Solving for $z_{t+1} = begin{bmatrix} -1 & -2\ -2& 2 end{bmatrix}z_{t} - begin{bmatrix} 24\ 9...
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The above equation is a nonhomogeneous difference equation. To find the general solution, I have to find the particular solution and a complementary solution.
I have obtained the complementary solution, that is, for the homogeneous difference equation $z_{t+1} = A z_{t}$, which is $$z_{c} = c_{1}left ( -2 right )^{n}begin{bmatrix} 2\ 1
end{bmatrix} + c_{2}left ( 3 right )^{n}begin{bmatrix}
1\
2
end{bmatrix}.$$
This is after calculating the eigenvalues and the eigenvalues for the matrix $A$.
Now for finding the particular solution, $z_{p}$ I was wondering if I can get some help in guessing a function.
The general solution for the nonhomogeneous difference equation will consist of $z_{n} = z_{c} + z_{p}$.
EDIT: For the particular solution, I have solved $z^{*} = A z^{*} + b$ and have solved for vector $$z^{*} = begin{bmatrix}
-1\
-11
end{bmatrix} = z_{p}.$$
Thus, for the general solution we have, $$z_{n} =
c_{1}left ( -2 right )^{n}begin{bmatrix} 2\ 1
end{bmatrix} + c_{2}left ( 3 right )^{n}begin{bmatrix}
1\
2
end{bmatrix} +
begin{bmatrix}
-1\
-11
end{bmatrix}.$$
Does this general solution for the nonhomogeneous difference equation make sense?
differential-equations proof-verification recurrence-relations systems-of-equations
asked Nov 29 at 4:01
OGC
1,41921228
add a comment |
up vote
0
down vote
favorite
The above equation is a nonhomogeneous difference equation. To find the general solution, I have to find the particular solution and a complementary solution.
I have obtained the complementary solution, that is, for the homogeneous difference equation $z_{t+1} = A z_{t}$, which is $$z_{c} = c_{1}left ( -2 right )^{n}begin{bmatrix} 2\ 1
end{bmatrix} + c_{2}left ( 3 right )^{n}begin{bmatrix}
1\
2
end{bmatrix}.$$
This is after calculating the eigenvalues and the eigenvalues for the matrix $A$.
Now for finding the particular solution, $z_{p}$ I was wondering if I can get some help in guessing a function.
The general solution for the nonhomogeneous difference equation will consist of $z_{n} = z_{c} + z_{p}$.
EDIT: For the particular solution, I have solved $z^{*} = A z^{*} + b$ and have solved for vector $$z^{*} = begin{bmatrix}
-1\
-11
end{bmatrix} = z_{p}.$$
Thus, for the general solution we have, $$z_{n} =
c_{1}left ( -2 right )^{n}begin{bmatrix} 2\ 1
end{bmatrix} + c_{2}left ( 3 right )^{n}begin{bmatrix}
1\
2
end{bmatrix} +
begin{bmatrix}
-1\
-11
end{bmatrix}.$$
Does this general solution for the nonhomogeneous difference equation make sense?
differential-equations proof-verification recurrence-relations systems-of-equations
asked Nov 29 at 4:01
OGC
1,41921228
1
Try guessing that $z_t$ is constant.
– saulspatz
Nov 29 at 4:11
@saulspatz I have made an edit.
– OGC
Nov 29 at 5:27
1
Your solution certainly makes sense. I haven't checked the actual calculations, though.
– saulspatz
Nov 29 at 13:37
add a comment |
up vote
0
down vote
favorite
up vote
0
down vote
favorite
The above equation is a nonhomogeneous difference equation. To find the general solution, I have to find the particular solution and a complementary solution.
I have obtained the complementary solution, that is, for the homogeneous difference equation $z_{t+1} = A z_{t}$, which is $$z_{c} = c_{1}left ( -2 right )^{n}begin{bmatrix} 2\ 1
end{bmatrix} + c_{2}left ( 3 right )^{n}begin{bmatrix}
1\
2
end{bmatrix}.$$
This is after calculating the eigenvalues and the eigenvalues for the matrix $A$.
Now for finding the particular solution, $z_{p}$ I was wondering if I can get some help in guessing a function.
