First order ODE and hyper geometric function Announcing the arrival of Valued Associate #679: Cesar Manara Planned maintenance scheduled April 23, 2019 at 23:30 UTC (7:30pm US/Eastern)Does every ODE have a first integral?First order ODE definitionFirst Order ODE NotationVector fields and first integrals of ODESolution methods for first order nonlinear ODE - populationSolution of a nonlinear first order ODECan we convert first order ode to second order ode?Solve first order ordinary differential equation (ODE).Clairaut ODE and first order high degree ODEadvance techniques for converting second-order linear ODEs with rational function coefficients to some known ODE types
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First order ODE and hyper geometric function
Announcing the arrival of Valued Associate #679: Cesar Manara
Planned maintenance scheduled April 23, 2019 at 23:30 UTC (7:30pm US/Eastern)Does every ODE have a first integral?First order ODE definitionFirst Order ODE NotationVector fields and first integrals of ODESolution methods for first order nonlinear ODE - populationSolution of a nonlinear first order ODECan we convert first order ode to second order ode?Solve first order ordinary differential equation (ODE).Clairaut ODE and first order high degree ODEadvance techniques for converting second-order linear ODEs with rational function coefficients to some known ODE types
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Is it possible that the solution to a first order differential equation involves a hyper geometric function? I thought that only second order ODE solution contains hypergeometric function. I am relying on Wolfram to find the solution.
ordinary-differential-equations
asked Apr 2 at 16:35
AshAsh
1
$endgroup$
add a comment |
$begingroup$
Is it possible that the solution to a first order differential equation involves a hyper geometric function? I thought that only second order ODE solution contains hypergeometric function. I am relying on Wolfram to find the solution.
ordinary-differential-equations
asked Apr 2 at 16:35
AshAsh
1
$endgroup$
$begingroup$
I think you need to put more conditions on your equation. I suspect your cited claim is only true for linear DE, and you are dealing with a non-linear Riccati DE or similar.
$endgroup$
– LutzL
Apr 2 at 16:45
$begingroup$
Which equation, which hypergeometric function? $exp,arcsin, ln$ are all hypergeometric functions and solutions of simple first order differential equations.
$endgroup$
– user647486
Apr 2 at 16:45
$begingroup$
Solve the linear equation -3/4 x^n ( dy(x))/( dx) - 3/20 x^(n - 1) ( dy(x))/( dx) + x^n (n - 1) - 3/20 x^(n - 2) (n - 1) y(x) - 3/4 x^(n - 1) (n - 1) y(x) = 0, such that y(0) = 0:
$endgroup$
– Ash
Apr 2 at 17:50
add a comment |
$begingroup$
Is it possible that the solution to a first order differential equation involves a hyper geometric function? I thought that only second order ODE solution contains hypergeometric function. I am relying on Wolfram to find the solution.
ordinary-differential-equations
asked Apr 2 at 16:35
AshAsh
1
$endgroup$
Is it possible that the solution to a first order differential equation involves a hyper geometric function? I thought that only second order ODE solution contains hypergeometric function. I am relying on Wolfram to find the solution.
ordinary-differential-equations
ordinary-differential-equations
asked Apr 2 at 16:35
AshAsh
1
asked Apr 2 at 16:35
AshAsh
1
asked Apr 2 at 16:35
AshAsh
1
asked Apr 2 at 16:35
AshAsh
1
asked Apr 2 at 16:35
AshAsh
1
1
$begingroup$
I think you need to put more conditions on your equation. I suspect your cited claim is only true for linear DE, and you are dealing with a non-linear Riccati DE or similar.
$endgroup$
– LutzL
Apr 2 at 16:45
$begingroup$
Which equation, which hypergeometric function? $exp,arcsin, ln$ are all hypergeometric functions and solutions of simple first order differential equations.
$endgroup$
– user647486
Apr 2 at 16:45
$begingroup$
Solve the linear equation -3/4 x^n ( dy(x))/( dx) - 3/20 x^(n - 1) ( dy(x))/( dx) + x^n (n - 1) - 3/20 x^(n - 2) (n - 1) y(x) - 3/4 x^(n - 1) (n - 1) y(x) = 0, such that y(0) = 0:
$endgroup$
– Ash
Apr 2 at 17:50
add a comment |
$begingroup$
I think you need to put more conditions on your equation. I suspect your cited claim is only true for linear DE, and you are dealing with a non-linear Riccati DE or similar.
$endgroup$
– LutzL
Apr 2 at 16:45
$begingroup$
Which equation, which hypergeometric function? $exp,arcsin, ln$ are all hypergeometric functions and solutions of simple first order differential equations.
$endgroup$
– user647486
Apr 2 at 16:45
$begingroup$
Solve the linear equation -3/4 x^n ( dy(x))/( dx) - 3/20 x^(n - 1) ( dy(x))/( dx) + x^n (n - 1) - 3/20 x^(n - 2) (n - 1) y(x) - 3/4 x^(n - 1) (n - 1) y(x) = 0, such that y(0) = 0:
$endgroup$
– Ash
Apr 2 at 17:50
$begingroup$
I think you need to put more conditions on your equation. I suspect your cited claim is only true for linear DE, and you are dealing with a non-linear Riccati DE or similar.
