How to solve a system of differential equations with complex numbers?
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So I have a general understanding of how to do these types of problems but I do not know how to get rid of the complex numbers. I found that the eigenvalues are 3i and -3i and the eigenvectors are [(-2/5 + 1/5 i), (1)] and [(-2/5 - 1/5 i), (1)].
The general equation would be v1*c1*cos(3t) + v2*c2*sin(3t) where I would then attempt to find the constants, however, when multiplying the vectors, more complex numbers are introduced and my answer must be real only.
How do I isolate the real part only?
linear-algebra ordinary-differential-equations eigenvalues-eigenvectors
asked Dec 13 '18 at 1:42
S. SnakeS. Snake
485
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add a comment |
$begingroup$
So I have a general understanding of how to do these types of problems but I do not know how to get rid of the complex numbers. I found that the eigenvalues are 3i and -3i and the eigenvectors are [(-2/5 + 1/5 i), (1)] and [(-2/5 - 1/5 i), (1)].
The general equation would be v1*c1*cos(3t) + v2*c2*sin(3t) where I would then attempt to find the constants, however, when multiplying the vectors, more complex numbers are introduced and my answer must be real only.
How do I isolate the real part only?
linear-algebra ordinary-differential-equations eigenvalues-eigenvectors
asked Dec 13 '18 at 1:42
S. SnakeS. Snake
485
$endgroup$
$begingroup$
Observe that the two eigenvectors are complex conjugates. Does this suggest any linear combinations of them that have all real components?
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– amd
Dec 13 '18 at 1:59
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amd, not sure what you mean by the linear combination of them? Am I suppose to find other vectors to use?
$endgroup$
– S. Snake
Dec 13 '18 at 2:19
1
$begingroup$
That’s it exactly. The real and imaginary parts of either of the eigenfunction solutions are also linearly-independent solutions of the differential equation (verify this). You can use a linear combination of those as the general solution. See math.stackexchange.com/questions/549838/… from the related questions at right if you don’t remember this from your course material.
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– amd
Dec 13 '18 at 3:30
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You literally took a screen shot of your homework?
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– orange
Dec 13 '18 at 11:11
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orange, is that against the rules? I'd rather avoid typos and formatting issues.
$endgroup$
– S. Snake
Dec 15 '18 at 4:51
add a comment |
$begingroup$
So I have a general understanding of how to do these types of problems but I do not know how to get rid of the complex numbers. I found that the eigenvalues are 3i and -3i and the eigenvectors are [(-2/5 + 1/5 i), (1)] and [(-2/5 - 1/5 i), (1)].
The general equation would be v1*c1*cos(3t) + v2*c2*sin(3t) where I would then attempt to find the constants, however, when multiplying the vectors, more complex numbers are introduced and my answer must be real only.
How do I isolate the real part only?
linear-algebra ordinary-differential-equations eigenvalues-eigenvectors
asked Dec 13 '18 at 1:42
S. SnakeS. Snake
485
$endgroup$
So I have a general understanding of how to do these types of problems but I do not know how to get rid of the complex numbers. I found that the eigenvalues are 3i and -3i and the eigenvectors are [(-2/5 + 1/5 i), (1)] and [(-2/5 - 1/5 i), (1)].
The general equation would be v1*c1*cos(3t) + v2*c2*sin(3t) where I would then attempt to find the constants, however, when multiplying the vectors, more complex numbers are introduced and my answer must be real only.
How do I isolate the real part only?
linear-algebra ordinary-differential-equations eigenvalues-eigenvectors
linear-algebra ordinary-differential-equations eigenvalues-eigenvectors
asked Dec 13 '18 at 1:42
S. SnakeS. Snake
485
asked Dec 13 '18 at 1:42
S. SnakeS. Snake
485
asked Dec 13 '18 at 1:42
S. SnakeS. Snake
485
asked Dec 13 '18 at 1:42
S. SnakeS. Snake
485
asked Dec 13 '18 at 1:42
S. SnakeS. Snake
485
485
$begingroup$
Observe that the two eigenvectors are complex conjugates. Does this suggest any linear combinations of them that have all real components?
$endgroup$
– amd
Dec 13 '18 at 1:59
$begingroup$
amd, not sure what you mean by the linear combination of them? Am I suppose to find other vectors to use?
$endgroup$
– S. Snake
Dec 13 '18 at 2:19
1
$begingroup$
That’s it exactly. The real and imaginary parts of either of the eigenfunction solutions are also linearly-independent solutions of the differential equation (verify this). You can use a linear combination of those as the general solution. See math.stackexchange.com/questions/549838/… from the related questions at right if you don’t remember this from your course material.
