How to find ODE , with constant-coefficient ,homogeneous , with minmum power such that $y_1(x) = xe^x$, $...
$begingroup$
Find ODE , with constant-coefficient ,homogeneous , with minmum power,linear such that $y_1(x) = xe^x$, $ y_2(x) = sin^2(x)$ is soultions for it.
how can one construct such ODE ?
my trial :
by finding the roots of the unique polynomal for the Homogeneous ODE :
since $y_1(x) = xe^x$ is solution $rightarrow$ $(r-1)(r-1)$
since $ y_2(x) = sin^2(x)$ is solution we rewrite as $ y_2(x) = frac{1-cos(2x)}{2}$ so $ 1 , cos(2x)$ is solutions .
we then write $(r)(r^2 + 4)$ so our polynomal is :
$F(r) = (r-1)(r-1)(r)(r^2+4)$ after simplfying :
$F(r) = r^5 - 2r^4 + 5r^3 - 8r^2 + 4r$
so $ y^{(5)}-2y^{(4)}+5y^{'''}-8y^{''}+4y^{'} = 0 $ is our ODE.
not sure if the solution is right , also not sure if its the minimum powers. much thanks for help
real-analysis ordinary-differential-equations polynomials
asked Jan 2 at 18:59
Mather Mather
3888
$endgroup$
|
show 2 more comments
$begingroup$
Find ODE , with constant-coefficient ,homogeneous , with minmum power,linear such that $y_1(x) = xe^x$, $ y_2(x) = sin^2(x)$ is soultions for it.
how can one construct such ODE ?
my trial :
by finding the roots of the unique polynomal for the Homogeneous ODE :
since $y_1(x) = xe^x$ is solution $rightarrow$ $(r-1)(r-1)$
since $ y_2(x) = sin^2(x)$ is solution we rewrite as $ y_2(x) = frac{1-cos(2x)}{2}$ so $ 1 , cos(2x)$ is solutions .
we then write $(r)(r^2 + 4)$ so our polynomal is :
$F(r) = (r-1)(r-1)(r)(r^2+4)$ after simplfying :
$F(r) = r^5 - 2r^4 + 5r^3 - 8r^2 + 4r$
so $ y^{(5)}-2y^{(4)}+5y^{'''}-8y^{''}+4y^{'} = 0 $ is our ODE.
not sure if the solution is right , also not sure if its the minimum powers. much thanks for help
real-analysis ordinary-differential-equations polynomials
asked Jan 2 at 18:59
Mather Mather
3888
$endgroup$
$begingroup$
That's the correct answer...
$endgroup$
– Mostafa Ayaz
Jan 2 at 19:53
1
$begingroup$
You need to word it a little differently to transform it into a rigorous answer. Start with assuming the minimal equation has characteristic polynomial $p(r).$ Then, show that $p(r)$ must be divisible by the two polynomials you have found and this would conclude the proof.
$endgroup$
– dezdichado
Jan 2 at 19:56
$begingroup$
thank you both , but how can one be sure that this ODE has minimum powers of $ ''''' $
$endgroup$
– Mather
Jan 2 at 19:58
1
$begingroup$
Maybe by showing that all solutions are linearly independent of each other? So you can conclude that you actually need all your factors since you cannot write of of these in terms of another one.
$endgroup$
– mrtaurho
Jan 2 at 20:01
$begingroup$
It's more of an algebra concept. $mathbb{Z}[x]$ is a Unique Factorization Domain, which basically means that if a polynomial is divisible by bunch of other polynomials that are pairwise - relatively prime, then it has to be divisible by the product of all divisors. But if this is ODE class then I doubt you need to provide proof of this fact. Just mention it.
$endgroup$
– dezdichado
Jan 2 at 20:06
|
show 2 more comments
$begingroup$
Find ODE , with constant-coefficient ,homogeneous , with minmum power,linear such that $y_1(x) = xe^x$, $ y_2(x) = sin^2(x)$ is soultions for it.
how can one construct such ODE ?
my trial :
by finding the roots of the unique polynomal for the Homogeneous ODE :
since $y_1(x) = xe^x$ is solution $rightarrow$ $(r-1)(r-1)$
since $ y_2(x) = sin^2(x)$ is solution we rewrite as $ y_2(x) = frac{1-cos(2x)}{2}$ so $ 1 , cos(2x)$ is solutions .
we then write $(r)(r^2 + 4)$ so our polynomal is :
$F(r) = (r-1)(r-1)(r)(r^2+4)$ after simplfying :
$F(r) = r^5 - 2r^4 + 5r^3 - 8r^2 + 4r$
so $ y^{(5)}-2y^{(4)}+5y^{'''}-8y^{''}+4y^{'} = 0 $ is our ODE.
