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Existence of closed form solution of the below DE
Announcing the arrival of Valued Associate #679: Cesar Manara
Planned maintenance scheduled April 17/18, 2019 at 00:00UTC (8:00pm US/Eastern)Closed form of the solution of a nonlinear differential equationWhat to do when regular approaches fail on linear, non-homogeneous ODES.Numerical solution to differential equationDoes this nonlinear differential equation have a closed form solution?Closed form for $intfracleft((x + i) betaright)^beta x^beta - 2(x^2 + 1)^beta expleft(-fracalphaA xright) , mathrmd x$A question on the interval of existence of IVPIs it possible to obtain the closed-form expression of this differential equation?Finding closed form of Frobenius solutionSolution to a 2nd order ODE with a Gaussian coefficientHow to find the exact solution of a Sturm Liouville form, 2nd order ODE?
$begingroup$
I have a problem finding an analytical closed form solution of the following DE.
$ displaystyle left( fracy(x)+23 right) fracdydx + y(x) +frac1x left( a y(x)^2 + b y(x) -c right) =0. \
a,b,c in mathbbR, qquad a,b,c>0.
$
I encountered this differential equation in my research in fluid dynamics. I failed to reduce it to some standard form of DE for which solutions are possible. Mathematica and Maple are unable to find analytical solutions for this.
Can anyone point out any method that can be used to know if a solution for this DE exists?
And if a solution exists, can anyone point out a method to solve the DE?
ordinary-differential-equations
asked Feb 26 at 12:00
Sunil JaiswalSunil Jaiswal
82
$endgroup$
add a comment |
$begingroup$
I have a problem finding an analytical closed form solution of the following DE.
$ displaystyle left( fracy(x)+23 right) fracdydx + y(x) +frac1x left( a y(x)^2 + b y(x) -c right) =0. \
a,b,c in mathbbR, qquad a,b,c>0.
$
I encountered this differential equation in my research in fluid dynamics. I failed to reduce it to some standard form of DE for which solutions are possible. Mathematica and Maple are unable to find analytical solutions for this.
Can anyone point out any method that can be used to know if a solution for this DE exists?
And if a solution exists, can anyone point out a method to solve the DE?
ordinary-differential-equations
asked Feb 26 at 12:00
Sunil JaiswalSunil Jaiswal
82
$endgroup$
$begingroup$
How about using a power series?
$endgroup$
– Mattos
Feb 26 at 14:01
$begingroup$
@Mattos Power series solution for this DE diverges. Also I am looking for a closed form solution if possible.
$endgroup$
– Sunil Jaiswal
Feb 27 at 9:43
$begingroup$
I'm confused as to why you write the 'power series solution for this DE diverges'. The power series solution is the solution, hence if the power series diverges, then the solution diverges. You've already computed a power series solution, so you've shown a solution exists, but you can check to see if a solution exists and is unique using the Picard-Lindelof theorem i.e show that $partial f(x,y)/partial y$ exists somewhere for the ODE $y' = f(x,y(x))$, though this existence theorem might be too strenuous.
$endgroup$
– Mattos
Feb 27 at 9:56
add a comment |
$begingroup$
I have a problem finding an analytical closed form solution of the following DE.
$ displaystyle left( fracy(x)+23 right) fracdydx + y(x) +frac1x left( a y(x)^2 + b y(x) -c right) =0. \
a,b,c in mathbbR, qquad a,b,c>0.
$
I encountered this differential equation in my research in fluid dynamics. I failed to reduce it to some standard form of DE for which solutions are possible. Mathematica and Maple are unable to find analytical solutions for this.
Can anyone point out any method that can be used to know if a solution for this DE exists?
And if a solution exists, can anyone point out a method to solve the DE?
ordinary-differential-equations
asked Feb 26 at 12:00
Sunil JaiswalSunil Jaiswal
82
$endgroup$
I have a problem finding an analytical closed form solution of the following DE.
$ displaystyle left( fracy(x)+23 right) fracdydx + y(x) +frac1x left( a y(x)^2 + b y(x) -c right) =0. \
a,b,c in mathbbR, qquad a,b,c>0.
$
I encountered this differential equation in my research in fluid dynamics. I failed to reduce it to some standard form of DE for which solutions are possible. Mathematica and Maple are unable to find analytical solutions for this.
Can anyone point out any method that can be used to know if a solution for this DE exists?
