How to solve this 4th order ODE with polynomial coefficients?












3












$begingroup$



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







share|cite|improve this question











$endgroup$












  • $begingroup$
    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
















3












$begingroup$



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







share|cite|improve this question











$endgroup$












  • $begingroup$
    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














3












3








3


3



$begingroup$



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







share|cite|improve this question











$endgroup$





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






share|cite|improve this question















share|cite|improve this question













share|cite|improve this question




share|cite|improve this question








edited Jan 2 at 15:40







Mather

















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


















  • $begingroup$
    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










2 Answers
2






active

oldest

votes


















3












$begingroup$

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 $e^xu$,
$$
[x^2D^2-2xD+2](e^xu)=0.
$$






share|cite|improve this answer











$endgroup$













  • $begingroup$
    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



















1












$begingroup$

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 $z'$.






share|cite|improve this answer









$endgroup$













  • $begingroup$
    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













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2 Answers
2






active

oldest

votes








2 Answers
2






active

oldest

votes









active

oldest

votes






active

oldest

votes









3












$begingroup$

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 $e^xu$,
$$
[x^2D^2-2xD+2](e^xu)=0.
$$






share|cite|improve this answer











$endgroup$













  • $begingroup$
    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
















3












$begingroup$

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 $e^xu$,
$$
[x^2D^2-2xD+2](e^xu)=0.
$$






share|cite|improve this answer











$endgroup$













  • $begingroup$
    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














3












3








3





$begingroup$

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 $e^xu$,
$$
[x^2D^2-2xD+2](e^xu)=0.
$$






share|cite|improve this answer











$endgroup$



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 $e^xu$,
$$
[x^2D^2-2xD+2](e^xu)=0.
$$







share|cite|improve this answer














share|cite|improve this answer



share|cite|improve this answer








edited Jan 2 at 15:21

























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


















  • $begingroup$
    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











1












$begingroup$

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 $z'$.






share|cite|improve this answer









$endgroup$













  • $begingroup$
    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


















1












$begingroup$

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 $z'$.






share|cite|improve this answer









$endgroup$













  • $begingroup$
    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
















1












1








1





$begingroup$

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 $z'$.






share|cite|improve this answer









$endgroup$



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 $z'$.







share|cite|improve this answer












share|cite|improve this answer



share|cite|improve this answer










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




















  • $begingroup$
    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




















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