Mixing
Differential Polynomials(?)
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1
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Consider an equation of the form:
cy"+cy'+cy
Or something of the form. Essentially, it's a polynomial but instead of powers, there are derivatives. Do these kind of things have a name? Or are they completely useless?
Note: I KNOW what Taylor Polynomials and the like are, but I mean something in the form of what I have shown.
calculus differential-equations polynomials taylor-expansion
asked Aug 6 at 1:53
Shadow Sniper
85
add a comment |Â
up vote
1
down vote
favorite
Consider an equation of the form:
cy"+cy'+cy
Or something of the form. Essentially, it's a polynomial but instead of powers, there are derivatives. Do these kind of things have a name? Or are they completely useless?
Note: I KNOW what Taylor Polynomials and the like are, but I mean something in the form of what I have shown.
calculus differential-equations polynomials taylor-expansion
asked Aug 6 at 1:53
Shadow Sniper
85
2
"cy"+cy'+cy" is not an equation. An equation is when you have some expression and another and an equals sign in between.
– vrugtehagel
Aug 6 at 2:06
You get what I mean...
– Shadow Sniper
Aug 6 at 12:03
I get what you mean. That's why I commented, so you can correct the post to say what you mean
– vrugtehagel
Aug 6 at 22:37
add a comment |Â
up vote
1
down vote
favorite
up vote
1
down vote
favorite
Consider an equation of the form:
cy"+cy'+cy
Or something of the form. Essentially, it's a polynomial but instead of powers, there are derivatives. Do these kind of things have a name? Or are they completely useless?
Note: I KNOW what Taylor Polynomials and the like are, but I mean something in the form of what I have shown.
calculus differential-equations polynomials taylor-expansion
asked Aug 6 at 1:53
Shadow Sniper
85
Consider an equation of the form:
cy"+cy'+cy
Or something of the form. Essentially, it's a polynomial but instead of powers, there are derivatives. Do these kind of things have a name? Or are they completely useless?
Note: I KNOW what Taylor Polynomials and the like are, but I mean something in the form of what I have shown.
calculus differential-equations polynomials taylor-expansion
asked Aug 6 at 1:53
Shadow Sniper
85
asked Aug 6 at 1:53
Shadow Sniper
85
asked Aug 6 at 1:53
Shadow Sniper
85
asked Aug 6 at 1:53
Shadow Sniper
85
85
2
"cy"+cy'+cy" is not an equation. An equation is when you have some expression and another and an equals sign in between.
– vrugtehagel
Aug 6 at 2:06
You get what I mean...
– Shadow Sniper
Aug 6 at 12:03
I get what you mean. That's why I commented, so you can correct the post to say what you mean
– vrugtehagel
Aug 6 at 22:37
add a comment |Â
2
"cy"+cy'+cy" is not an equation. An equation is when you have some expression and another and an equals sign in between.
– vrugtehagel
Aug 6 at 2:06
You get what I mean...
– Shadow Sniper
Aug 6 at 12:03
I get what you mean. That's why I commented, so you can correct the post to say what you mean
– vrugtehagel
Aug 6 at 22:37
2
2
"cy"+cy'+cy" is not an equation. An equation is when you have some expression and another and an equals sign in between.
– vrugtehagel
Aug 6 at 2:06
"cy"+cy'+cy" is not an equation. An equation is when you have some expression and another and an equals sign in between.
– vrugtehagel
Aug 6 at 2:06
You get what I mean...
– Shadow Sniper
Aug 6 at 12:03
You get what I mean...
– Shadow Sniper
Aug 6 at 12:03
I get what you mean. That's why I commented, so you can correct the post to say what you mean
– vrugtehagel
Aug 6 at 22:37
I get what you mean. That's why I commented, so you can correct the post to say what you mean
– vrugtehagel
Aug 6 at 22:37
add a comment |Â
3 Answers
3
active
oldest
votes
up vote
0
down vote
accepted
There is a rich study of so-called "differential algebra."
https://en.wikipedia.org/wiki/Differential_algebra
However, what you're realizing is the connection between linear algebra and differential equations.
