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How do I linearise the rational function to analyze the critical points?
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For the system $$fracdxdt=frac3xy1+x^2+y^2-frac1+x^21+y^2\fracdydt=x^2-y^2,$$ the point $beginpmatrix1\1endpmatrix$ is
A. an unstable node
B. a stable node
C. a saddle point
D. a stable spiral point
E. an unstable spiral point.
I understand I need to shift the point to the origin and just eliminate $x^2$ and $y^2$, but how do I deal with denominators?
Thank you
calculus differential-equations linearization
edited Jul 31 at 18:49
TheSimpliFire
9,51451851
asked Jul 31 at 18:28
Jeff
161
 |Â
show 1 more comment
up vote
3
down vote
favorite
For the system $$fracdxdt=frac3xy1+x^2+y^2-frac1+x^21+y^2\fracdydt=x^2-y^2,$$ the point $beginpmatrix1\1endpmatrix$ is
A. an unstable node
B. a stable node
C. a saddle point
D. a stable spiral point
E. an unstable spiral point.
I understand I need to shift the point to the origin and just eliminate $x^2$ and $y^2$, but how do I deal with denominators?
Thank you
calculus differential-equations linearization
edited Jul 31 at 18:49
TheSimpliFire
9,51451851
asked Jul 31 at 18:28
Jeff
161
Image of the question is in vQuestion. Thank you!
– Jeff
Jul 31 at 18:28
Please use Mathjax...
– TheSimpliFire
Jul 31 at 18:39
can you not see the question?
– Jeff
Jul 31 at 18:40
1
Yes, but for future readers, this question would be easier to find, rather than just a .png file.
– TheSimpliFire
Jul 31 at 18:42
I will keep it in mind. Thank you! It's my first time posting a question here. I'm sorry about that!
– Jeff
Jul 31 at 18:44
 |Â
show 1 more comment
up vote
3
down vote
favorite
up vote
3
down vote
favorite
For the system $$fracdxdt=frac3xy1+x^2+y^2-frac1+x^21+y^2\fracdydt=x^2-y^2,$$ the point $beginpmatrix1\1endpmatrix$ is
A. an unstable node
B. a stable node
C. a saddle point
D. a stable spiral point
E. an unstable spiral point.
I understand I need to shift the point to the origin and just eliminate $x^2$ and $y^2$, but how do I deal with denominators?
Thank you
calculus differential-equations linearization
edited Jul 31 at 18:49
TheSimpliFire
9,51451851
asked Jul 31 at 18:28
Jeff
161
For the system $$fracdxdt=frac3xy1+x^2+y^2-frac1+x^21+y^2\fracdydt=x^2-y^2,$$ the point $beginpmatrix1\1endpmatrix$ is
A. an unstable node
B. a stable node
C. a saddle point
D. a stable spiral point
E. an unstable spiral point.
I understand I need to shift the point to the origin and just eliminate $x^2$ and $y^2$, but how do I deal with denominators?
Thank you
calculus differential-equations linearization
edited Jul 31 at 18:49
TheSimpliFire
9,51451851
asked Jul 31 at 18:28
Jeff
161
edited Jul 31 at 18:49
TheSimpliFire
9,51451851
edited Jul 31 at 18:49
TheSimpliFire
9,51451851
edited Jul 31 at 18:49
TheSimpliFire
9,51451851
9,51451851
asked Jul 31 at 18:28
Jeff
161
asked Jul 31 at 18:28
Jeff
161
asked Jul 31 at 18:28
Jeff
161
161
Image of the question is in vQuestion. Thank you!
– Jeff
Jul 31 at 18:28
Please use Mathjax...
– TheSimpliFire
Jul 31 at 18:39
can you not see the question?
– Jeff
Jul 31 at 18:40
1
Yes, but for future readers, this question would be easier to find, rather than just a .png file.
– TheSimpliFire
Jul 31 at 18:42
I will keep it in mind. Thank you! It's my first time posting a question here. I'm sorry about that!
– Jeff
Jul 31 at 18:44
 |Â
show 1 more comment
Image of the question is in vQuestion. Thank you!
– Jeff
Jul 31 at 18:28
Please use Mathjax...
– TheSimpliFire
Jul 31 at 18:39
can you not see the question?
– Jeff
Jul 31 at 18:40
1
Yes, but for future readers, this question would be easier to find, rather than just a .png file.
– TheSimpliFire
Jul 31 at 18:42
I will keep it in mind. Thank you! It's my first time posting a question here. I'm sorry about that!
– Jeff
Jul 31 at 18:44
Image of the question is in vQuestion. Thank you!
– Jeff
Jul 31 at 18:28
Image of the question is in vQuestion. Thank you!
– Jeff
Jul 31 at 18:28
Please use Mathjax...
– TheSimpliFire
Jul 31 at 18:39
Please use Mathjax...
– TheSimpliFire
Jul 31 at 18:39
can you not see the question?
– Jeff
Jul 31 at 18:40
can you not see the question?
– Jeff
Jul 31 at 18:40
1
1
Yes, but for future readers, this question would be easier to find, rather than just a .png file.