not sure if the solution is right , also not sure if its the minimum powers. much thanks for help
real-analysis ordinary-differential-equations polynomials
asked Jan 2 at 18:59
Mather Mather
3888
$endgroup$
Find ODE , with constant-coefficient ,homogeneous , with minmum power,linear such that $y_1(x) = xe^x$, $ y_2(x) = sin^2(x)$ is soultions for it.
how can one construct such ODE ?
my trial :
by finding the roots of the unique polynomal for the Homogeneous ODE :
since $y_1(x) = xe^x$ is solution $rightarrow$ $(r-1)(r-1)$
since $ y_2(x) = sin^2(x)$ is solution we rewrite as $ y_2(x) = frac{1-cos(2x)}{2}$ so $ 1 , cos(2x)$ is solutions .
we then write $(r)(r^2 + 4)$ so our polynomal is :
$F(r) = (r-1)(r-1)(r)(r^2+4)$ after simplfying :
$F(r) = r^5 - 2r^4 + 5r^3 - 8r^2 + 4r$
so $ y^{(5)}-2y^{(4)}+5y^{'''}-8y^{''}+4y^{'} = 0 $ is our ODE.
not sure if the solution is right , also not sure if its the minimum powers. much thanks for help
real-analysis ordinary-differential-equations polynomials
real-analysis ordinary-differential-equations polynomials
asked Jan 2 at 18:59
Mather Mather
3888
asked Jan 2 at 18:59
Mather Mather
3888
edited Jan 2 at 19:49
Mather
asked Jan 2 at 18:59
Mather Mather
3888
asked Jan 2 at 18:59
Mather Mather
3888
asked Jan 2 at 18:59
Mather Mather
3888
3888
$begingroup$
That's the correct answer...
$endgroup$
– Mostafa Ayaz
Jan 2 at 19:53
1
$begingroup$
You need to word it a little differently to transform it into a rigorous answer. Start with assuming the minimal equation has characteristic polynomial $p(r).$ Then, show that $p(r)$ must be divisible by the two polynomials you have found and this would conclude the proof.
$endgroup$
– dezdichado
Jan 2 at 19:56
$begingroup$
thank you both , but how can one be sure that this ODE has minimum powers of $ ''''' $
$endgroup$
– Mather
Jan 2 at 19:58
1
$begingroup$
Maybe by showing that all solutions are linearly independent of each other? So you can conclude that you actually need all your factors since you cannot write of of these in terms of another one.
$endgroup$
– mrtaurho
Jan 2 at 20:01
$begingroup$
It's more of an algebra concept. $mathbb{Z}[x]$ is a Unique Factorization Domain, which basically means that if a polynomial is divisible by bunch of other polynomials that are pairwise - relatively prime, then it has to be divisible by the product of all divisors. But if this is ODE class then I doubt you need to provide proof of this fact. Just mention it.
$endgroup$
– dezdichado
Jan 2 at 20:06
|
show 2 more comments
$begingroup$
That's the correct answer...
$endgroup$
– Mostafa Ayaz
Jan 2 at 19:53
1
$begingroup$
You need to word it a little differently to transform it into a rigorous answer. Start with assuming the minimal equation has characteristic polynomial $p(r).$ Then, show that $p(r)$ must be divisible by the two polynomials you have found and this would conclude the proof.
$endgroup$
– dezdichado
Jan 2 at 19:56
$begingroup$
thank you both , but how can one be sure that this ODE has minimum powers of $ ''''' $
$endgroup$
– Mather
Jan 2 at 19:58
1
$begingroup$
Maybe by showing that all solutions are linearly independent of each other? So you can conclude that you actually need all your factors since you cannot write of of these in terms of another one.
$endgroup$
– mrtaurho
Jan 2 at 20:01
$begingroup$
It's more of an algebra concept. $mathbb{Z}[x]$ is a Unique Factorization Domain, which basically means that if a polynomial is divisible by bunch of other polynomials that are pairwise - relatively prime, then it has to be divisible by the product of all divisors. But if this is ODE class then I doubt you need to provide proof of this fact. Just mention it.
$endgroup$
– dezdichado
Jan 2 at 20:06
$begingroup$
That's the correct answer...
$endgroup$
– Mostafa Ayaz
Jan 2 at 19:53
$begingroup$
That's the correct answer...
$endgroup$
– Mostafa Ayaz
Jan 2 at 19:53
1
1
$begingroup$
You need to word it a little differently to transform it into a rigorous answer. Start with assuming the minimal equation has characteristic polynomial $p(r).$ Then, show that $p(r)$ must be divisible by the two polynomials you have found and this would conclude the proof.
$endgroup$
– dezdichado
Jan 2 at 19:56
$begingroup$
You need to word it a little differently to transform it into a rigorous answer. Start with assuming the minimal equation has characteristic polynomial $p(r).$ Then, show that $p(r)$ must be divisible by the two polynomials you have found and this would conclude the proof.