Namely, if $D: C^infty(X) longrightarrow C^infty(X)$ is the derivative operator, and $1$ represents the identity map, then the vectors $y$ which satisfy the following polynomial equation
$$(a_n D^n + cdots+ a_1D + a_01)y=0$$
are said to be solutions to the differential equation
$$a_n y^(n) + cdots+ a_2y''+ a_1y' + a_0y=0.$$
The missing connection would be the Cayley Hamilton Theorem. Which would say that if $T:Vlongrightarrow V$ is a linear operator, with characteristic equation
$$a_n lambda^n + cdots+ a_1lambda + a_0=0$$
Then $T$ satisfies this characteristic equation
$$a_n T^nv + cdots+ a_1Tv + a_0v=0$$
for all $vin V.$
answered Aug 6 at 1:55
Chickenmancer
3,017622
add a comment |Â
up vote
1
down vote
$$L= a_0(x) + a_1(x)fracddx+ ldots + a_n(x) fracd^ndx^n $$
is known as a linear differential operator.
We have
$$Ly= a_0(x)y + a_1(x)fracdydx+ ldots + a_n(x) fracd^nydx^n $$
$$a_0(x)y+a_1(x)y'+ldots a_n(x)y^(n)+b(x)=0$$ is a linear differential equation.
when the $a_i(x)$ is independent of $x$, we describe them as constant coefficients.
answered Aug 6 at 2:15
Siong Thye Goh
78.1k134997
add a comment |Â
up vote
0
down vote
Yes the polynomial associated to $$ ay'' + by' +cy =0$$ is $$P(lambda )= a lambda ^2 + b lambda +c$$ which is called the charateristic polynomial.
This polynomial plays a very important role in finding the solutions to your differential equation.
The genera solution to the differential equation is $$ y=C_1 e^lambda _1 +C_2 e^lambda _2 $$ where $lambda _1$ and $lambda _2$ are solutions to $P(lambda)$
answered Aug 6 at 2:13
Mohammad Riazi-Kermani
27.8k41852
add a comment |Â
3 Answers
3
active
oldest
votes
3 Answers
3
active
oldest
votes
active
oldest
votes
active
oldest
votes
up vote
0
down vote
accepted
There is a rich study of so-called "differential algebra."
https://en.wikipedia.org/wiki/Differential_algebra
However, what you're realizing is the connection between linear algebra and differential equations.
Namely, if $D: C^infty(X) longrightarrow C^infty(X)$ is the derivative operator, and $1$ represents the identity map, then the vectors $y$ which satisfy the following polynomial equation
$$(a_n D^n + cdots+ a_1D + a_01)y=0$$
are said to be solutions to the differential equation
$$a_n y^(n) + cdots+ a_2y''+ a_1y' + a_0y=0.$$
The missing connection would be the Cayley Hamilton Theorem. Which would say that if $T:Vlongrightarrow V$ is a linear operator, with characteristic equation
$$a_n lambda^n + cdots+ a_1lambda + a_0=0$$
Then $T$ satisfies this characteristic equation
$$a_n T^nv + cdots+ a_1Tv + a_0v=0$$
for all $vin V.$
answered Aug 6 at 1:55
Chickenmancer
3,017622
add a comment |Â
up vote
0
down vote
accepted
There is a rich study of so-called "differential algebra."
https://en.wikipedia.org/wiki/Differential_algebra
However, what you're realizing is the connection between linear algebra and differential equations.
Namely, if $D: C^infty(X) longrightarrow C^infty(X)$ is the derivative operator, and $1$ represents the identity map, then the vectors $y$ which satisfy the following polynomial equation
$$(a_n D^n + cdots+ a_1D + a_01)y=0$$
are said to be solutions to the differential equation
$$a_n y^(n) + cdots+ a_2y''+ a_1y' + a_0y=0.$$
The missing connection would be the Cayley Hamilton Theorem. Which would say that if $T:Vlongrightarrow V$ is a linear operator, with characteristic equation
$$a_n lambda^n + cdots+ a_1lambda + a_0=0$$
Then $T$ satisfies this characteristic equation
$$a_n T^nv + cdots+ a_1Tv + a_0v=0$$
for all $vin V.$
answered Aug 6 at 1:55
Chickenmancer
3,017622
add a comment |Â
up vote
0
down vote
accepted
up vote
0
down vote
accepted
There is a rich study of so-called "differential algebra."
https://en.wikipedia.org/wiki/Differential_algebra
However, what you're realizing is the connection between linear algebra and differential equations.
