How to solve this 4th order ODE with polynomial coefficients?
$begingroup$
Write the general solution
for $ (x^2) y'''' + (3x^2-2x)y''' + (3x^2-4x+2)y'' +(x^2-2x+2)y'
= 0 $
I tried to guess a solution and use the fact that i can decrease the ODE to less power ( to $y'''$ ) by using the Wronskian.
I guessed that $ e^{-x} $ is a solution.
Is it the way we solve this kind of equations? It's a homework question so i guess (i / you) can solve it.
Euler ODE doesn't work here .
Can you help me with the solutions i got 4 solutions : ( after solving Euler equation and moving to $ v''' $ .
$y'(x) = { e^{-x} ,~ frac{x^3}{3}e^{-x} ,~ frac{x^2}{2}e^{-x} }$ or any linear combination of those, uniqueness theorem doesn't apply here near $x=0$
ordinary-differential-equations
asked Jan 2 at 13:37
Mather Mather
3047
$endgroup$
|
show 1 more comment
$begingroup$
Write the general solution
for $ (x^2) y'''' + (3x^2-2x)y''' + (3x^2-4x+2)y'' +(x^2-2x+2)y'
= 0 $
I tried to guess a solution and use the fact that i can decrease the ODE to less power ( to $y'''$ ) by using the Wronskian.
I guessed that $ e^{-x} $ is a solution.
Is it the way we solve this kind of equations? It's a homework question so i guess (i / you) can solve it.
Euler ODE doesn't work here .
Can you help me with the solutions i got 4 solutions : ( after solving Euler equation and moving to $ v''' $ .
$y'(x) = { e^{-x} ,~ frac{x^3}{3}e^{-x} ,~ frac{x^2}{2}e^{-x} }$ or any linear combination of those, uniqueness theorem doesn't apply here near $x=0$
ordinary-differential-equations
asked Jan 2 at 13:37
Mather Mather
3047
$endgroup$
$begingroup$
Your coefficients are polynomials, so it might be reasonable to assume $y$ takes the form of a polynomial. Else, if you have found a solution (say $y_{1} = e^{-x}$), then you can search for another solution of the form $y_{2} = v(x)e^{-x}$.
$endgroup$
– Mattos
Jan 2 at 13:41
$begingroup$
Solution $0$ won't help. But the constant $1$ is also a solution, and is linearly independent of the others, so it will help.
$endgroup$
– GEdgar
Jan 2 at 15:04
$begingroup$
true i will integrate all and get a linear Space of those but how can i know that the answer is right its impossible to start puting each one at a time
$endgroup$
– Mather
Jan 2 at 15:08
$begingroup$
not possible to integrate some of them like $ frac{x^3}{3} e^{-x} $
$endgroup$
– Mather
Jan 2 at 15:10
$begingroup$
Are you familiar with this theorem? encyclopediaofmath.org/index.php/Liouville-Ostrogradski_formula
$endgroup$
– Zacky
Jan 2 at 15:13
|
show 1 more comment
$begingroup$
Write the general solution
for $ (x^2) y'''' + (3x^2-2x)y''' + (3x^2-4x+2)y'' +(x^2-2x+2)y'
= 0 $
I tried to guess a solution and use the fact that i can decrease the ODE to less power ( to $y'''$ ) by using the Wronskian.
I guessed that $ e^{-x} $ is a solution.
Is it the way we solve this kind of equations? It's a homework question so i guess (i / you) can solve it.
Euler ODE doesn't work here .
Can you help me with the solutions i got 4 solutions : ( after solving Euler equation and moving to $ v''' $ .
$y'(x) = { e^{-x} ,~ frac{x^3}{3}e^{-x} ,~ frac{x^2}{2}e^{-x} }$ or any linear combination of those, uniqueness theorem doesn't apply here near $x=0$
ordinary-differential-equations
asked Jan 2 at 13:37
Mather Mather
3047
$endgroup$
Write the general solution
for $ (x^2) y'''' + (3x^2-2x)y''' + (3x^2-4x+2)y'' +(x^2-2x+2)y'
= 0 $
I tried to guess a solution and use the fact that i can decrease the ODE to less power ( to $y'''$ ) by using the Wronskian.
I guessed that $ e^{-x} $ is a solution.
Is it the way we solve this kind of equations? It's a homework question so i guess (i / you) can solve it.
Euler ODE doesn't work here .