And if a solution exists, can anyone point out a method to solve the DE?
ordinary-differential-equations
ordinary-differential-equations
asked Feb 26 at 12:00
Sunil JaiswalSunil Jaiswal
82
asked Feb 26 at 12:00
Sunil JaiswalSunil Jaiswal
82
asked Feb 26 at 12:00
Sunil JaiswalSunil Jaiswal
82
asked Feb 26 at 12:00
Sunil JaiswalSunil Jaiswal
82
asked Feb 26 at 12:00
Sunil JaiswalSunil Jaiswal
82
82
$begingroup$
How about using a power series?
$endgroup$
– Mattos
Feb 26 at 14:01
$begingroup$
@Mattos Power series solution for this DE diverges. Also I am looking for a closed form solution if possible.
$endgroup$
– Sunil Jaiswal
Feb 27 at 9:43
$begingroup$
I'm confused as to why you write the 'power series solution for this DE diverges'. The power series solution is the solution, hence if the power series diverges, then the solution diverges. You've already computed a power series solution, so you've shown a solution exists, but you can check to see if a solution exists and is unique using the Picard-Lindelof theorem i.e show that $partial f(x,y)/partial y$ exists somewhere for the ODE $y' = f(x,y(x))$, though this existence theorem might be too strenuous.
$endgroup$
– Mattos
Feb 27 at 9:56
add a comment |
$begingroup$
How about using a power series?
$endgroup$
– Mattos
Feb 26 at 14:01
$begingroup$
@Mattos Power series solution for this DE diverges. Also I am looking for a closed form solution if possible.
$endgroup$
– Sunil Jaiswal
Feb 27 at 9:43
$begingroup$
I'm confused as to why you write the 'power series solution for this DE diverges'. The power series solution is the solution, hence if the power series diverges, then the solution diverges. You've already computed a power series solution, so you've shown a solution exists, but you can check to see if a solution exists and is unique using the Picard-Lindelof theorem i.e show that $partial f(x,y)/partial y$ exists somewhere for the ODE $y' = f(x,y(x))$, though this existence theorem might be too strenuous.
$endgroup$
– Mattos
Feb 27 at 9:56
$begingroup$
How about using a power series?
$endgroup$
– Mattos
Feb 26 at 14:01
$begingroup$
How about using a power series?
$endgroup$
– Mattos
Feb 26 at 14:01
$begingroup$
@Mattos Power series solution for this DE diverges. Also I am looking for a closed form solution if possible.
$endgroup$
– Sunil Jaiswal
Feb 27 at 9:43
$begingroup$
@Mattos Power series solution for this DE diverges. Also I am looking for a closed form solution if possible.
$endgroup$
– Sunil Jaiswal
Feb 27 at 9:43
$begingroup$
I'm confused as to why you write the 'power series solution for this DE diverges'. The power series solution is the solution, hence if the power series diverges, then the solution diverges. You've already computed a power series solution, so you've shown a solution exists, but you can check to see if a solution exists and is unique using the Picard-Lindelof theorem i.e show that $partial f(x,y)/partial y$ exists somewhere for the ODE $y' = f(x,y(x))$, though this existence theorem might be too strenuous.
$endgroup$
– Mattos
Feb 27 at 9:56
$begingroup$
I'm confused as to why you write the 'power series solution for this DE diverges'. The power series solution is the solution, hence if the power series diverges, then the solution diverges. You've already computed a power series solution, so you've shown a solution exists, but you can check to see if a solution exists and is unique using the Picard-Lindelof theorem i.e show that $partial f(x,y)/partial y$ exists somewhere for the ODE $y' = f(x,y(x))$, though this existence theorem might be too strenuous.
$endgroup$
– Mattos
Feb 27 at 9:56
add a comment |
1 Answer
1
active
oldest
votes
$begingroup$
Hint:
Approach $1$:
This belongs to an Abel equation of the second kind.