Namely, if $D: C^infty(X) longrightarrow C^infty(X)$ is the derivative operator, and $1$ represents the identity map, then the vectors $y$ which satisfy the following polynomial equation
$$(a_n D^n + cdots+ a_1D + a_01)y=0$$
are said to be solutions to the differential equation
$$a_n y^(n) + cdots+ a_2y''+ a_1y' + a_0y=0.$$
The missing connection would be the Cayley Hamilton Theorem. Which would say that if $T:Vlongrightarrow V$ is a linear operator, with characteristic equation
$$a_n lambda^n + cdots+ a_1lambda + a_0=0$$
Then $T$ satisfies this characteristic equation
$$a_n T^nv + cdots+ a_1Tv + a_0v=0$$
for all $vin V.$
answered Aug 6 at 1:55
Chickenmancer
3,017622
There is a rich study of so-called "differential algebra."
https://en.wikipedia.org/wiki/Differential_algebra
However, what you're realizing is the connection between linear algebra and differential equations.
Namely, if $D: C^infty(X) longrightarrow C^infty(X)$ is the derivative operator, and $1$ represents the identity map, then the vectors $y$ which satisfy the following polynomial equation
$$(a_n D^n + cdots+ a_1D + a_01)y=0$$
are said to be solutions to the differential equation
$$a_n y^(n) + cdots+ a_2y''+ a_1y' + a_0y=0.$$
The missing connection would be the Cayley Hamilton Theorem. Which would say that if $T:Vlongrightarrow V$ is a linear operator, with characteristic equation
$$a_n lambda^n + cdots+ a_1lambda + a_0=0$$
Then $T$ satisfies this characteristic equation
$$a_n T^nv + cdots+ a_1Tv + a_0v=0$$
for all $vin V.$
answered Aug 6 at 1:55
Chickenmancer
3,017622
edited Aug 6 at 2:13
answered Aug 6 at 1:55
Chickenmancer
3,017622
answered Aug 6 at 1:55
Chickenmancer
3,017622
answered Aug 6 at 1:55
Chickenmancer
3,017622
3,017622
add a comment |Â
add a comment |Â
up vote
1
down vote
$$L= a_0(x) + a_1(x)fracddx+ ldots + a_n(x) fracd^ndx^n $$
is known as a linear differential operator.
We have
$$Ly= a_0(x)y + a_1(x)fracdydx+ ldots + a_n(x) fracd^nydx^n $$
$$a_0(x)y+a_1(x)y'+ldots a_n(x)y^(n)+b(x)=0$$ is a linear differential equation.
when the $a_i(x)$ is independent of $x$, we describe them as constant coefficients.
answered Aug 6 at 2:15
Siong Thye Goh
78.1k134997
add a comment |Â
up vote
1
down vote
$$L= a_0(x) + a_1(x)fracddx+ ldots + a_n(x) fracd^ndx^n $$
is known as a linear differential operator.
We have
$$Ly= a_0(x)y + a_1(x)fracdydx+ ldots + a_n(x) fracd^nydx^n $$
$$a_0(x)y+a_1(x)y'+ldots a_n(x)y^(n)+b(x)=0$$ is a linear differential equation.
when the $a_i(x)$ is independent of $x$, we describe them as constant coefficients.
answered Aug 6 at 2:15
Siong Thye Goh
78.1k134997
add a comment |Â
up vote
1
down vote
up vote
1
down vote
$$L= a_0(x) + a_1(x)fracddx+ ldots + a_n(x) fracd^ndx^n $$
is known as a linear differential operator.