$endgroup$
– amd
Dec 13 '18 at 3:30
$begingroup$
You literally took a screen shot of your homework?
$endgroup$
– orange
Dec 13 '18 at 11:11
$begingroup$
orange, is that against the rules? I'd rather avoid typos and formatting issues.
$endgroup$
– S. Snake
Dec 15 '18 at 4:51
add a comment |
$begingroup$
Observe that the two eigenvectors are complex conjugates. Does this suggest any linear combinations of them that have all real components?
$endgroup$
– amd
Dec 13 '18 at 1:59
$begingroup$
amd, not sure what you mean by the linear combination of them? Am I suppose to find other vectors to use?
$endgroup$
– S. Snake
Dec 13 '18 at 2:19
1
$begingroup$
That’s it exactly. The real and imaginary parts of either of the eigenfunction solutions are also linearly-independent solutions of the differential equation (verify this). You can use a linear combination of those as the general solution. See math.stackexchange.com/questions/549838/… from the related questions at right if you don’t remember this from your course material.
$endgroup$
– amd
Dec 13 '18 at 3:30
$begingroup$
You literally took a screen shot of your homework?
$endgroup$
– orange
Dec 13 '18 at 11:11
$begingroup$
orange, is that against the rules? I'd rather avoid typos and formatting issues.
$endgroup$
– S. Snake
Dec 15 '18 at 4:51
$begingroup$
Observe that the two eigenvectors are complex conjugates. Does this suggest any linear combinations of them that have all real components?
$endgroup$
– amd
Dec 13 '18 at 1:59
$begingroup$
Observe that the two eigenvectors are complex conjugates. Does this suggest any linear combinations of them that have all real components?
$endgroup$
– amd
Dec 13 '18 at 1:59
$begingroup$
amd, not sure what you mean by the linear combination of them? Am I suppose to find other vectors to use?
$endgroup$
– S. Snake
Dec 13 '18 at 2:19
$begingroup$
amd, not sure what you mean by the linear combination of them? Am I suppose to find other vectors to use?
$endgroup$
– S. Snake
Dec 13 '18 at 2:19
1
1
$begingroup$
That’s it exactly. The real and imaginary parts of either of the eigenfunction solutions are also linearly-independent solutions of the differential equation (verify this). You can use a linear combination of those as the general solution. See math.stackexchange.com/questions/549838/… from the related questions at right if you don’t remember this from your course material.
$endgroup$
– amd
Dec 13 '18 at 3:30
$begingroup$
That’s it exactly. The real and imaginary parts of either of the eigenfunction solutions are also linearly-independent solutions of the differential equation (verify this). You can use a linear combination of those as the general solution. See math.stackexchange.com/questions/549838/… from the related questions at right if you don’t remember this from your course material.
$endgroup$
– amd
Dec 13 '18 at 3:30
$begingroup$
You literally took a screen shot of your homework?
$endgroup$
– orange
Dec 13 '18 at 11:11
$begingroup$
You literally took a screen shot of your homework?
$endgroup$
– orange
Dec 13 '18 at 11:11
$begingroup$
orange, is that against the rules? I'd rather avoid typos and formatting issues.
$endgroup$
– S. Snake
Dec 15 '18 at 4:51
$begingroup$
orange, is that against the rules? I'd rather avoid typos and formatting issues.
$endgroup$
– S. Snake
Dec 15 '18 at 4:51
add a comment |
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$begingroup$
Observe that the two eigenvectors are complex conjugates. Does this suggest any linear combinations of them that have all real components?
$endgroup$
– amd
Dec 13 '18 at 1:59
$begingroup$
amd, not sure what you mean by the linear combination of them? Am I suppose to find other vectors to use?
$endgroup$
– S. Snake
Dec 13 '18 at 2:19
1
$begingroup$
That’s it exactly. The real and imaginary parts of either of the eigenfunction solutions are also linearly-independent solutions of the differential equation (verify this). You can use a linear combination of those as the general solution. See math.stackexchange.com/questions/549838/… from the related questions at right if you don’t remember this from your course material.
$endgroup$
– amd
Dec 13 '18 at 3:30
$begingroup$
You literally took a screen shot of your homework?
$endgroup$
– orange
Dec 13 '18 at 11:11
$begingroup$
orange, is that against the rules? I'd rather avoid typos and formatting issues.
$endgroup$
– S. Snake
Dec 15 '18 at 4:51