Can you help me with the solutions i got 4 solutions : ( after solving Euler equation and moving to $ v''' $ .
$y'(x) = { e^{-x} ,~ frac{x^3}{3}e^{-x} ,~ frac{x^2}{2}e^{-x} }$ or any linear combination of those, uniqueness theorem doesn't apply here near $x=0$
ordinary-differential-equations
ordinary-differential-equations
asked Jan 2 at 13:37
Mather Mather
3047
asked Jan 2 at 13:37
Mather Mather
3047
edited Jan 2 at 15:40
Mather
asked Jan 2 at 13:37
Mather Mather
3047
asked Jan 2 at 13:37
Mather Mather
3047
asked Jan 2 at 13:37
Mather Mather
3047
3047
$begingroup$
Your coefficients are polynomials, so it might be reasonable to assume $y$ takes the form of a polynomial. Else, if you have found a solution (say $y_{1} = e^{-x}$), then you can search for another solution of the form $y_{2} = v(x)e^{-x}$.
$endgroup$
– Mattos
Jan 2 at 13:41
$begingroup$
Solution $0$ won't help. But the constant $1$ is also a solution, and is linearly independent of the others, so it will help.
$endgroup$
– GEdgar
Jan 2 at 15:04
$begingroup$
true i will integrate all and get a linear Space of those but how can i know that the answer is right its impossible to start puting each one at a time
$endgroup$
– Mather
Jan 2 at 15:08
$begingroup$
not possible to integrate some of them like $ frac{x^3}{3} e^{-x} $
$endgroup$
– Mather
Jan 2 at 15:10
$begingroup$
Are you familiar with this theorem? encyclopediaofmath.org/index.php/Liouville-Ostrogradski_formula
$endgroup$
– Zacky
Jan 2 at 15:13
|
show 1 more comment
$begingroup$
Your coefficients are polynomials, so it might be reasonable to assume $y$ takes the form of a polynomial. Else, if you have found a solution (say $y_{1} = e^{-x}$), then you can search for another solution of the form $y_{2} = v(x)e^{-x}$.
$endgroup$
– Mattos
Jan 2 at 13:41
$begingroup$
Solution $0$ won't help. But the constant $1$ is also a solution, and is linearly independent of the others, so it will help.
$endgroup$
– GEdgar
Jan 2 at 15:04
$begingroup$
true i will integrate all and get a linear Space of those but how can i know that the answer is right its impossible to start puting each one at a time
$endgroup$
– Mather
Jan 2 at 15:08
$begingroup$
not possible to integrate some of them like $ frac{x^3}{3} e^{-x} $
$endgroup$
– Mather
Jan 2 at 15:10
$begingroup$
Are you familiar with this theorem? encyclopediaofmath.org/index.php/Liouville-Ostrogradski_formula
$endgroup$
– Zacky
Jan 2 at 15:13
$begingroup$
Your coefficients are polynomials, so it might be reasonable to assume $y$ takes the form of a polynomial. Else, if you have found a solution (say $y_{1} = e^{-x}$), then you can search for another solution of the form $y_{2} = v(x)e^{-x}$.
$endgroup$
– Mattos
Jan 2 at 13:41
$begingroup$
Your coefficients are polynomials, so it might be reasonable to assume $y$ takes the form of a polynomial. Else, if you have found a solution (say $y_{1} = e^{-x}$), then you can search for another solution of the form $y_{2} = v(x)e^{-x}$.
$endgroup$
– Mattos
Jan 2 at 13:41
$begingroup$
Solution $0$ won't help. But the constant $1$ is also a solution, and is linearly independent of the others, so it will help.
$endgroup$
– GEdgar
Jan 2 at 15:04
$begingroup$
Solution $0$ won't help. But the constant $1$ is also a solution, and is linearly independent of the others, so it will help.