Let $u=y+2$ ,
Then $dfracdudx=dfracdydx$
$thereforedfracu3dfracdudx+u-2+dfrac1x(a(u-2)^2+b(u-2)-c)=0$
$dfracu3dfracdudx+u-2+dfrac1x(au^2-(4a-b)u+4a-2b-c)=0$
$udfracdudx=-dfrac3au^2x-3left(1-dfrac4a-bxright)u+6-dfrac3(4a-2b-c)x$
Let $u=x^-3av$ ,
Then $dfracdudx=x^-3adfracdvdx-3ax^-3a-1v$
$therefore x^-3avleft(x^-3adfracdvdx-3ax^-3a-1vright)=-dfrac3ax^-6av^2x-3left(1-dfrac4a-bxright)x^-3av+6-dfrac3(4a-2b-c)x$
$x^-6avdfracdvdx-3ax^-6a-1v^2=-3ax^-6a-1v^2-3left(1-dfrac4a-bxright)x^-3av+6-dfrac3(4a-2b-c)x$
$x^-6avdfracdvdx=-3left(1-dfrac4a-bxright)x^-3av+6-dfrac3(4a-2b-c)x$
$vdfracdvdx=-3(x^3a-(4a-b)x^3a-1)v+6x^6a-3(4a-2b-c)x^6a-1$
Approach $2$:
$dfracy+23dfracdydx+y+dfracay^2+by-cx=0$
$y+dfracay^2+by-cx=-dfracy+23dfracdydx$
$left(x+ay+b-dfraccyright)dfracdxdy=-left(dfrac13+dfrac23yright)x$
This belongs to an Abel equation of the second kind.
Let $u=x+ay+b-dfraccy$ ,
Then $x=u-ay-b+dfraccy$
$dfracdxdy=dfracdudy-a-dfraccy^2$
$therefore uleft(dfracdudy-a-dfraccy^2right)=-left(dfrac13+dfrac23yright)left(u-ay-b+dfraccyright)$
$udfracdudy-left(a+dfraccy^2right)u=-left(dfrac13+dfrac23yright)u+dfrac(y+2)(ay^2+by-c)3y^2$
$udfracdudy=dfrac(3a-1)y^2-2y+3c3y^2u+dfrac(y+2)(ay^2+by-c)3y^2$
answered Mar 1 at 12:35
doraemonpauldoraemonpaul
13k31761
$endgroup$
$begingroup$
Thank you! That's extremely helpful.
$endgroup$
– Sunil Jaiswal
Mar 5 at 10:08
$begingroup$
I tried reducing the form given in approach 1 in normal form. Doing a variable transform, $$z=x^3 gamma left( frac4 a - ba - frac31+3 a x right)$$ to the result obtained from approach 1, the DE reduces to- $$v fracdvdz -v = - x^3a frac(c + 2b -4a) +x(b - 4a) + x.$$ RHS can't be made a function of just z. I am confused here because I thought Able equation of second kind can always be reduced to canonical form. The value of constants in my case is $a=4/3, b=10/21, c=4/15$.
$endgroup$
– Sunil Jaiswal
2 days ago
add a comment |
Your Answer
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1 Answer
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1 Answer
1
active
oldest
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active
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active
oldest
votes
$begingroup$
Hint:
Approach $1$:
This belongs to an Abel equation of the second kind.
Let $u=y+2$ ,
Then $dfracdudx=dfracdydx$
$thereforedfracu3dfracdudx+u-2+dfrac1x(a(u-2)^2+b(u-2)-c)=0$
$dfracu3dfracdudx+u-2+dfrac1x(au^2-(4a-b)u+4a-2b-c)=0$
$udfracdudx=-dfrac3au^2x-3left(1-dfrac4a-bxright)u+6-dfrac3(4a-2b-c)x$
Let $u=x^-3av$ ,
Then $dfracdudx=x^-3adfracdvdx-3ax^-3a-1v$
$therefore x^-3avleft(x^-3adfracdvdx-3ax^-3a-1vright)=-dfrac3ax^-6av^2x-3left(1-dfrac4a-bxright)x^-3av+6-dfrac3(4a-2b-c)x$
$x^-6avdfracdvdx-3ax^-6a-1v^2=-3ax^-6a-1v^2-3left(1-dfrac4a-bxright)x^-3av+6-dfrac3(4a-2b-c)x$
$x^-6avdfracdvdx=-3left(1-dfrac4a-bxright)x^-3av+6-dfrac3(4a-2b-c)x$
$vdfracdvdx=-3(x^3a-(4a-b)x^3a-1)v+6x^6a-3(4a-2b-c)x^6a-1$
Approach $2$:
$dfracy+23dfracdydx+y+dfracay^2+by-cx=0$
$y+dfracay^2+by-cx=-dfracy+23dfracdydx$
$left(x+ay+b-dfraccyright)dfracdxdy=-left(dfrac13+dfrac23yright)x$
This belongs to an Abel equation of the second kind.