$endgroup$
– dezdichado
Jan 2 at 19:56
$begingroup$
thank you both , but how can one be sure that this ODE has minimum powers of $ ''''' $
$endgroup$
– Mather
Jan 2 at 19:58
$begingroup$
thank you both , but how can one be sure that this ODE has minimum powers of $ ''''' $
$endgroup$
– Mather
Jan 2 at 19:58
1
1
$begingroup$
Maybe by showing that all solutions are linearly independent of each other? So you can conclude that you actually need all your factors since you cannot write of of these in terms of another one.
$endgroup$
– mrtaurho
Jan 2 at 20:01
$begingroup$
Maybe by showing that all solutions are linearly independent of each other? So you can conclude that you actually need all your factors since you cannot write of of these in terms of another one.
$endgroup$
– mrtaurho
Jan 2 at 20:01
$begingroup$
It's more of an algebra concept. $mathbb{Z}[x]$ is a Unique Factorization Domain, which basically means that if a polynomial is divisible by bunch of other polynomials that are pairwise - relatively prime, then it has to be divisible by the product of all divisors. But if this is ODE class then I doubt you need to provide proof of this fact. Just mention it.
$endgroup$
– dezdichado
Jan 2 at 20:06
$begingroup$
It's more of an algebra concept. $mathbb{Z}[x]$ is a Unique Factorization Domain, which basically means that if a polynomial is divisible by bunch of other polynomials that are pairwise - relatively prime, then it has to be divisible by the product of all divisors. But if this is ODE class then I doubt you need to provide proof of this fact. Just mention it.
$endgroup$
– dezdichado
Jan 2 at 20:06
|
show 2 more comments
1 Answer
1
active
oldest
votes
$begingroup$
Assuming $mathcal{D}$ linear we have
$$
mathcal{D}y_1 = 0\
mathcal{D}y_2 = 0
$$
and
$$
mathcal{D}(y_1+y_2) = mathcal{D}y_1 +mathcal{D}y_2 = 0
$$
then $y = y_1+y_2$ satisfies as well the DE. Considering now the more general problem with $mathcal{D}$ invertible
$$
mathcal{D}(y) = uRightarrow y = mathcal{D}^{-1}u
$$
or using the Laplace transform
$$
Y(s) = mathcal{L}left(xe^x+sin^2 xright) = frac{2}{s left(s^2+4right)}+frac{1}{(s-1)^2} = frac{s^3+2 s^2+2}{(s-1)^2 s left(s^2+4right)}=G(s) U(s)
$$
making now $u = delta(t)$ associated to initial conditions we have
$$
G(s) = mathcal{L}left(mathcal{D}^{-1}right)
$$
hence
$$
mathcal{D} = partial^{(5)}_t-2partial^{(4)}_t+5partial^{(3)}_t-8partial^{(2)}_t+4partial_t
$$
so your answer is correct.
answered Jan 3 at 12:24
CesareoCesareo
9,2213517
$endgroup$
add a comment |
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1 Answer
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active
oldest
votes
1 Answer
1
active
oldest
votes
active
oldest
votes
active
oldest
votes
$begingroup$
Assuming $mathcal{D}$ linear we have
$$
mathcal{D}y_1 = 0\
mathcal{D}y_2 = 0
$$
and
$$
mathcal{D}(y_1+y_2) = mathcal{D}y_1 +mathcal{D}y_2 = 0
$$
then $y = y_1+y_2$ satisfies as well the DE. Considering now the more general problem with $mathcal{D}$ invertible
$$
mathcal{D}(y) = uRightarrow y = mathcal{D}^{-1}u
$$
or using the Laplace transform
$$
Y(s) = mathcal{L}left(xe^x+sin^2 xright) = frac{2}{s left(s^2+4right)}+frac{1}{(s-1)^2} = frac{s^3+2 s^2+2}{(s-1)^2 s left(s^2+4right)}=G(s) U(s)
$$
making now $u = delta(t)$ associated to initial conditions we have
$$
G(s) = mathcal{L}left(mathcal{D}^{-1}right)
$$
hence
$$
mathcal{D} = partial^{(5)}_t-2partial^{(4)}_t+5partial^{(3)}_t-8partial^{(2)}_t+4partial_t
$$
so your answer is correct.