$endgroup$
– GEdgar
Jan 2 at 15:04
$begingroup$
true i will integrate all and get a linear Space of those but how can i know that the answer is right its impossible to start puting each one at a time
$endgroup$
– Mather
Jan 2 at 15:08
$begingroup$
true i will integrate all and get a linear Space of those but how can i know that the answer is right its impossible to start puting each one at a time
$endgroup$
– Mather
Jan 2 at 15:08
$begingroup$
not possible to integrate some of them like $ frac{x^3}{3} e^{-x} $
$endgroup$
– Mather
Jan 2 at 15:10
$begingroup$
not possible to integrate some of them like $ frac{x^3}{3} e^{-x} $
$endgroup$
– Mather
Jan 2 at 15:10
$begingroup$
Are you familiar with this theorem? encyclopediaofmath.org/index.php/Liouville-Ostrogradski_formula
$endgroup$
– Zacky
Jan 2 at 15:13
$begingroup$
Are you familiar with this theorem? encyclopediaofmath.org/index.php/Liouville-Ostrogradski_formula
$endgroup$
– Zacky
Jan 2 at 15:13
|
show 1 more comment
2 Answers
2
active
oldest
votes
$begingroup$
Take a look at the coefficients. You have the sequence $1,3,3,1$ in the quadratic terms, remembering the binomial theorem you get thus $x^2(D+1)^3Dy$, $D=frac{d}{dx}$. Then in the linear terms coefficients $2,4,2$ which gives you $-2x(D+1)^2Dy$ and similarly for the constant terms, so that
$$
0=[x^2(D+1)^3-2x(D+1)^2+2(D+1)]Dy
$$
which means that you get to solve the factorized system
$$
(D+1)Dy=u,\
[x^2(D+1)^2-2x(D+1)+2]u=0.
$$
The last equation is an Euler-Cauchy equation for $e^xu$,
$$
[x^2D^2-2xD+2](e^xu)=0.
$$
answered Jan 2 at 15:16
LutzLLutzL
57.4k42054
$endgroup$
$begingroup$
not sure what did you do but i got the same unique polynomal equation ! $ (r-1)(r-2) $ and from there i got the semi solutions writing up there after integrating
$endgroup$
– Mather
Jan 2 at 15:27
$begingroup$
Well, then working backwards you get $e^xu=span{x,x^2}$, $e^xy'in span{1,x^2,x^3}$, $yin span{1,e^{-x},frac{d^2}{da^2}(e^{ax}/a)|_{a=-1}, frac{d^3}{da^3}(e^{ax}/a)|_{a=-1}}$.
$endgroup$
– LutzL
Jan 2 at 15:34
$begingroup$
how can one obtain general solution ?
$endgroup$
– Mather
Jan 2 at 15:34
$begingroup$
yes this is what i got without the $1$ how did you find the $ 1 $ in the span , did you guess it ?
$endgroup$
– Mather
Jan 2 at 15:38
$begingroup$
In every integration you get an integration constant. These are the multiples of the constant function with value $1$.
$endgroup$
– LutzL
Jan 2 at 15:40
|
show 4 more comments
$begingroup$
Hint. Once you have guessed a solution, namely $y_1(x)=mathrm{e}^{-x}$, set $y=zmathrm{e}^{-x}$ and obtain an third order equation for $z'$.
answered Jan 2 at 13:43
Yiorgos S. SmyrlisYiorgos S. Smyrlis
63k1384163
$endgroup$
$begingroup$
Indeed but is it the the best type of solving this equation , there is no other easy ways to solve this right ?
$endgroup$
– Mather
Jan 2 at 13:44
$begingroup$
You can substitue $$frac{dy(x)}{dx}=v(x)$$ and then use the Laplace transformation
$endgroup$
– Dr. Sonnhard Graubner
Jan 2 at 13:47
$begingroup$
can you please help out
$endgroup$
– Mather
Jan 2 at 14:55
$begingroup$
i have found 4 solutions using a long way
$endgroup$
– Mather
Jan 2 at 14:55
$begingroup$
do you know a good site to check it out , wolfram doesn't give me a good one , here is my solutions : {$ e^{-x}$ , $frac{x^3}{3}e^{-x}$ , $frac{x^2}{2}e^{-x}$ , $ 0 $}
$endgroup$
– Mather
Jan 2 at 14:56
add a comment |
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2 Answers
2
active
oldest
votes
2 Answers
2
active
oldest
votes
active
oldest
votes
active
oldest
votes
$begingroup$
Take a look at the coefficients. You have the sequence $1,3,3,1$ in the quadratic terms, remembering the binomial theorem you get thus $x^2(D+1)^3Dy$, $D=frac{d}{dx}$. Then in the linear terms coefficients $2,4,2$ which gives you $-2x(D+1)^2Dy$ and similarly for the constant terms, so that
$$
0=[x^2(D+1)^3-2x(D+1)^2+2(D+1)]Dy
$$
which means that you get to solve the factorized system
$$
(D+1)Dy=u,\
[x^2(D+1)^2-2x(D+1)+2]u=0.