Let $u=x+ay+b-dfraccy$ ,
Then $x=u-ay-b+dfraccy$
$dfracdxdy=dfracdudy-a-dfraccy^2$
$therefore uleft(dfracdudy-a-dfraccy^2right)=-left(dfrac13+dfrac23yright)left(u-ay-b+dfraccyright)$
$udfracdudy-left(a+dfraccy^2right)u=-left(dfrac13+dfrac23yright)u+dfrac(y+2)(ay^2+by-c)3y^2$
$udfracdudy=dfrac(3a-1)y^2-2y+3c3y^2u+dfrac(y+2)(ay^2+by-c)3y^2$
answered Mar 1 at 12:35
doraemonpauldoraemonpaul
13k31761
$endgroup$
$begingroup$
Thank you! That's extremely helpful.
$endgroup$
– Sunil Jaiswal
Mar 5 at 10:08
$begingroup$
I tried reducing the form given in approach 1 in normal form. Doing a variable transform, $$z=x^3 gamma left( frac4 a - ba - frac31+3 a x right)$$ to the result obtained from approach 1, the DE reduces to- $$v fracdvdz -v = - x^3a frac(c + 2b -4a) +x(b - 4a) + x.$$ RHS can't be made a function of just z. I am confused here because I thought Able equation of second kind can always be reduced to canonical form. The value of constants in my case is $a=4/3, b=10/21, c=4/15$.
$endgroup$
– Sunil Jaiswal
2 days ago
add a comment |
$begingroup$
Hint:
Approach $1$:
This belongs to an Abel equation of the second kind.
Let $u=y+2$ ,
Then $dfracdudx=dfracdydx$
$thereforedfracu3dfracdudx+u-2+dfrac1x(a(u-2)^2+b(u-2)-c)=0$
$dfracu3dfracdudx+u-2+dfrac1x(au^2-(4a-b)u+4a-2b-c)=0$
$udfracdudx=-dfrac3au^2x-3left(1-dfrac4a-bxright)u+6-dfrac3(4a-2b-c)x$
Let $u=x^-3av$ ,
Then $dfracdudx=x^-3adfracdvdx-3ax^-3a-1v$
$therefore x^-3avleft(x^-3adfracdvdx-3ax^-3a-1vright)=-dfrac3ax^-6av^2x-3left(1-dfrac4a-bxright)x^-3av+6-dfrac3(4a-2b-c)x$
$x^-6avdfracdvdx-3ax^-6a-1v^2=-3ax^-6a-1v^2-3left(1-dfrac4a-bxright)x^-3av+6-dfrac3(4a-2b-c)x$
$x^-6avdfracdvdx=-3left(1-dfrac4a-bxright)x^-3av+6-dfrac3(4a-2b-c)x$
$vdfracdvdx=-3(x^3a-(4a-b)x^3a-1)v+6x^6a-3(4a-2b-c)x^6a-1$
Approach $2$:
$dfracy+23dfracdydx+y+dfracay^2+by-cx=0$
$y+dfracay^2+by-cx=-dfracy+23dfracdydx$
$left(x+ay+b-dfraccyright)dfracdxdy=-left(dfrac13+dfrac23yright)x$
This belongs to an Abel equation of the second kind.
Let $u=x+ay+b-dfraccy$ ,
Then $x=u-ay-b+dfraccy$
$dfracdxdy=dfracdudy-a-dfraccy^2$
$therefore uleft(dfracdudy-a-dfraccy^2right)=-left(dfrac13+dfrac23yright)left(u-ay-b+dfraccyright)$
$udfracdudy-left(a+dfraccy^2right)u=-left(dfrac13+dfrac23yright)u+dfrac(y+2)(ay^2+by-c)3y^2$
$udfracdudy=dfrac(3a-1)y^2-2y+3c3y^2u+dfrac(y+2)(ay^2+by-c)3y^2$
answered Mar 1 at 12:35
doraemonpauldoraemonpaul
13k31761
$endgroup$
$begingroup$
Thank you! That's extremely helpful.
$endgroup$
– Sunil Jaiswal
Mar 5 at 10:08
$begingroup$
I tried reducing the form given in approach 1 in normal form. Doing a variable transform, $$z=x^3 gamma left( frac4 a - ba - frac31+3 a x right)$$ to the result obtained from approach 1, the DE reduces to- $$v fracdvdz -v = - x^3a frac(c + 2b -4a) +x(b - 4a) + x.$$ RHS can't be made a function of just z. I am confused here because I thought Able equation of second kind can always be reduced to canonical form. The value of constants in my case is $a=4/3, b=10/21, c=4/15$.