– TheSimpliFire
Jul 31 at 18:42
Yes, but for future readers, this question would be easier to find, rather than just a .png file.
– TheSimpliFire
Jul 31 at 18:42
I will keep it in mind. Thank you! It's my first time posting a question here. I'm sorry about that!
– Jeff
Jul 31 at 18:44
I will keep it in mind. Thank you! It's my first time posting a question here. I'm sorry about that!
– Jeff
Jul 31 at 18:44
 |Â
show 1 more comment
1 Answer
1
active
oldest
votes
up vote
2
down vote
This is a nonlinear system, so we can analyze behavior at critical points by finding the Jacobian matrix evaluated at those points and then computing eigenvalues of that matrix. Recall, the Jacobian matrix is:
$$beginbmatrixfracpartial fpartial x&fracpartial fpartial y\fracpartial gpartial x&fracpartial gpartial yendbmatrix$$
Where $f(x,y)=fracdxdt$ and $g(x,y)=fracdydt$.
So we have the following Jacobian matrix:
$$beginbmatrixfrac3y(1-x^2+y^2)(1+x^2+y^2)^2-frac2x1+y^2&frac3x(1-y^2+x^2)(1+x^2+y^2)^2-frac2y(1+x^2)(1+y^2)^2\2x&-2yendbmatrix$$
Plugging in $(x,y)=(1,1)$, we get:
$$beginbmatrix-frac23&frac43\2&-2endbmatrix$$
Now we need to find the eigenvalues of this matrix. We end up with the characteristic polynomial $frac3lambda^2+8lambda-43$. The roots of this polynomial are our eigenvalues, namely:
$$frac-4+2sqrt73, frac-4-2sqrt73$$
These are both real eigenvalues, one positive and one negative. We therefore have an (unstable) saddle point. You can verify this by graphing the system in pplane or some equivalent software.
answered Jul 31 at 19:59
Sriram Gopalakrishnan
1564
I thought I would need to linearize it first and shift the critical point. Thank you!
– Jeff
Jul 31 at 20:20
No problem! Finding the Jacobian actually linearizes and evaluating it at a point shifts.
– Sriram Gopalakrishnan
Jul 31 at 20:23
add a comment |Â
1 Answer
1
active
oldest
votes
1 Answer
1
active
oldest
votes
active
oldest
votes
active
oldest
votes
up vote
2
down vote
This is a nonlinear system, so we can analyze behavior at critical points by finding the Jacobian matrix evaluated at those points and then computing eigenvalues of that matrix. Recall, the Jacobian matrix is:
$$beginbmatrixfracpartial fpartial x&fracpartial fpartial y\fracpartial gpartial x&fracpartial gpartial yendbmatrix$$
Where $f(x,y)=fracdxdt$ and $g(x,y)=fracdydt$.
So we have the following Jacobian matrix:
$$beginbmatrixfrac3y(1-x^2+y^2)(1+x^2+y^2)^2-frac2x1+y^2&frac3x(1-y^2+x^2)(1+x^2+y^2)^2-frac2y(1+x^2)(1+y^2)^2\2x&-2yendbmatrix$$
Plugging in $(x,y)=(1,1)$, we get:
$$beginbmatrix-frac23&frac43\2&-2endbmatrix$$
Now we need to find the eigenvalues of this matrix. We end up with the characteristic polynomial $frac3lambda^2+8lambda-43$. The roots of this polynomial are our eigenvalues, namely:
$$frac-4+2sqrt73, frac-4-2sqrt73$$
These are both real eigenvalues, one positive and one negative. We therefore have an (unstable) saddle point. You can verify this by graphing the system in pplane or some equivalent software.
answered Jul 31 at 19:59
Sriram Gopalakrishnan
1564
I thought I would need to linearize it first and shift the critical point. Thank you!
– Jeff
Jul 31 at 20:20
No problem! Finding the Jacobian actually linearizes and evaluating it at a point shifts.
– Sriram Gopalakrishnan
Jul 31 at 20:23
add a comment |Â
up vote
2
down vote
This is a nonlinear system, so we can analyze behavior at critical points by finding the Jacobian matrix evaluated at those points and then computing eigenvalues of that matrix. Recall, the Jacobian matrix is:
$$beginbmatrixfracpartial fpartial x&fracpartial fpartial y\fracpartial gpartial x&fracpartial gpartial yendbmatrix$$
Where $f(x,y)=fracdxdt$ and $g(x,y)=fracdydt$.
So we have the following Jacobian matrix:
$$beginbmatrixfrac3y(1-x^2+y^2)(1+x^2+y^2)^2-frac2x1+y^2&frac3x(1-y^2+x^2)(1+x^2+y^2)^2-frac2y(1+x^2)(1+y^2)^2\2x&-2yendbmatrix$$
Plugging in $(x,y)=(1,1)$, we get:
$$beginbmatrix-frac23&frac43\2&-2endbmatrix$$
Now we need to find the eigenvalues of this matrix. We end up with the characteristic polynomial $frac3lambda^2+8lambda-43$. The roots of this polynomial are our eigenvalues, namely:
$$frac-4+2sqrt73, frac-4-2sqrt73$$
These are both real eigenvalues, one positive and one negative. We therefore have an (unstable) saddle point. You can verify this by graphing the system in pplane or some equivalent software.
answered Jul 31 at 19:59
Sriram Gopalakrishnan
1564
I thought I would need to linearize it first and shift the critical point. Thank you!