$$
The last equation is an Euler-Cauchy equation for $e^xu$,
$$
[x^2D^2-2xD+2](e^xu)=0.
$$
answered Jan 2 at 15:16
LutzLLutzL
57.4k42054
$endgroup$
$begingroup$
not sure what did you do but i got the same unique polynomal equation ! $ (r-1)(r-2) $ and from there i got the semi solutions writing up there after integrating
$endgroup$
– Mather
Jan 2 at 15:27
$begingroup$
Well, then working backwards you get $e^xu=span{x,x^2}$, $e^xy'in span{1,x^2,x^3}$, $yin span{1,e^{-x},frac{d^2}{da^2}(e^{ax}/a)|_{a=-1}, frac{d^3}{da^3}(e^{ax}/a)|_{a=-1}}$.
$endgroup$
– LutzL
Jan 2 at 15:34
$begingroup$
how can one obtain general solution ?
$endgroup$
– Mather
Jan 2 at 15:34
$begingroup$
yes this is what i got without the $1$ how did you find the $ 1 $ in the span , did you guess it ?
$endgroup$
– Mather
Jan 2 at 15:38
$begingroup$
In every integration you get an integration constant. These are the multiples of the constant function with value $1$.
$endgroup$
– LutzL
Jan 2 at 15:40
|
show 4 more comments
$begingroup$
Take a look at the coefficients. You have the sequence $1,3,3,1$ in the quadratic terms, remembering the binomial theorem you get thus $x^2(D+1)^3Dy$, $D=frac{d}{dx}$. Then in the linear terms coefficients $2,4,2$ which gives you $-2x(D+1)^2Dy$ and similarly for the constant terms, so that
$$
0=[x^2(D+1)^3-2x(D+1)^2+2(D+1)]Dy
$$
which means that you get to solve the factorized system
$$
(D+1)Dy=u,\
[x^2(D+1)^2-2x(D+1)+2]u=0.
$$
The last equation is an Euler-Cauchy equation for $e^xu$,
$$
[x^2D^2-2xD+2](e^xu)=0.
$$
answered Jan 2 at 15:16
LutzLLutzL
57.4k42054
$endgroup$
$begingroup$
not sure what did you do but i got the same unique polynomal equation ! $ (r-1)(r-2) $ and from there i got the semi solutions writing up there after integrating
$endgroup$
– Mather
Jan 2 at 15:27
$begingroup$
Well, then working backwards you get $e^xu=span{x,x^2}$, $e^xy'in span{1,x^2,x^3}$, $yin span{1,e^{-x},frac{d^2}{da^2}(e^{ax}/a)|_{a=-1}, frac{d^3}{da^3}(e^{ax}/a)|_{a=-1}}$.
$endgroup$
– LutzL
Jan 2 at 15:34
$begingroup$
how can one obtain general solution ?
$endgroup$
– Mather
Jan 2 at 15:34
$begingroup$
yes this is what i got without the $1$ how did you find the $ 1 $ in the span , did you guess it ?
$endgroup$
– Mather
Jan 2 at 15:38
$begingroup$
In every integration you get an integration constant. These are the multiples of the constant function with value $1$.
$endgroup$
– LutzL
Jan 2 at 15:40
|
show 4 more comments
$begingroup$
Take a look at the coefficients. You have the sequence $1,3,3,1$ in the quadratic terms, remembering the binomial theorem you get thus $x^2(D+1)^3Dy$, $D=frac{d}{dx}$. Then in the linear terms coefficients $2,4,2$ which gives you $-2x(D+1)^2Dy$ and similarly for the constant terms, so that
$$
0=[x^2(D+1)^3-2x(D+1)^2+2(D+1)]Dy
$$
which means that you get to solve the factorized system
$$
(D+1)Dy=u,\
[x^2(D+1)^2-2x(D+1)+2]u=0.
$$
The last equation is an Euler-Cauchy equation for $e^xu$,
$$
[x^2D^2-2xD+2](e^xu)=0.
$$
answered Jan 2 at 15:16
LutzLLutzL
57.4k42054
$endgroup$
Take a look at the coefficients. You have the sequence $1,3,3,1$ in the quadratic terms, remembering the binomial theorem you get thus $x^2(D+1)^3Dy$, $D=frac{d}{dx}$. Then in the linear terms coefficients $2,4,2$ which gives you $-2x(D+1)^2Dy$ and similarly for the constant terms, so that
$$
0=[x^2(D+1)^3-2x(D+1)^2+2(D+1)]Dy
$$
which means that you get to solve the factorized system
$$
(D+1)Dy=u,\
[x^2(D+1)^2-2x(D+1)+2]u=0.
