Solving a linear system of differential equations
$begingroup$
Given that $v_1 = begin{bmatrix}1&1end{bmatrix}$ and $v_2 = begin{bmatrix}2 &1end{bmatrix}$ are eigenvectors of the matrix
$$
begin{bmatrix}-1&-2\1&-4end{bmatrix}
$$
which is a $2times 2$ matrix.
Find the solution to the linear system of differential equations
begin{align*}
x' &= -x - 2y\
y' &= x - 4y
end{align*}
satisfying the initial conditions $x(0)=7$ and $y(0)=5$.
So I already found the eigenvalues, $-3$ and $-2$ and I know that you need to plug the eigenvalues into the matrix you get from doing $det(It - A)$ but I'm not sure where to go from there in terms of making it into an equation?
linear-algebra ordinary-differential-equations eigenvalues-eigenvectors
asked Dec 8 '18 at 20:56
S. SnakeS. Snake
485
$endgroup$
add a comment |
$begingroup$
Given that $v_1 = begin{bmatrix}1&1end{bmatrix}$ and $v_2 = begin{bmatrix}2 &1end{bmatrix}$ are eigenvectors of the matrix
$$
begin{bmatrix}-1&-2\1&-4end{bmatrix}
$$
which is a $2times 2$ matrix.
Find the solution to the linear system of differential equations
begin{align*}
x' &= -x - 2y\
y' &= x - 4y
end{align*}
satisfying the initial conditions $x(0)=7$ and $y(0)=5$.
So I already found the eigenvalues, $-3$ and $-2$ and I know that you need to plug the eigenvalues into the matrix you get from doing $det(It - A)$ but I'm not sure where to go from there in terms of making it into an equation?
linear-algebra ordinary-differential-equations eigenvalues-eigenvectors
asked Dec 8 '18 at 20:56
S. SnakeS. Snake
485
$endgroup$
add a comment |
$begingroup$
Given that $v_1 = begin{bmatrix}1&1end{bmatrix}$ and $v_2 = begin{bmatrix}2 &1end{bmatrix}$ are eigenvectors of the matrix
$$
begin{bmatrix}-1&-2\1&-4end{bmatrix}
$$
which is a $2times 2$ matrix.
Find the solution to the linear system of differential equations
begin{align*}
x' &= -x - 2y\
y' &= x - 4y
end{align*}
satisfying the initial conditions $x(0)=7$ and $y(0)=5$.
So I already found the eigenvalues, $-3$ and $-2$ and I know that you need to plug the eigenvalues into the matrix you get from doing $det(It - A)$ but I'm not sure where to go from there in terms of making it into an equation?
linear-algebra ordinary-differential-equations eigenvalues-eigenvectors
asked Dec 8 '18 at 20:56
S. SnakeS. Snake
485
$endgroup$
Given that $v_1 = begin{bmatrix}1&1end{bmatrix}$ and $v_2 = begin{bmatrix}2 &1end{bmatrix}$ are eigenvectors of the matrix
$$
begin{bmatrix}-1&-2\1&-4end{bmatrix}
$$
which is a $2times 2$ matrix.
Find the solution to the linear system of differential equations
begin{align*}
x' &= -x - 2y\
y' &= x - 4y
end{align*}
satisfying the initial conditions $x(0)=7$ and $y(0)=5$.
