Mixing
Euler-Ansatz for hom. ODEs with constant coefficients
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Given is an hom. ODE with constant constant coefficients:
$A_0y(x)+A_1y'(x)+A_2y''(x) + dots + A_ny^{(n)}=0 tag{1}$
Now it's clear to me that the solution space $yinmathbb L$ is a vector space.
We can solve (1) using the Ansatz: $y(x)=e^{kx}$. We get:
$chi(k)=A_0 + A_1k + A_2k^2 + dots + A_nk^n=0 tag{2}$
Now with $k_i$ being a solution to $chi$ with multiplicity $m_i$, the solution to the ODE is:
$y(x)=sum_i y_i(x) tag{3}$
with
$y_i(x)=sum_{j=0}^{m_i} C_j x^j e^{k_i x}, quad C_jinmathbb R tag{4}$
Question: Since I expect $mathbb L$ to be a vector space, (4) kind of makes sense. I mean it doesn't look wrong and I can work with it and solve such ODEs, but I can't derive it. So (1), (2), (3) is clear but (4) isn't. How exactly do we get the $x^j$ part in (4)?
calculus differential-equations
asked yesterday
xotix
32629
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0
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Given is an hom. ODE with constant constant coefficients:
$A_0y(x)+A_1y'(x)+A_2y''(x) + dots + A_ny^{(n)}=0 tag{1}$
Now it's clear to me that the solution space $yinmathbb L$ is a vector space.
We can solve (1) using the Ansatz: $y(x)=e^{kx}$. We get:
$chi(k)=A_0 + A_1k + A_2k^2 + dots + A_nk^n=0 tag{2}$
Now with $k_i$ being a solution to $chi$ with multiplicity $m_i$, the solution to the ODE is:
$y(x)=sum_i y_i(x) tag{3}$
with
$y_i(x)=sum_{j=0}^{m_i} C_j x^j e^{k_i x}, quad C_jinmathbb R tag{4}$
Question: Since I expect $mathbb L$ to be a vector space, (4) kind of makes sense. I mean it doesn't look wrong and I can work with it and solve such ODEs, but I can't derive it. So (1), (2), (3) is clear but (4) isn't. How exactly do we get the $x^j$ part in (4)?
calculus differential-equations
asked yesterday
xotix
32629
add a comment |
up vote
0
down vote
favorite
up vote
0
down vote
favorite
Given is an hom. ODE with constant constant coefficients:
$A_0y(x)+A_1y'(x)+A_2y''(x) + dots + A_ny^{(n)}=0 tag{1}$
Now it's clear to me that the solution space $yinmathbb L$ is a vector space.
We can solve (1) using the Ansatz: $y(x)=e^{kx}$. We get:
$chi(k)=A_0 + A_1k + A_2k^2 + dots + A_nk^n=0 tag{2}$
Now with $k_i$ being a solution to $chi$ with multiplicity $m_i$, the solution to the ODE is:
$y(x)=sum_i y_i(x) tag{3}$
with
$y_i(x)=sum_{j=0}^{m_i} C_j x^j e^{k_i x}, quad C_jinmathbb R tag{4}$
Question: Since I expect $mathbb L$ to be a vector space, (4) kind of makes sense. I mean it doesn't look wrong and I can work with it and solve such ODEs, but I can't derive it. So (1), (2), (3) is clear but (4) isn't. How exactly do we get the $x^j$ part in (4)?
calculus differential-equations
asked yesterday
xotix
32629
Given is an hom. ODE with constant constant coefficients:
$A_0y(x)+A_1y'(x)+A_2y''(x) + dots + A_ny^{(n)}=0 tag{1}$
Now it's clear to me that the solution space $yinmathbb L$ is a vector space.
We can solve (1) using the Ansatz: $y(x)=e^{kx}$. We get:
$chi(k)=A_0 + A_1k + A_2k^2 + dots + A_nk^n=0 tag{2}$
Now with $k_i$ being a solution to $chi$ with multiplicity $m_i$, the solution to the ODE is:
$y(x)=sum_i y_i(x) tag{3}$
with
$y_i(x)=sum_{j=0}^{m_i} C_j x^j e^{k_i x}, quad C_jinmathbb R tag{4}$
Question: Since I expect $mathbb L$ to be a vector space, (4) kind of makes sense. I mean it doesn't look wrong and I can work with it and solve such ODEs, but I can't derive it. So (1), (2), (3) is clear but (4) isn't. How exactly do we get the $x^j$ part in (4)?
calculus differential-equations
calculus differential-equations
asked yesterday
xotix
32629
asked yesterday
xotix
32629
asked yesterday
xotix
32629
asked yesterday
xotix
32629
asked yesterday
xotix
32629
32629
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1 Answer
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Defining $v = (y, y^{(1)}, ..., y^{(n-1)})$, you can interpret your ODE as a linear system of differential equations,
$frac{dv}{dt} = Av$
$v(0) = v_0$
where $A$ is an appropriate constant matrix. Now, you can show that
$v(t) = (I + At + frac{(At)^2}{2!} + ...)v_0$
is a well defined analytic solution of this problem (in fact, the only solution). This infinite sum defines the exponential of a matrix, $e^{At}$.