We have
$$Ly= a_0(x)y + a_1(x)fracdydx+ ldots + a_n(x) fracd^nydx^n $$
$$a_0(x)y+a_1(x)y'+ldots a_n(x)y^(n)+b(x)=0$$ is a linear differential equation.
when the $a_i(x)$ is independent of $x$, we describe them as constant coefficients.
answered Aug 6 at 2:15
Siong Thye Goh
78.1k134997
$$L= a_0(x) + a_1(x)fracddx+ ldots + a_n(x) fracd^ndx^n $$
is known as a linear differential operator.
We have
$$Ly= a_0(x)y + a_1(x)fracdydx+ ldots + a_n(x) fracd^nydx^n $$
$$a_0(x)y+a_1(x)y'+ldots a_n(x)y^(n)+b(x)=0$$ is a linear differential equation.
when the $a_i(x)$ is independent of $x$, we describe them as constant coefficients.
answered Aug 6 at 2:15
Siong Thye Goh
78.1k134997
edited Aug 6 at 2:30
answered Aug 6 at 2:15
Siong Thye Goh
78.1k134997
answered Aug 6 at 2:15
Siong Thye Goh
78.1k134997
answered Aug 6 at 2:15
Siong Thye Goh
78.1k134997
78.1k134997
add a comment |Â
add a comment |Â
up vote
0
down vote
Yes the polynomial associated to $$ ay'' + by' +cy =0$$ is $$P(lambda )= a lambda ^2 + b lambda +c$$ which is called the charateristic polynomial.
This polynomial plays a very important role in finding the solutions to your differential equation.
The genera solution to the differential equation is $$ y=C_1 e^lambda _1 +C_2 e^lambda _2 $$ where $lambda _1$ and $lambda _2$ are solutions to $P(lambda)$
answered Aug 6 at 2:13
Mohammad Riazi-Kermani
27.8k41852
add a comment |Â
up vote
0
down vote
Yes the polynomial associated to $$ ay'' + by' +cy =0$$ is $$P(lambda )= a lambda ^2 + b lambda +c$$ which is called the charateristic polynomial.
This polynomial plays a very important role in finding the solutions to your differential equation.
The genera solution to the differential equation is $$ y=C_1 e^lambda _1 +C_2 e^lambda _2 $$ where $lambda _1$ and $lambda _2$ are solutions to $P(lambda)$
answered Aug 6 at 2:13
Mohammad Riazi-Kermani
27.8k41852
add a comment |Â
up vote
0
down vote
up vote
0
down vote
Yes the polynomial associated to $$ ay'' + by' +cy =0$$ is $$P(lambda )= a lambda ^2 + b lambda +c$$ which is called the charateristic polynomial.
This polynomial plays a very important role in finding the solutions to your differential equation.
The genera solution to the differential equation is $$ y=C_1 e^lambda _1 +C_2 e^lambda _2 $$ where $lambda _1$ and $lambda _2$ are solutions to $P(lambda)$
answered Aug 6 at 2:13
Mohammad Riazi-Kermani
27.8k41852
Yes the polynomial associated to $$ ay'' + by' +cy =0$$ is $$P(lambda )= a lambda ^2 + b lambda +c$$ which is called the charateristic polynomial.
This polynomial plays a very important role in finding the solutions to your differential equation.
The genera solution to the differential equation is $$ y=C_1 e^lambda _1 +C_2 e^lambda _2 $$ where $lambda _1$ and $lambda _2$ are solutions to $P(lambda)$
answered Aug 6 at 2:13
Mohammad Riazi-Kermani
27.8k41852
answered Aug 6 at 2:13
Mohammad Riazi-Kermani
27.8k41852
answered Aug 6 at 2:13
Mohammad Riazi-Kermani
27.8k41852
answered Aug 6 at 2:13
Mohammad Riazi-Kermani
27.8k41852
27.8k41852
add a comment |Â
add a comment |Â
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2
"cy"+cy'+cy" is not an equation. An equation is when you have some expression and another and an equals sign in between.
– vrugtehagel
Aug 6 at 2:06
You get what I mean...
– Shadow Sniper
Aug 6 at 12:03
I get what you mean. That's why I commented, so you can correct the post to say what you mean
– vrugtehagel
Aug 6 at 22:37