Find a particular solution for this linear system
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1
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$mathbf{x'}=begin{bmatrix}3 & 1 \ 0 & 3end{bmatrix}mathbf{x}+begin{bmatrix} e^{3t}\e^{3t}end{bmatrix}$
My attempt:
We first find a solution to the associated homogeneous system $mathbf{x'}=begin{bmatrix}3 & 1 \ 0 & 3end{bmatrix}mathbf{x}$.
My answer is $mathbf{x}_c(t)=a_0begin{bmatrix}1 \ 0end{bmatrix}e^{3t}+b_0left(begin{bmatrix}1 \ 0end{bmatrix}t+begin{bmatrix}0 \ 1end{bmatrix}right)e^{3t}$
In order to eliminate duplication, let $mathbf{x}_p(t)=begin{bmatrix}a_1 \ a_2end{bmatrix}t^2e^{3t}+begin{bmatrix}b_1 \ b_2end{bmatrix}te^{3t}+begin{bmatrix}c_1 \ c_2end{bmatrix}e^{3t}$
When I substitute this into the given ODE, I get 6 equations in 6 variables but I do not manage to obtain values for all of the variables. How do
I proceed?
linear-algebra differential-equations
add a comment |
up vote
1
down vote
favorite
$mathbf{x'}=begin{bmatrix}3 & 1 \ 0 & 3end{bmatrix}mathbf{x}+begin{bmatrix} e^{3t}\e^{3t}end{bmatrix}$
My attempt:
We first find a solution to the associated homogeneous system $mathbf{x'}=begin{bmatrix}3 & 1 \ 0 & 3end{bmatrix}mathbf{x}$.
My answer is $mathbf{x}_c(t)=a_0begin{bmatrix}1 \ 0end{bmatrix}e^{3t}+b_0left(begin{bmatrix}1 \ 0end{bmatrix}t+begin{bmatrix}0 \ 1end{bmatrix}right)e^{3t}$
In order to eliminate duplication, let $mathbf{x}_p(t)=begin{bmatrix}a_1 \ a_2end{bmatrix}t^2e^{3t}+begin{bmatrix}b_1 \ b_2end{bmatrix}te^{3t}+begin{bmatrix}c_1 \ c_2end{bmatrix}e^{3t}$
When I substitute this into the given ODE, I get 6 equations in 6 variables but I do not manage to obtain values for all of the variables. How do
I proceed?
linear-algebra differential-equations
But are my steps right? I'm curious to know where I'm going wrong.
– Thomas
Nov 15 at 23:24
add a comment |
up vote
1
down vote
favorite
up vote
1
down vote
favorite
$mathbf{x'}=begin{bmatrix}3 & 1 \ 0 & 3end{bmatrix}mathbf{x}+begin{bmatrix} e^{3t}\e^{3t}end{bmatrix}$
My attempt:
We first find a solution to the associated homogeneous system $mathbf{x'}=begin{bmatrix}3 & 1 \ 0 & 3end{bmatrix}mathbf{x}$.
My answer is $mathbf{x}_c(t)=a_0begin{bmatrix}1 \ 0end{bmatrix}e^{3t}+b_0left(begin{bmatrix}1 \ 0end{bmatrix}t+begin{bmatrix}0 \ 1end{bmatrix}right)e^{3t}$
In order to eliminate duplication, let $mathbf{x}_p(t)=begin{bmatrix}a_1 \ a_2end{bmatrix}t^2e^{3t}+begin{bmatrix}b_1 \ b_2end{bmatrix}te^{3t}+begin{bmatrix}c_1 \ c_2end{bmatrix}e^{3t}$
When I substitute this into the given ODE, I get 6 equations in 6 variables but I do not manage to obtain values for all of the variables. How do
I proceed?
linear-algebra differential-equations
$mathbf{x'}=begin{bmatrix}3 & 1 \ 0 & 3end{bmatrix}mathbf{x}+begin{bmatrix} e^{3t}\e^{3t}end{bmatrix}$
My attempt:
We first find a solution to the associated homogeneous system $mathbf{x'}=begin{bmatrix}3 & 1 \ 0 & 3end{bmatrix}mathbf{x}$.
My answer is $mathbf{x}_c(t)=a_0begin{bmatrix}1 \ 0end{bmatrix}e^{3t}+b_0left(begin{bmatrix}1 \ 0end{bmatrix}t+begin{bmatrix}0 \ 1end{bmatrix}right)e^{3t}$
In order to eliminate duplication, let $mathbf{x}_p(t)=begin{bmatrix}a_1 \ a_2end{bmatrix}t^2e^{3t}+begin{bmatrix}b_1 \ b_2end{bmatrix}te^{3t}+begin{bmatrix}c_1 \ c_2end{bmatrix}e^{3t}$
When I substitute this into the given ODE, I get 6 equations in 6 variables but I do not manage to obtain values for all of the variables. How do
I proceed?
linear-algebra differential-equations
linear-algebra differential-equations
asked Nov 15 at 23:11
Thomas
695415
695415
But are my steps right? I'm curious to know where I'm going wrong.
– Thomas
Nov 15 at 23:24
add a comment |
But are my steps right? I'm curious to know where I'm going wrong.
