A problem of a linear ODE with order $n$
$begingroup$
Is there an algorithm to solve this EDO in $mathbb{R}$?
begin{equation}
x^{(n)}(t) + x^{(n-1)}(t) + cdots + x''(t) + x'(t) + x(t) = 0
end{equation}
Or a more general ODE:
begin{equation}
sum_{k=0}^{n}a_kx^{(k)}(t) = f(t)
end{equation}
With ${a_k}_{kinmathbb{N}}$ a sucession belonging to $mathbb{R}$. I don't how to start.
ordinary-differential-equations
asked Jan 1 at 3:01
El PastaEl Pasta
46015
$endgroup$
add a comment |
$begingroup$
Is there an algorithm to solve this EDO in $mathbb{R}$?
begin{equation}
x^{(n)}(t) + x^{(n-1)}(t) + cdots + x''(t) + x'(t) + x(t) = 0
end{equation}
Or a more general ODE:
begin{equation}
sum_{k=0}^{n}a_kx^{(k)}(t) = f(t)
end{equation}
With ${a_k}_{kinmathbb{N}}$ a sucession belonging to $mathbb{R}$. I don't how to start.
ordinary-differential-equations
asked Jan 1 at 3:01
El PastaEl Pasta
46015
$endgroup$
1
$begingroup$
This is probably the most "standard" $n$-th order ODE. Why is it weird?
$endgroup$
– Arctic Char
Jan 1 at 3:27
$begingroup$
@ArcticChar You are right, fixed
$endgroup$
– El Pasta
Jan 1 at 3:28
add a comment |
$begingroup$
Is there an algorithm to solve this EDO in $mathbb{R}$?
begin{equation}
x^{(n)}(t) + x^{(n-1)}(t) + cdots + x''(t) + x'(t) + x(t) = 0
end{equation}
Or a more general ODE:
begin{equation}
sum_{k=0}^{n}a_kx^{(k)}(t) = f(t)
end{equation}
With ${a_k}_{kinmathbb{N}}$ a sucession belonging to $mathbb{R}$. I don't how to start.
ordinary-differential-equations
asked Jan 1 at 3:01
El PastaEl Pasta
46015
$endgroup$
Is there an algorithm to solve this EDO in $mathbb{R}$?
begin{equation}
x^{(n)}(t) + x^{(n-1)}(t) + cdots + x''(t) + x'(t) + x(t) = 0
end{equation}
Or a more general ODE:
begin{equation}
sum_{k=0}^{n}a_kx^{(k)}(t) = f(t)
end{equation}
With ${a_k}_{kinmathbb{N}}$ a sucession belonging to $mathbb{R}$. I don't how to start.
ordinary-differential-equations
ordinary-differential-equations
asked Jan 1 at 3:01
El PastaEl Pasta
46015
asked Jan 1 at 3:01
El PastaEl Pasta
46015
edited Jan 1 at 3:29
El Pasta
asked Jan 1 at 3:01
El PastaEl Pasta
46015
asked Jan 1 at 3:01
El PastaEl Pasta
46015
asked Jan 1 at 3:01
El PastaEl Pasta
46015
46015
1
$begingroup$
This is probably the most "standard" $n$-th order ODE. Why is it weird?
$endgroup$
– Arctic Char
Jan 1 at 3:27
$begingroup$
@ArcticChar You are right, fixed
$endgroup$
– El Pasta
Jan 1 at 3:28
add a comment |
1
$begingroup$
This is probably the most "standard" $n$-th order ODE. Why is it weird?
$endgroup$
– Arctic Char
Jan 1 at 3:27
$begingroup$
@ArcticChar You are right, fixed
$endgroup$
– El Pasta
Jan 1 at 3:28
1
1
$begingroup$
This is probably the most "standard" $n$-th order ODE. Why is it weird?
$endgroup$
– Arctic Char
Jan 1 at 3:27
$begingroup$
This is probably the most "standard" $n$-th order ODE. Why is it weird?