$$
The last equation is an Euler-Cauchy equation for $e^xu$,
$$
[x^2D^2-2xD+2](e^xu)=0.
$$
answered Jan 2 at 15:16
LutzLLutzL
57.4k42054
edited Jan 2 at 15:21
answered Jan 2 at 15:16
LutzLLutzL
57.4k42054
answered Jan 2 at 15:16
LutzLLutzL
57.4k42054
answered Jan 2 at 15:16
LutzLLutzL
57.4k42054
57.4k42054
$begingroup$
not sure what did you do but i got the same unique polynomal equation ! $ (r-1)(r-2) $ and from there i got the semi solutions writing up there after integrating
$endgroup$
– Mather
Jan 2 at 15:27
$begingroup$
Well, then working backwards you get $e^xu=span{x,x^2}$, $e^xy'in span{1,x^2,x^3}$, $yin span{1,e^{-x},frac{d^2}{da^2}(e^{ax}/a)|_{a=-1}, frac{d^3}{da^3}(e^{ax}/a)|_{a=-1}}$.
$endgroup$
– LutzL
Jan 2 at 15:34
$begingroup$
how can one obtain general solution ?
$endgroup$
– Mather
Jan 2 at 15:34
$begingroup$
yes this is what i got without the $1$ how did you find the $ 1 $ in the span , did you guess it ?
$endgroup$
– Mather
Jan 2 at 15:38
$begingroup$
In every integration you get an integration constant. These are the multiples of the constant function with value $1$.
$endgroup$
– LutzL
Jan 2 at 15:40
|
show 4 more comments
$begingroup$
not sure what did you do but i got the same unique polynomal equation ! $ (r-1)(r-2) $ and from there i got the semi solutions writing up there after integrating
$endgroup$
– Mather
Jan 2 at 15:27
$begingroup$
Well, then working backwards you get $e^xu=span{x,x^2}$, $e^xy'in span{1,x^2,x^3}$, $yin span{1,e^{-x},frac{d^2}{da^2}(e^{ax}/a)|_{a=-1}, frac{d^3}{da^3}(e^{ax}/a)|_{a=-1}}$.
$endgroup$
– LutzL
Jan 2 at 15:34
$begingroup$
how can one obtain general solution ?
$endgroup$
– Mather
Jan 2 at 15:34
$begingroup$
yes this is what i got without the $1$ how did you find the $ 1 $ in the span , did you guess it ?
$endgroup$
– Mather
Jan 2 at 15:38
$begingroup$
In every integration you get an integration constant. These are the multiples of the constant function with value $1$.
$endgroup$
– LutzL
Jan 2 at 15:40
$begingroup$
not sure what did you do but i got the same unique polynomal equation ! $ (r-1)(r-2) $ and from there i got the semi solutions writing up there after integrating
$endgroup$
– Mather
Jan 2 at 15:27
$begingroup$
not sure what did you do but i got the same unique polynomal equation ! $ (r-1)(r-2) $ and from there i got the semi solutions writing up there after integrating
$endgroup$
– Mather
Jan 2 at 15:27
$begingroup$
Well, then working backwards you get $e^xu=span{x,x^2}$, $e^xy'in span{1,x^2,x^3}$, $yin span{1,e^{-x},frac{d^2}{da^2}(e^{ax}/a)|_{a=-1}, frac{d^3}{da^3}(e^{ax}/a)|_{a=-1}}$.
$endgroup$
– LutzL
Jan 2 at 15:34
$begingroup$
Well, then working backwards you get $e^xu=span{x,x^2}$, $e^xy'in span{1,x^2,x^3}$, $yin span{1,e^{-x},frac{d^2}{da^2}(e^{ax}/a)|_{a=-1}, frac{d^3}{da^3}(e^{ax}/a)|_{a=-1}}$.
$endgroup$
– LutzL
Jan 2 at 15:34
$begingroup$
how can one obtain general solution ?
$endgroup$
– Mather
Jan 2 at 15:34
$begingroup$
how can one obtain general solution ?