– Jeff
Jul 31 at 20:20
No problem! Finding the Jacobian actually linearizes and evaluating it at a point shifts.
– Sriram Gopalakrishnan
Jul 31 at 20:23
add a comment |Â
up vote
2
down vote
up vote
2
down vote
This is a nonlinear system, so we can analyze behavior at critical points by finding the Jacobian matrix evaluated at those points and then computing eigenvalues of that matrix. Recall, the Jacobian matrix is:
$$beginbmatrixfracpartial fpartial x&fracpartial fpartial y\fracpartial gpartial x&fracpartial gpartial yendbmatrix$$
Where $f(x,y)=fracdxdt$ and $g(x,y)=fracdydt$.
So we have the following Jacobian matrix:
$$beginbmatrixfrac3y(1-x^2+y^2)(1+x^2+y^2)^2-frac2x1+y^2&frac3x(1-y^2+x^2)(1+x^2+y^2)^2-frac2y(1+x^2)(1+y^2)^2\2x&-2yendbmatrix$$
Plugging in $(x,y)=(1,1)$, we get:
$$beginbmatrix-frac23&frac43\2&-2endbmatrix$$
Now we need to find the eigenvalues of this matrix. We end up with the characteristic polynomial $frac3lambda^2+8lambda-43$. The roots of this polynomial are our eigenvalues, namely:
$$frac-4+2sqrt73, frac-4-2sqrt73$$
These are both real eigenvalues, one positive and one negative. We therefore have an (unstable) saddle point. You can verify this by graphing the system in pplane or some equivalent software.
answered Jul 31 at 19:59
Sriram Gopalakrishnan
1564
This is a nonlinear system, so we can analyze behavior at critical points by finding the Jacobian matrix evaluated at those points and then computing eigenvalues of that matrix. Recall, the Jacobian matrix is:
$$beginbmatrixfracpartial fpartial x&fracpartial fpartial y\fracpartial gpartial x&fracpartial gpartial yendbmatrix$$
Where $f(x,y)=fracdxdt$ and $g(x,y)=fracdydt$.
So we have the following Jacobian matrix:
$$beginbmatrixfrac3y(1-x^2+y^2)(1+x^2+y^2)^2-frac2x1+y^2&frac3x(1-y^2+x^2)(1+x^2+y^2)^2-frac2y(1+x^2)(1+y^2)^2\2x&-2yendbmatrix$$
Plugging in $(x,y)=(1,1)$, we get:
$$beginbmatrix-frac23&frac43\2&-2endbmatrix$$
Now we need to find the eigenvalues of this matrix. We end up with the characteristic polynomial $frac3lambda^2+8lambda-43$. The roots of this polynomial are our eigenvalues, namely:
$$frac-4+2sqrt73, frac-4-2sqrt73$$
These are both real eigenvalues, one positive and one negative. We therefore have an (unstable) saddle point. You can verify this by graphing the system in pplane or some equivalent software.
answered Jul 31 at 19:59
Sriram Gopalakrishnan
1564
answered Jul 31 at 19:59
Sriram Gopalakrishnan
1564
answered Jul 31 at 19:59
Sriram Gopalakrishnan
1564
answered Jul 31 at 19:59
Sriram Gopalakrishnan
1564
1564
I thought I would need to linearize it first and shift the critical point. Thank you!
– Jeff
Jul 31 at 20:20
No problem! Finding the Jacobian actually linearizes and evaluating it at a point shifts.
– Sriram Gopalakrishnan
Jul 31 at 20:23
add a comment |Â
I thought I would need to linearize it first and shift the critical point. Thank you!
– Jeff
Jul 31 at 20:20
No problem! Finding the Jacobian actually linearizes and evaluating it at a point shifts.
– Sriram Gopalakrishnan
Jul 31 at 20:23
I thought I would need to linearize it first and shift the critical point. Thank you!
– Jeff
Jul 31 at 20:20
I thought I would need to linearize it first and shift the critical point. Thank you!
– Jeff
Jul 31 at 20:20
No problem! Finding the Jacobian actually linearizes and evaluating it at a point shifts.
– Sriram Gopalakrishnan
Jul 31 at 20:23
No problem! Finding the Jacobian actually linearizes and evaluating it at a point shifts.
– Sriram Gopalakrishnan
Jul 31 at 20:23
add a comment |Â
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Image of the question is in vQuestion. Thank you!
– Jeff
Jul 31 at 18:28
Please use Mathjax...
– TheSimpliFire
Jul 31 at 18:39
can you not see the question?
– Jeff
Jul 31 at 18:40
1
Yes, but for future readers, this question would be easier to find, rather than just a .png file.
– TheSimpliFire
Jul 31 at 18:42
I will keep it in mind. Thank you! It's my first time posting a question here. I'm sorry about that!
– Jeff
Jul 31 at 18:44