– Thomas
Nov 15 at 23:24
But are my steps right? I'm curious to know where I'm going wrong.
– Thomas
Nov 15 at 23:24
But are my steps right? I'm curious to know where I'm going wrong.
– Thomas
Nov 15 at 23:24
add a comment |
2 Answers
2
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oldest
votes
up vote
1
down vote
accepted
$$X'=AX+g(t)$$
Your particular solution is correct but since the matrix A is already in Jordan form.
$$pmatrix {x \y}'=pmatrix{3 & 1 \ 0 & 3}pmatrix {x \ y}+begin{bmatrix} e^{3t}\e^{3t}end{bmatrix}$$
$$y'=3y+e^{3t} implies (ye^{-3t})'=1$$
$$y(t)=e^{3t}(t+K)$$
$$x'=3x+y+e^{3t}$$
$$x'=3x+e^{3t}(t+K+1) implies (xe^{-3t})'=t+K+1$$
$$x(t)=(frac {t^2}2+(K+1)t+C)e^{3t}$$
add a comment |
up vote
-1
down vote
It appears that the variables that don't have values can take on arbitrary values that are consistent with the 6 equations. I checked one case using MATLAB and it works.
add a comment |
2 Answers
2
active
oldest
votes
2 Answers
2
active
oldest
votes
active
oldest
votes
active
oldest
votes
up vote
1
down vote
accepted
$$X'=AX+g(t)$$
Your particular solution is correct but since the matrix A is already in Jordan form.
$$pmatrix {x \y}'=pmatrix{3 & 1 \ 0 & 3}pmatrix {x \ y}+begin{bmatrix} e^{3t}\e^{3t}end{bmatrix}$$
$$y'=3y+e^{3t} implies (ye^{-3t})'=1$$
$$y(t)=e^{3t}(t+K)$$
$$x'=3x+y+e^{3t}$$
$$x'=3x+e^{3t}(t+K+1) implies (xe^{-3t})'=t+K+1$$
$$x(t)=(frac {t^2}2+(K+1)t+C)e^{3t}$$
add a comment |
up vote
1
down vote
accepted
$$X'=AX+g(t)$$
Your particular solution is correct but since the matrix A is already in Jordan form.
$$pmatrix {x \y}'=pmatrix{3 & 1 \ 0 & 3}pmatrix {x \ y}+begin{bmatrix} e^{3t}\e^{3t}end{bmatrix}$$
$$y'=3y+e^{3t} implies (ye^{-3t})'=1$$
$$y(t)=e^{3t}(t+K)$$
$$x'=3x+y+e^{3t}$$
$$x'=3x+e^{3t}(t+K+1) implies (xe^{-3t})'=t+K+1$$
$$x(t)=(frac {t^2}2+(K+1)t+C)e^{3t}$$
add a comment |
up vote
1
down vote
accepted
up vote
1
down vote
accepted
$$X'=AX+g(t)$$
Your particular solution is correct but since the matrix A is already in Jordan form.
$$pmatrix {x \y}'=pmatrix{3 & 1 \ 0 & 3}pmatrix {x \ y}+begin{bmatrix} e^{3t}\e^{3t}end{bmatrix}$$
$$y'=3y+e^{3t} implies (ye^{-3t})'=1$$
$$y(t)=e^{3t}(t+K)$$
$$x'=3x+y+e^{3t}$$
$$x'=3x+e^{3t}(t+K+1) implies (xe^{-3t})'=t+K+1$$
$$x(t)=(frac {t^2}2+(K+1)t+C)e^{3t}$$
$$X'=AX+g(t)$$
Your particular solution is correct but since the matrix A is already in Jordan form.
$$pmatrix {x \y}'=pmatrix{3 & 1 \ 0 & 3}pmatrix {x \ y}+begin{bmatrix} e^{3t}\e^{3t}end{bmatrix}$$
$$y'=3y+e^{3t} implies (ye^{-3t})'=1$$
$$y(t)=e^{3t}(t+K)$$
$$x'=3x+y+e^{3t}$$
$$x'=3x+e^{3t}(t+K+1) implies (xe^{-3t})'=t+K+1$$
$$x(t)=(frac {t^2}2+(K+1)t+C)e^{3t}$$
answered Nov 16 at 5:27
Isham
12.7k3929
12.7k3929
add a comment |
add a comment |
up vote
-1
down vote
It appears that the variables that don't have values can take on arbitrary values that are consistent with the 6 equations. I checked one case using MATLAB and it works.
add a comment |
up vote
-1
down vote
It appears that the variables that don't have values can take on arbitrary values that are consistent with the 6 equations. I checked one case using MATLAB and it works.
add a comment |
up vote
-1
down vote
up vote
-1
down vote
It appears that the variables that don't have values can take on arbitrary values that are consistent with the 6 equations. I checked one case using MATLAB and it works.
It appears that the variables that don't have values can take on arbitrary values that are consistent with the 6 equations. I checked one case using MATLAB and it works.
answered Nov 15 at 23:50
Thomas
695415
695415
add a comment |
add a comment |
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But are my steps right? I'm curious to know where I'm going wrong.
– Thomas
Nov 15 at 23:24