So I already found the eigenvalues, $-3$ and $-2$ and I know that you need to plug the eigenvalues into the matrix you get from doing $det(It - A)$ but I'm not sure where to go from there in terms of making it into an equation?
linear-algebra ordinary-differential-equations eigenvalues-eigenvectors
linear-algebra ordinary-differential-equations eigenvalues-eigenvectors
asked Dec 8 '18 at 20:56
S. SnakeS. Snake
485
asked Dec 8 '18 at 20:56
S. SnakeS. Snake
485
edited Dec 8 '18 at 23:19
S. Snake
asked Dec 8 '18 at 20:56
S. SnakeS. Snake
485
asked Dec 8 '18 at 20:56
S. SnakeS. Snake
485
asked Dec 8 '18 at 20:56
S. SnakeS. Snake
485
485
add a comment |
add a comment |
1 Answer
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$begingroup$
We can write the solution to the system as
$$X(t) = begin{bmatrix} x(t) \ y(t)end{bmatrix} = c_1 e^{lambda_1 t} v_1 + c_2 e^{lambda_2 t} v_2$$
From the given information, we have
$$X(t) = c_1 e^{-3 t}begin{bmatrix} 1 \ 1 end{bmatrix} + c_2 e^{-2 t}begin{bmatrix} 2 \ 1 end{bmatrix}$$
Now, use the initial conditions to solve for $c_1$ and $c_2$. You can see examples here.
answered Dec 9 '18 at 0:18
MooMoo
5,63131020
$endgroup$
add a comment |
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1 Answer
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1 Answer
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$begingroup$
We can write the solution to the system as
$$X(t) = begin{bmatrix} x(t) \ y(t)end{bmatrix} = c_1 e^{lambda_1 t} v_1 + c_2 e^{lambda_2 t} v_2$$
From the given information, we have
$$X(t) = c_1 e^{-3 t}begin{bmatrix} 1 \ 1 end{bmatrix} + c_2 e^{-2 t}begin{bmatrix} 2 \ 1 end{bmatrix}$$
Now, use the initial conditions to solve for $c_1$ and $c_2$. You can see examples here.
answered Dec 9 '18 at 0:18
MooMoo
5,63131020
$endgroup$
add a comment |
$begingroup$
We can write the solution to the system as
$$X(t) = begin{bmatrix} x(t) \ y(t)end{bmatrix} = c_1 e^{lambda_1 t} v_1 + c_2 e^{lambda_2 t} v_2$$
From the given information, we have
$$X(t) = c_1 e^{-3 t}begin{bmatrix} 1 \ 1 end{bmatrix} + c_2 e^{-2 t}begin{bmatrix} 2 \ 1 end{bmatrix}$$
Now, use the initial conditions to solve for $c_1$ and $c_2$. You can see examples here.
answered Dec 9 '18 at 0:18
MooMoo
5,63131020
$endgroup$
add a comment |
$begingroup$
We can write the solution to the system as
$$X(t) = begin{bmatrix} x(t) \ y(t)end{bmatrix} = c_1 e^{lambda_1 t} v_1 + c_2 e^{lambda_2 t} v_2$$
From the given information, we have
$$X(t) = c_1 e^{-3 t}begin{bmatrix} 1 \ 1 end{bmatrix} + c_2 e^{-2 t}begin{bmatrix} 2 \ 1 end{bmatrix}$$
Now, use the initial conditions to solve for $c_1$ and $c_2$. You can see examples here.
answered Dec 9 '18 at 0:18
MooMoo
5,63131020
$endgroup$
We can write the solution to the system as
$$X(t) = begin{bmatrix} x(t) \ y(t)end{bmatrix} = c_1 e^{lambda_1 t} v_1 + c_2 e^{lambda_2 t} v_2$$
From the given information, we have
$$X(t) = c_1 e^{-3 t}begin{bmatrix} 1 \ 1 end{bmatrix} + c_2 e^{-2 t}begin{bmatrix} 2 \ 1 end{bmatrix}$$
Now, use the initial conditions to solve for $c_1$ and $c_2$. You can see examples here.
answered Dec 9 '18 at 0:18
MooMoo
5,63131020
edited Dec 9 '18 at 17:38
answered Dec 9 '18 at 0:18
MooMoo
5,63131020
answered Dec 9 '18 at 0:18
MooMoo
5,63131020
answered Dec 9 '18 at 0:18
MooMoo
5,63131020
5,63131020
add a comment |
add a comment |
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