,0,1)+be^t(1,0,1)+ce^t(0,1,0) ?
$endgroup$
– MathGuyForLife
Nov 25 '18 at 5:22
$begingroup$
so then, is the solution just the same format as in the case of 3 distinct eigenvalues? i.e. ae^-t(-1,0,1)+be^t(1,0,1)+ce^t(0,1,0) ?
$endgroup$
– MathGuyForLife
Nov 25 '18 at 5:22
$begingroup$
so then, is the solution just the same format as in the case of 3 distinct eigenvalues? i.e. ae^-t(-1,0,1)+be^t(1,0,1)+ce^t(0,1,0) ?
$endgroup$
– MathGuyForLife
Nov 25 '18 at 5:22
add a comment |
1 Answer
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I'll just post what Moo explained above. The solution is similar format as in the case of 3 distinct eigenvalues, because the eigenvectors are linearly independent.
$$ae^{-t}(-1,0,1)+be^t(1,0,1)+ce^t(0,1,0)$$
answered Nov 25 '18 at 5:29
MathGuyForLifeMathGuyForLife
1007
$endgroup$
1
$begingroup$
Thank you, edited.
$endgroup$
– MathGuyForLife
Nov 25 '18 at 5:41
add a comment |
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1 Answer
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1 Answer
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$begingroup$
I'll just post what Moo explained above. The solution is similar format as in the case of 3 distinct eigenvalues, because the eigenvectors are linearly independent.
$$ae^{-t}(-1,0,1)+be^t(1,0,1)+ce^t(0,1,0)$$
answered Nov 25 '18 at 5:29
MathGuyForLifeMathGuyForLife
1007
$endgroup$
1
$begingroup$
Thank you, edited.
$endgroup$
– MathGuyForLife
Nov 25 '18 at 5:41
add a comment |
$begingroup$
I'll just post what Moo explained above. The solution is similar format as in the case of 3 distinct eigenvalues, because the eigenvectors are linearly independent.
$$ae^{-t}(-1,0,1)+be^t(1,0,1)+ce^t(0,1,0)$$
answered Nov 25 '18 at 5:29
MathGuyForLifeMathGuyForLife
1007
$endgroup$
1
$begingroup$
Thank you, edited.
$endgroup$
– MathGuyForLife
Nov 25 '18 at 5:41
add a comment |
$begingroup$
I'll just post what Moo explained above. The solution is similar format as in the case of 3 distinct eigenvalues, because the eigenvectors are linearly independent.
$$ae^{-t}(-1,0,1)+be^t(1,0,1)+ce^t(0,1,0)$$
answered Nov 25 '18 at 5:29
MathGuyForLifeMathGuyForLife
1007
$endgroup$
I'll just post what Moo explained above. The solution is similar format as in the case of 3 distinct eigenvalues, because the eigenvectors are linearly independent.
$$ae^{-t}(-1,0,1)+be^t(1,0,1)+ce^t(0,1,0)$$
answered Nov 25 '18 at 5:29
MathGuyForLifeMathGuyForLife
1007
edited Nov 25 '18 at 5:41
answered Nov 25 '18 at 5:29
MathGuyForLifeMathGuyForLife
1007
answered Nov 25 '18 at 5:29
MathGuyForLifeMathGuyForLife
1007
answered Nov 25 '18 at 5:29
MathGuyForLifeMathGuyForLife
1007
1007
1
$begingroup$
Thank you, edited.
$endgroup$
– MathGuyForLife
Nov 25 '18 at 5:41
add a comment |
1
$begingroup$
Thank you, edited.
$endgroup$
– MathGuyForLife
Nov 25 '18 at 5:41
1
1
$begingroup$
Thank you, edited.
$endgroup$
– MathGuyForLife
Nov 25 '18 at 5:41
$begingroup$
Thank you, edited.
$endgroup$
– MathGuyForLife
Nov 25 '18 at 5:41
add a comment |
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so then, is the solution just the same format as in the case of 3 distinct eigenvalues? i.e. ae^-t(-1,0,1)+be^t(1,0,1)+ce^t(0,1,0) ?
$endgroup$
– MathGuyForLife
Nov 25 '18 at 5:22