$endgroup$
– Mather
Jan 2 at 15:34
$begingroup$
yes this is what i got without the $1$ how did you find the $ 1 $ in the span , did you guess it ?
$endgroup$
– Mather
Jan 2 at 15:38
$begingroup$
yes this is what i got without the $1$ how did you find the $ 1 $ in the span , did you guess it ?
$endgroup$
– Mather
Jan 2 at 15:38
$begingroup$
In every integration you get an integration constant. These are the multiples of the constant function with value $1$.
$endgroup$
– LutzL
Jan 2 at 15:40
$begingroup$
In every integration you get an integration constant. These are the multiples of the constant function with value $1$.
$endgroup$
– LutzL
Jan 2 at 15:40
|
show 4 more comments
$begingroup$
Hint. Once you have guessed a solution, namely $y_1(x)=mathrm{e}^{-x}$, set $y=zmathrm{e}^{-x}$ and obtain an third order equation for $z'$.
answered Jan 2 at 13:43
Yiorgos S. SmyrlisYiorgos S. Smyrlis
63k1384163
$endgroup$
$begingroup$
Indeed but is it the the best type of solving this equation , there is no other easy ways to solve this right ?
$endgroup$
– Mather
Jan 2 at 13:44
$begingroup$
You can substitue $$frac{dy(x)}{dx}=v(x)$$ and then use the Laplace transformation
$endgroup$
– Dr. Sonnhard Graubner
Jan 2 at 13:47
$begingroup$
can you please help out
$endgroup$
– Mather
Jan 2 at 14:55
$begingroup$
i have found 4 solutions using a long way
$endgroup$
– Mather
Jan 2 at 14:55
$begingroup$
do you know a good site to check it out , wolfram doesn't give me a good one , here is my solutions : {$ e^{-x}$ , $frac{x^3}{3}e^{-x}$ , $frac{x^2}{2}e^{-x}$ , $ 0 $}
$endgroup$
– Mather
Jan 2 at 14:56
add a comment |
$begingroup$
Hint. Once you have guessed a solution, namely $y_1(x)=mathrm{e}^{-x}$, set $y=zmathrm{e}^{-x}$ and obtain an third order equation for $z'$.
answered Jan 2 at 13:43
Yiorgos S. SmyrlisYiorgos S. Smyrlis
63k1384163
$endgroup$
$begingroup$
Indeed but is it the the best type of solving this equation , there is no other easy ways to solve this right ?
$endgroup$
– Mather
Jan 2 at 13:44
$begingroup$
You can substitue $$frac{dy(x)}{dx}=v(x)$$ and then use the Laplace transformation
$endgroup$
– Dr. Sonnhard Graubner
Jan 2 at 13:47
$begingroup$
can you please help out
$endgroup$
– Mather
Jan 2 at 14:55
$begingroup$
i have found 4 solutions using a long way
$endgroup$
– Mather
Jan 2 at 14:55
$begingroup$
do you know a good site to check it out , wolfram doesn't give me a good one , here is my solutions : {$ e^{-x}$ , $frac{x^3}{3}e^{-x}$ , $frac{x^2}{2}e^{-x}$ , $ 0 $}
$endgroup$
– Mather
Jan 2 at 14:56
add a comment |
$begingroup$
Hint. Once you have guessed a solution, namely $y_1(x)=mathrm{e}^{-x}$, set $y=zmathrm{e}^{-x}$ and obtain an third order equation for $z'$.
answered Jan 2 at 13:43
Yiorgos S. SmyrlisYiorgos S. Smyrlis
63k1384163
$endgroup$
Hint. Once you have guessed a solution, namely $y_1(x)=mathrm{e}^{-x}$, set $y=zmathrm{e}^{-x}$ and obtain an third order equation for $z'$.
answered Jan 2 at 13:43
Yiorgos S. SmyrlisYiorgos S. Smyrlis
63k1384163
answered Jan 2 at 13:43
Yiorgos S. SmyrlisYiorgos S. Smyrlis
63k1384163
answered Jan 2 at 13:43
Yiorgos S. SmyrlisYiorgos S. Smyrlis
63k1384163
answered Jan 2 at 13:43
Yiorgos S. SmyrlisYiorgos S. Smyrlis
63k1384163
63k1384163
$begingroup$
Indeed but is it the the best type of solving this equation , there is no other easy ways to solve this right ?