The general solution for the nonhomogeneous difference equation will consist of $z_{n} = z_{c} + z_{p}$.
EDIT: For the particular solution, I have solved $z^{*} = A z^{*} + b$ and have solved for vector $$z^{*} = begin{bmatrix}
-1\
-11
end{bmatrix} = z_{p}.$$
Thus, for the general solution we have, $$z_{n} =
c_{1}left ( -2 right )^{n}begin{bmatrix} 2\ 1
end{bmatrix} + c_{2}left ( 3 right )^{n}begin{bmatrix}
1\
2
end{bmatrix} +
begin{bmatrix}
-1\
-11
end{bmatrix}.$$
Does this general solution for the nonhomogeneous difference equation make sense?
differential-equations proof-verification recurrence-relations systems-of-equations
asked Nov 29 at 4:01
OGC
1,41921228
The above equation is a nonhomogeneous difference equation. To find the general solution, I have to find the particular solution and a complementary solution.
I have obtained the complementary solution, that is, for the homogeneous difference equation $z_{t+1} = A z_{t}$, which is $$z_{c} = c_{1}left ( -2 right )^{n}begin{bmatrix} 2\ 1
end{bmatrix} + c_{2}left ( 3 right )^{n}begin{bmatrix}
1\
2
end{bmatrix}.$$
This is after calculating the eigenvalues and the eigenvalues for the matrix $A$.
Now for finding the particular solution, $z_{p}$ I was wondering if I can get some help in guessing a function.
The general solution for the nonhomogeneous difference equation will consist of $z_{n} = z_{c} + z_{p}$.
EDIT: For the particular solution, I have solved $z^{*} = A z^{*} + b$ and have solved for vector $$z^{*} = begin{bmatrix}
-1\
-11
end{bmatrix} = z_{p}.$$
Thus, for the general solution we have, $$z_{n} =
c_{1}left ( -2 right )^{n}begin{bmatrix} 2\ 1
end{bmatrix} + c_{2}left ( 3 right )^{n}begin{bmatrix}
1\
2
end{bmatrix} +
begin{bmatrix}
-1\
-11
end{bmatrix}.$$
Does this general solution for the nonhomogeneous difference equation make sense?
differential-equations proof-verification recurrence-relations systems-of-equations
differential-equations proof-verification recurrence-relations systems-of-equations
asked Nov 29 at 4:01
OGC
1,41921228
asked Nov 29 at 4:01
OGC
1,41921228
edited Nov 29 at 5:40
asked Nov 29 at 4:01
OGC
1,41921228
asked Nov 29 at 4:01
OGC
1,41921228
asked Nov 29 at 4:01
OGC
1,41921228
1,41921228
1
Try guessing that $z_t$ is constant.
– saulspatz
Nov 29 at 4:11
@saulspatz I have made an edit.
– OGC
Nov 29 at 5:27
1
Your solution certainly makes sense. I haven't checked the actual calculations, though.
– saulspatz
Nov 29 at 13:37
add a comment |
1
Try guessing that $z_t$ is constant.
– saulspatz
Nov 29 at 4:11
@saulspatz I have made an edit.
– OGC
Nov 29 at 5:27
1
Your solution certainly makes sense. I haven't checked the actual calculations, though.
– saulspatz
Nov 29 at 13:37
1
1
Try guessing that $z_t$ is constant.
– saulspatz
Nov 29 at 4:11
Try guessing that $z_t$ is constant.
– saulspatz
Nov 29 at 4:11
@saulspatz I have made an edit.
– OGC
Nov 29 at 5:27
@saulspatz I have made an edit.
– OGC
Nov 29 at 5:27
1
1
Your solution certainly makes sense. I haven't checked the actual calculations, though.
– saulspatz
Nov 29 at 13:37
Your solution certainly makes sense. I haven't checked the actual calculations, though.
– saulspatz
Nov 29 at 13:37
add a comment |
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1
Try guessing that $z_t$ is constant.
– saulspatz
Nov 29 at 4:11
@saulspatz I have made an edit.
– OGC
Nov 29 at 5:27
1
Your solution certainly makes sense. I haven't checked the actual calculations, though.
– saulspatz
Nov 29 at 13:37