$endgroup$
– Sunil Jaiswal
2 days ago
add a comment |
$begingroup$
Hint:
Approach $1$:
This belongs to an Abel equation of the second kind.
Let $u=y+2$ ,
Then $dfracdudx=dfracdydx$
$thereforedfracu3dfracdudx+u-2+dfrac1x(a(u-2)^2+b(u-2)-c)=0$
$dfracu3dfracdudx+u-2+dfrac1x(au^2-(4a-b)u+4a-2b-c)=0$
$udfracdudx=-dfrac3au^2x-3left(1-dfrac4a-bxright)u+6-dfrac3(4a-2b-c)x$
Let $u=x^-3av$ ,
Then $dfracdudx=x^-3adfracdvdx-3ax^-3a-1v$
$therefore x^-3avleft(x^-3adfracdvdx-3ax^-3a-1vright)=-dfrac3ax^-6av^2x-3left(1-dfrac4a-bxright)x^-3av+6-dfrac3(4a-2b-c)x$
$x^-6avdfracdvdx-3ax^-6a-1v^2=-3ax^-6a-1v^2-3left(1-dfrac4a-bxright)x^-3av+6-dfrac3(4a-2b-c)x$
$x^-6avdfracdvdx=-3left(1-dfrac4a-bxright)x^-3av+6-dfrac3(4a-2b-c)x$
$vdfracdvdx=-3(x^3a-(4a-b)x^3a-1)v+6x^6a-3(4a-2b-c)x^6a-1$
Approach $2$:
$dfracy+23dfracdydx+y+dfracay^2+by-cx=0$
$y+dfracay^2+by-cx=-dfracy+23dfracdydx$
$left(x+ay+b-dfraccyright)dfracdxdy=-left(dfrac13+dfrac23yright)x$
This belongs to an Abel equation of the second kind.
Let $u=x+ay+b-dfraccy$ ,
Then $x=u-ay-b+dfraccy$
$dfracdxdy=dfracdudy-a-dfraccy^2$
$therefore uleft(dfracdudy-a-dfraccy^2right)=-left(dfrac13+dfrac23yright)left(u-ay-b+dfraccyright)$
$udfracdudy-left(a+dfraccy^2right)u=-left(dfrac13+dfrac23yright)u+dfrac(y+2)(ay^2+by-c)3y^2$
$udfracdudy=dfrac(3a-1)y^2-2y+3c3y^2u+dfrac(y+2)(ay^2+by-c)3y^2$
answered Mar 1 at 12:35
doraemonpauldoraemonpaul
13k31761
$endgroup$
Hint:
Approach $1$:
This belongs to an Abel equation of the second kind.
Let $u=y+2$ ,
Then $dfracdudx=dfracdydx$
$thereforedfracu3dfracdudx+u-2+dfrac1x(a(u-2)^2+b(u-2)-c)=0$
$dfracu3dfracdudx+u-2+dfrac1x(au^2-(4a-b)u+4a-2b-c)=0$
$udfracdudx=-dfrac3au^2x-3left(1-dfrac4a-bxright)u+6-dfrac3(4a-2b-c)x$
Let $u=x^-3av$ ,
Then $dfracdudx=x^-3adfracdvdx-3ax^-3a-1v$
$therefore x^-3avleft(x^-3adfracdvdx-3ax^-3a-1vright)=-dfrac3ax^-6av^2x-3left(1-dfrac4a-bxright)x^-3av+6-dfrac3(4a-2b-c)x$
$x^-6avdfracdvdx-3ax^-6a-1v^2=-3ax^-6a-1v^2-3left(1-dfrac4a-bxright)x^-3av+6-dfrac3(4a-2b-c)x$
$x^-6avdfracdvdx=-3left(1-dfrac4a-bxright)x^-3av+6-dfrac3(4a-2b-c)x$
$vdfracdvdx=-3(x^3a-(4a-b)x^3a-1)v+6x^6a-3(4a-2b-c)x^6a-1$
Approach $2$:
$dfracy+23dfracdydx+y+dfracay^2+by-cx=0$
$y+dfracay^2+by-cx=-dfracy+23dfracdydx$
$left(x+ay+b-dfraccyright)dfracdxdy=-left(dfrac13+dfrac23yright)x$
This belongs to an Abel equation of the second kind.