answered Jan 3 at 12:24
CesareoCesareo
9,2213517
$endgroup$
add a comment |
$begingroup$
Assuming $mathcal{D}$ linear we have
$$
mathcal{D}y_1 = 0\
mathcal{D}y_2 = 0
$$
and
$$
mathcal{D}(y_1+y_2) = mathcal{D}y_1 +mathcal{D}y_2 = 0
$$
then $y = y_1+y_2$ satisfies as well the DE. Considering now the more general problem with $mathcal{D}$ invertible
$$
mathcal{D}(y) = uRightarrow y = mathcal{D}^{-1}u
$$
or using the Laplace transform
$$
Y(s) = mathcal{L}left(xe^x+sin^2 xright) = frac{2}{s left(s^2+4right)}+frac{1}{(s-1)^2} = frac{s^3+2 s^2+2}{(s-1)^2 s left(s^2+4right)}=G(s) U(s)
$$
making now $u = delta(t)$ associated to initial conditions we have
$$
G(s) = mathcal{L}left(mathcal{D}^{-1}right)
$$
hence
$$
mathcal{D} = partial^{(5)}_t-2partial^{(4)}_t+5partial^{(3)}_t-8partial^{(2)}_t+4partial_t
$$
so your answer is correct.
answered Jan 3 at 12:24
CesareoCesareo
9,2213517
$endgroup$
add a comment |
$begingroup$
Assuming $mathcal{D}$ linear we have
$$
mathcal{D}y_1 = 0\
mathcal{D}y_2 = 0
$$
and
$$
mathcal{D}(y_1+y_2) = mathcal{D}y_1 +mathcal{D}y_2 = 0
$$
then $y = y_1+y_2$ satisfies as well the DE. Considering now the more general problem with $mathcal{D}$ invertible
$$
mathcal{D}(y) = uRightarrow y = mathcal{D}^{-1}u
$$
or using the Laplace transform
$$
Y(s) = mathcal{L}left(xe^x+sin^2 xright) = frac{2}{s left(s^2+4right)}+frac{1}{(s-1)^2} = frac{s^3+2 s^2+2}{(s-1)^2 s left(s^2+4right)}=G(s) U(s)
$$
making now $u = delta(t)$ associated to initial conditions we have
$$
G(s) = mathcal{L}left(mathcal{D}^{-1}right)
$$
hence
$$
mathcal{D} = partial^{(5)}_t-2partial^{(4)}_t+5partial^{(3)}_t-8partial^{(2)}_t+4partial_t
$$
so your answer is correct.
answered Jan 3 at 12:24
CesareoCesareo
9,2213517
$endgroup$
Assuming $mathcal{D}$ linear we have
$$
mathcal{D}y_1 = 0\
mathcal{D}y_2 = 0
$$
and
$$
mathcal{D}(y_1+y_2) = mathcal{D}y_1 +mathcal{D}y_2 = 0
$$
then $y = y_1+y_2$ satisfies as well the DE. Considering now the more general problem with $mathcal{D}$ invertible
$$
mathcal{D}(y) = uRightarrow y = mathcal{D}^{-1}u
$$
or using the Laplace transform
$$
Y(s) = mathcal{L}left(xe^x+sin^2 xright) = frac{2}{s left(s^2+4right)}+frac{1}{(s-1)^2} = frac{s^3+2 s^2+2}{(s-1)^2 s left(s^2+4right)}=G(s) U(s)
$$
making now $u = delta(t)$ associated to initial conditions we have
$$
G(s) = mathcal{L}left(mathcal{D}^{-1}right)
$$
hence
$$
mathcal{D} = partial^{(5)}_t-2partial^{(4)}_t+5partial^{(3)}_t-8partial^{(2)}_t+4partial_t
$$
so your answer is correct.
answered Jan 3 at 12:24
CesareoCesareo
9,2213517
edited Jan 3 at 14:04
answered Jan 3 at 12:24
CesareoCesareo
9,2213517
answered Jan 3 at 12:24
CesareoCesareo
9,2213517
answered Jan 3 at 12:24
CesareoCesareo
9,2213517
9,2213517
add a comment |
add a comment |
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$begingroup$
That's the correct answer...
$endgroup$
– Mostafa Ayaz
Jan 2 at 19:53
1
$begingroup$
You need to word it a little differently to transform it into a rigorous answer. Start with assuming the minimal equation has characteristic polynomial $p(r).$ Then, show that $p(r)$ must be divisible by the two polynomials you have found and this would conclude the proof.
$endgroup$
– dezdichado
Jan 2 at 19:56
$begingroup$
thank you both , but how can one be sure that this ODE has minimum powers of $ ''''' $
$endgroup$
– Mather
Jan 2 at 19:58
1
$begingroup$
Maybe by showing that all solutions are linearly independent of each other? So you can conclude that you actually need all your factors since you cannot write of of these in terms of another one.
$endgroup$
– mrtaurho
Jan 2 at 20:01
$begingroup$
It's more of an algebra concept. $mathbb{Z}[x]$ is a Unique Factorization Domain, which basically means that if a polynomial is divisible by bunch of other polynomials that are pairwise - relatively prime, then it has to be divisible by the product of all divisors. But if this is ODE class then I doubt you need to provide proof of this fact. Just mention it.
$endgroup$
– dezdichado
Jan 2 at 20:06