$endgroup$
– Arctic Char
Jan 1 at 3:27
$begingroup$
@ArcticChar You are right, fixed
$endgroup$
– El Pasta
Jan 1 at 3:28
$begingroup$
@ArcticChar You are right, fixed
$endgroup$
– El Pasta
Jan 1 at 3:28
add a comment |
1 Answer
1
active
oldest
votes
$begingroup$
A common method for linear ODE like this is to assume that the solution takes the form $y(t) = e^{lambda t}$. This will give you a polynomial in $lambda$, the solutions of which are called the eigenvalues. The solution will be something like
$$y(t) = c_1e^{lambda_1 t}+c_2e^{lambda_2 t} + ... + c_ne^{lambda_n t}.$$
For any complex $lambda$, you will have some portion of your solution being sinusoidal. You also need to adjust your solution for any repeated $lambda$.
answered Jan 1 at 3:16
D.B.D.B.
1,2028
$endgroup$
add a comment |
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1 Answer
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1 Answer
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$begingroup$
A common method for linear ODE like this is to assume that the solution takes the form $y(t) = e^{lambda t}$. This will give you a polynomial in $lambda$, the solutions of which are called the eigenvalues. The solution will be something like
$$y(t) = c_1e^{lambda_1 t}+c_2e^{lambda_2 t} + ... + c_ne^{lambda_n t}.$$
For any complex $lambda$, you will have some portion of your solution being sinusoidal. You also need to adjust your solution for any repeated $lambda$.
answered Jan 1 at 3:16
D.B.D.B.
1,2028
$endgroup$
add a comment |
$begingroup$
A common method for linear ODE like this is to assume that the solution takes the form $y(t) = e^{lambda t}$. This will give you a polynomial in $lambda$, the solutions of which are called the eigenvalues. The solution will be something like
$$y(t) = c_1e^{lambda_1 t}+c_2e^{lambda_2 t} + ... + c_ne^{lambda_n t}.$$
For any complex $lambda$, you will have some portion of your solution being sinusoidal. You also need to adjust your solution for any repeated $lambda$.
answered Jan 1 at 3:16
D.B.D.B.
1,2028
$endgroup$
add a comment |
$begingroup$
A common method for linear ODE like this is to assume that the solution takes the form $y(t) = e^{lambda t}$. This will give you a polynomial in $lambda$, the solutions of which are called the eigenvalues. The solution will be something like
$$y(t) = c_1e^{lambda_1 t}+c_2e^{lambda_2 t} + ... + c_ne^{lambda_n t}.$$
For any complex $lambda$, you will have some portion of your solution being sinusoidal. You also need to adjust your solution for any repeated $lambda$.
answered Jan 1 at 3:16
D.B.D.B.
1,2028
$endgroup$
A common method for linear ODE like this is to assume that the solution takes the form $y(t) = e^{lambda t}$. This will give you a polynomial in $lambda$, the solutions of which are called the eigenvalues. The solution will be something like
$$y(t) = c_1e^{lambda_1 t}+c_2e^{lambda_2 t} + ... + c_ne^{lambda_n t}.$$
For any complex $lambda$, you will have some portion of your solution being sinusoidal. You also need to adjust your solution for any repeated $lambda$.
answered Jan 1 at 3:16
D.B.D.B.
1,2028
answered Jan 1 at 3:16
D.B.D.B.
1,2028
answered Jan 1 at 3:16
D.B.D.B.
1,2028
answered Jan 1 at 3:16
D.B.D.B.
1,2028
1,2028
add a comment |
add a comment |
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1
$begingroup$
This is probably the most "standard" $n$-th order ODE. Why is it weird?
$endgroup$
– Arctic Char
Jan 1 at 3:27
$begingroup$
@ArcticChar You are right, fixed
$endgroup$
– El Pasta
Jan 1 at 3:28