Let $u=x+ay+b-dfraccy$ ,
Then $x=u-ay-b+dfraccy$
$dfracdxdy=dfracdudy-a-dfraccy^2$
$therefore uleft(dfracdudy-a-dfraccy^2right)=-left(dfrac13+dfrac23yright)left(u-ay-b+dfraccyright)$
$udfracdudy-left(a+dfraccy^2right)u=-left(dfrac13+dfrac23yright)u+dfrac(y+2)(ay^2+by-c)3y^2$
$udfracdudy=dfrac(3a-1)y^2-2y+3c3y^2u+dfrac(y+2)(ay^2+by-c)3y^2$
answered Mar 1 at 12:35
doraemonpauldoraemonpaul
13k31761
edited Mar 26 at 8:57
answered Mar 1 at 12:35
doraemonpauldoraemonpaul
13k31761
answered Mar 1 at 12:35
doraemonpauldoraemonpaul
13k31761
answered Mar 1 at 12:35
doraemonpauldoraemonpaul
13k31761
13k31761
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Thank you! That's extremely helpful.
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– Sunil Jaiswal
Mar 5 at 10:08
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I tried reducing the form given in approach 1 in normal form. Doing a variable transform, $$z=x^3 gamma left( frac4 a - ba - frac31+3 a x right)$$ to the result obtained from approach 1, the DE reduces to- $$v fracdvdz -v = - x^3a frac(c + 2b -4a) +x(b - 4a) + x.$$ RHS can't be made a function of just z. I am confused here because I thought Able equation of second kind can always be reduced to canonical form. The value of constants in my case is $a=4/3, b=10/21, c=4/15$.
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– Sunil Jaiswal
2 days ago
add a comment |
$begingroup$
Thank you! That's extremely helpful.
$endgroup$
– Sunil Jaiswal
Mar 5 at 10:08
$begingroup$
I tried reducing the form given in approach 1 in normal form. Doing a variable transform, $$z=x^3 gamma left( frac4 a - ba - frac31+3 a x right)$$ to the result obtained from approach 1, the DE reduces to- $$v fracdvdz -v = - x^3a frac(c + 2b -4a) +x(b - 4a) + x.$$ RHS can't be made a function of just z. I am confused here because I thought Able equation of second kind can always be reduced to canonical form. The value of constants in my case is $a=4/3, b=10/21, c=4/15$.
$endgroup$
– Sunil Jaiswal
2 days ago
$begingroup$
Thank you! That's extremely helpful.
$endgroup$
– Sunil Jaiswal
Mar 5 at 10:08
$begingroup$
Thank you! That's extremely helpful.
$endgroup$
– Sunil Jaiswal
Mar 5 at 10:08
$begingroup$
I tried reducing the form given in approach 1 in normal form. Doing a variable transform, $$z=x^3 gamma left( frac4 a - ba - frac31+3 a x right)$$ to the result obtained from approach 1, the DE reduces to- $$v fracdvdz -v = - x^3a frac(c + 2b -4a) +x(b - 4a) + x.$$ RHS can't be made a function of just z. I am confused here because I thought Able equation of second kind can always be reduced to canonical form. The value of constants in my case is $a=4/3, b=10/21, c=4/15$.
$endgroup$
– Sunil Jaiswal
2 days ago
$begingroup$
I tried reducing the form given in approach 1 in normal form. Doing a variable transform, $$z=x^3 gamma left( frac4 a - ba - frac31+3 a x right)$$ to the result obtained from approach 1, the DE reduces to- $$v fracdvdz -v = - x^3a frac(c + 2b -4a) +x(b - 4a) + x.$$ RHS can't be made a function of just z. I am confused here because I thought Able equation of second kind can always be reduced to canonical form. The value of constants in my case is $a=4/3, b=10/21, c=4/15$.
$endgroup$
– Sunil Jaiswal
2 days ago
add a comment |
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How about using a power series?
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– Mattos
Feb 26 at 14:01
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@Mattos Power series solution for this DE diverges. Also I am looking for a closed form solution if possible.
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– Sunil Jaiswal
Feb 27 at 9:43
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I'm confused as to why you write the 'power series solution for this DE diverges'. The power series solution is the solution, hence if the power series diverges, then the solution diverges. You've already computed a power series solution, so you've shown a solution exists, but you can check to see if a solution exists and is unique using the Picard-Lindelof theorem i.e show that $partial f(x,y)/partial y$ exists somewhere for the ODE $y' = f(x,y(x))$, though this existence theorem might be too strenuous.
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– Mattos
Feb 27 at 9:56