$endgroup$
– Mather
Jan 2 at 13:44
$begingroup$
You can substitue $$frac{dy(x)}{dx}=v(x)$$ and then use the Laplace transformation
$endgroup$
– Dr. Sonnhard Graubner
Jan 2 at 13:47
$begingroup$
can you please help out
$endgroup$
– Mather
Jan 2 at 14:55
$begingroup$
i have found 4 solutions using a long way
$endgroup$
– Mather
Jan 2 at 14:55
$begingroup$
do you know a good site to check it out , wolfram doesn't give me a good one , here is my solutions : {$ e^{-x}$ , $frac{x^3}{3}e^{-x}$ , $frac{x^2}{2}e^{-x}$ , $ 0 $}
$endgroup$
– Mather
Jan 2 at 14:56
add a comment |
$begingroup$
Indeed but is it the the best type of solving this equation , there is no other easy ways to solve this right ?
$endgroup$
– Mather
Jan 2 at 13:44
$begingroup$
You can substitue $$frac{dy(x)}{dx}=v(x)$$ and then use the Laplace transformation
$endgroup$
– Dr. Sonnhard Graubner
Jan 2 at 13:47
$begingroup$
can you please help out
$endgroup$
– Mather
Jan 2 at 14:55
$begingroup$
i have found 4 solutions using a long way
$endgroup$
– Mather
Jan 2 at 14:55
$begingroup$
do you know a good site to check it out , wolfram doesn't give me a good one , here is my solutions : {$ e^{-x}$ , $frac{x^3}{3}e^{-x}$ , $frac{x^2}{2}e^{-x}$ , $ 0 $}
$endgroup$
– Mather
Jan 2 at 14:56
$begingroup$
Indeed but is it the the best type of solving this equation , there is no other easy ways to solve this right ?
$endgroup$
– Mather
Jan 2 at 13:44
$begingroup$
Indeed but is it the the best type of solving this equation , there is no other easy ways to solve this right ?
$endgroup$
– Mather
Jan 2 at 13:44
$begingroup$
You can substitue $$frac{dy(x)}{dx}=v(x)$$ and then use the Laplace transformation
$endgroup$
– Dr. Sonnhard Graubner
Jan 2 at 13:47
$begingroup$
You can substitue $$frac{dy(x)}{dx}=v(x)$$ and then use the Laplace transformation
$endgroup$
– Dr. Sonnhard Graubner
Jan 2 at 13:47
$begingroup$
can you please help out
$endgroup$
– Mather
Jan 2 at 14:55
$begingroup$
can you please help out
$endgroup$
– Mather
Jan 2 at 14:55
$begingroup$
i have found 4 solutions using a long way
$endgroup$
– Mather
Jan 2 at 14:55
$begingroup$
i have found 4 solutions using a long way
$endgroup$
– Mather
Jan 2 at 14:55
$begingroup$
do you know a good site to check it out , wolfram doesn't give me a good one , here is my solutions : {$ e^{-x}$ , $frac{x^3}{3}e^{-x}$ , $frac{x^2}{2}e^{-x}$ , $ 0 $}
$endgroup$
– Mather
Jan 2 at 14:56
$begingroup$
do you know a good site to check it out , wolfram doesn't give me a good one , here is my solutions : {$ e^{-x}$ , $frac{x^3}{3}e^{-x}$ , $frac{x^2}{2}e^{-x}$ , $ 0 $}
$endgroup$
– Mather
Jan 2 at 14:56
add a comment |
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$begingroup$
Your coefficients are polynomials, so it might be reasonable to assume $y$ takes the form of a polynomial. Else, if you have found a solution (say $y_{1} = e^{-x}$), then you can search for another solution of the form $y_{2} = v(x)e^{-x}$.
$endgroup$
– Mattos
Jan 2 at 13:41
$begingroup$
Solution $0$ won't help. But the constant $1$ is also a solution, and is linearly independent of the others, so it will help.
$endgroup$
– GEdgar
Jan 2 at 15:04
$begingroup$
true i will integrate all and get a linear Space of those but how can i know that the answer is right its impossible to start puting each one at a time
$endgroup$
– Mather
Jan 2 at 15:08
$begingroup$
not possible to integrate some of them like $ frac{x^3}{3} e^{-x} $
$endgroup$
– Mather
Jan 2 at 15:10
$begingroup$
Are you familiar with this theorem? encyclopediaofmath.org/index.php/Liouville-Ostrogradski_formula
$endgroup$
– Zacky
Jan 2 at 15:13