Example: $y'' - 2y' + y = 0$. Defining $v = (y, y')$, we have
$v' = begin{pmatrix} 0 & 1 \ -1 & 2 end{pmatrix} v$
The characteristic polynomial of this matrix (and of the associated ODE) has only one root, and for that reason the exponential of $At$ will be something of the form
$e^{At} = Qe^tbegin{pmatrix} 1 & 0 \ t & 1 end{pmatrix}Q^{-1}$, where Q is some constant matrix (search for Jordan Decomposition, in the context of the exponential of a matrix). And that's where the $t^j$ will come from :)
answered yesterday
M. Santos
314
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1 Answer
1
active
oldest
votes
1 Answer
1
active
oldest
votes
active
oldest
votes
active
oldest
votes
up vote
1
down vote
accepted
Defining $v = (y, y^{(1)}, ..., y^{(n-1)})$, you can interpret your ODE as a linear system of differential equations,
$frac{dv}{dt} = Av$
$v(0) = v_0$
where $A$ is an appropriate constant matrix. Now, you can show that
$v(t) = (I + At + frac{(At)^2}{2!} + ...)v_0$
is a well defined analytic solution of this problem (in fact, the only solution). This infinite sum defines the exponential of a matrix, $e^{At}$.
Example: $y'' - 2y' + y = 0$. Defining $v = (y, y')$, we have
$v' = begin{pmatrix} 0 & 1 \ -1 & 2 end{pmatrix} v$
The characteristic polynomial of this matrix (and of the associated ODE) has only one root, and for that reason the exponential of $At$ will be something of the form
$e^{At} = Qe^tbegin{pmatrix} 1 & 0 \ t & 1 end{pmatrix}Q^{-1}$, where Q is some constant matrix (search for Jordan Decomposition, in the context of the exponential of a matrix). And that's where the $t^j$ will come from :)
answered yesterday
M. Santos
314
add a comment |
up vote
1
down vote
accepted
Defining $v = (y, y^{(1)}, ..., y^{(n-1)})$, you can interpret your ODE as a linear system of differential equations,
$frac{dv}{dt} = Av$
$v(0) = v_0$
where $A$ is an appropriate constant matrix. Now, you can show that
$v(t) = (I + At + frac{(At)^2}{2!} + ...)v_0$
is a well defined analytic solution of this problem (in fact, the only solution). This infinite sum defines the exponential of a matrix, $e^{At}$.
Example: $y'' - 2y' + y = 0$. Defining $v = (y, y')$, we have
$v' = begin{pmatrix} 0 & 1 \ -1 & 2 end{pmatrix} v$
The characteristic polynomial of this matrix (and of the associated ODE) has only one root, and for that reason the exponential of $At$ will be something of the form
$e^{At} = Qe^tbegin{pmatrix} 1 & 0 \ t & 1 end{pmatrix}Q^{-1}$, where Q is some constant matrix (search for Jordan Decomposition, in the context of the exponential of a matrix). And that's where the $t^j$ will come from :)
answered yesterday
M. Santos
314
add a comment |
up vote
1
down vote
accepted
up vote
1
down vote
accepted
Defining $v = (y, y^{(1)}, ..., y^{(n-1)})$, you can interpret your ODE as a linear system of differential equations,
$frac{dv}{dt} = Av$
$v(0) = v_0$
where $A$ is an appropriate constant matrix. Now, you can show that
$v(t) = (I + At + frac{(At)^2}{2!} + ...)v_0$
is a well defined analytic solution of this problem (in fact, the only solution). This infinite sum defines the exponential of a matrix, $e^{At}$.
Example: $y'' - 2y' + y = 0$. Defining $v = (y, y')$, we have
$v' = begin{pmatrix} 0 & 1 \ -1 & 2 end{pmatrix} v$
The characteristic polynomial of this matrix (and of the associated ODE) has only one root, and for that reason the exponential of $At$ will be something of the form
$e^{At} = Qe^tbegin{pmatrix} 1 & 0 \ t & 1 end{pmatrix}Q^{-1}$, where Q is some constant matrix (search for Jordan Decomposition, in the context of the exponential of a matrix). And that's where the $t^j$ will come from :)
answered yesterday
M. Santos
314
Defining $v = (y, y^{(1)}, ..., y^{(n-1)})$, you can interpret your ODE as a linear system of differential equations,
$frac{dv}{dt} = Av$
$v(0) = v_0$
where $A$ is an appropriate constant matrix. Now, you can show that
$v(t) = (I + At + frac{(At)^2}{2!} + ...)v_0$
is a well defined analytic solution of this problem (in fact, the only solution). This infinite sum defines the exponential of a matrix, $e^{At}$.
Example: $y'' - 2y' + y = 0$. Defining $v = (y, y')$, we have
$v' = begin{pmatrix} 0 & 1 \ -1 & 2 end{pmatrix} v$
The characteristic polynomial of this matrix (and of the associated ODE) has only one root, and for that reason the exponential of $At$ will be something of the form
$e^{At} = Qe^tbegin{pmatrix} 1 & 0 \ t & 1 end{pmatrix}Q^{-1}$, where Q is some constant matrix (search for Jordan Decomposition, in the context of the exponential of a matrix). And that's where the $t^j$ will come from :)
answered yesterday
M. Santos
314
edited yesterday
answered yesterday
M. Santos
314
answered yesterday
M. Santos
314
answered yesterday
M. Santos
314
314
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