$endgroup$
– LutzL
Apr 2 at 16:45
$begingroup$
I think you need to put more conditions on your equation. I suspect your cited claim is only true for linear DE, and you are dealing with a non-linear Riccati DE or similar.
$endgroup$
– LutzL
Apr 2 at 16:45
$begingroup$
Which equation, which hypergeometric function? $exp,arcsin, ln$ are all hypergeometric functions and solutions of simple first order differential equations.
$endgroup$
– user647486
Apr 2 at 16:45
$begingroup$
Which equation, which hypergeometric function? $exp,arcsin, ln$ are all hypergeometric functions and solutions of simple first order differential equations.
$endgroup$
– user647486
Apr 2 at 16:45
$begingroup$
Solve the linear equation -3/4 x^n ( dy(x))/( dx) - 3/20 x^(n - 1) ( dy(x))/( dx) + x^n (n - 1) - 3/20 x^(n - 2) (n - 1) y(x) - 3/4 x^(n - 1) (n - 1) y(x) = 0, such that y(0) = 0:
$endgroup$
– Ash
Apr 2 at 17:50
$begingroup$
Solve the linear equation -3/4 x^n ( dy(x))/( dx) - 3/20 x^(n - 1) ( dy(x))/( dx) + x^n (n - 1) - 3/20 x^(n - 2) (n - 1) y(x) - 3/4 x^(n - 1) (n - 1) y(x) = 0, such that y(0) = 0:
$endgroup$
– Ash
Apr 2 at 17:50
add a comment |
1 Answer
1
active
oldest
votes
$begingroup$
A first-order Riccati equation "reduces" to a second-order linear equation, which might be hypergeometric. For example, the Riccati equation
$$ frac rm drm dxv left( x right) = left( v left( x
right) right) ^2-frac left( left( a+b+1 right) x-c
right) v left( x right) x^2-x+frac abx^2-x=0
$$
has solutions $v = -u'/u$ where $u$ satisfies the hypergeometric equation
$$ (x^2-x) u'' + ((a+b+1) x - c) u' + ab u = 0$$
and so one of its solutions is
$$ - ;_2F_1(a,b; c; x)^-1 dfracddx _2F_1(a,b; c; x) = -frac ab;mbox$_2$F$_1$(b+1,a+1;,c+1;,x)
c; mbox$_2$F$_1$(a,b;,c;,x)
$$
answered Apr 2 at 16:48
Robert IsraelRobert Israel
332k23222482
$endgroup$
add a comment |
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1 Answer
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1 Answer
1
active
oldest
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active
oldest
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active
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votes
$begingroup$
A first-order Riccati equation "reduces" to a second-order linear equation, which might be hypergeometric. For example, the Riccati equation
$$ frac rm drm dxv left( x right) = left( v left( x
right) right) ^2-frac left( left( a+b+1 right) x-c
right) v left( x right) x^2-x+frac abx^2-x=0
$$
has solutions $v = -u'/u$ where $u$ satisfies the hypergeometric equation
$$ (x^2-x) u'' + ((a+b+1) x - c) u' + ab u = 0$$
and so one of its solutions is
$$ - ;_2F_1(a,b; c; x)^-1 dfracddx _2F_1(a,b; c; x) = -frac ab;mbox$_2$F$_1$(b+1,a+1;,c+1;,x)
c; mbox$_2$F$_1$(a,b;,c;,x)
$$
answered Apr 2 at 16:48
Robert IsraelRobert Israel
332k23222482
$endgroup$
add a comment |
$begingroup$
A first-order Riccati equation "reduces" to a second-order linear equation, which might be hypergeometric. For example, the Riccati equation
$$ frac rm drm dxv left( x right) = left( v left( x
right) right) ^2-frac left( left( a+b+1 right) x-c
right) v left( x right) x^2-x+frac abx^2-x=0
$$
has solutions $v = -u'/u$ where $u$ satisfies the hypergeometric equation
$$ (x^2-x) u'' + ((a+b+1) x - c) u' + ab u = 0$$
and so one of its solutions is
$$ - ;_2F_1(a,b; c; x)^-1 dfracddx _2F_1(a,b; c; x) = -frac ab;mbox$_2$F$_1$(b+1,a+1;,c+1;,x)
c; mbox$_2$F$_1$(a,b;,c;,x)
$$
answered Apr 2 at 16:48
Robert IsraelRobert Israel
332k23222482
$endgroup$
add a comment |
$begingroup$
A first-order Riccati equation "reduces" to a second-order linear equation, which might be hypergeometric. For example, the Riccati equation
$$ frac rm drm dxv left( x right) = left( v left( x
right) right) ^2-frac left( left( a+b+1 right) x-c
right) v left( x right) x^2-x+frac abx^2-x=0
$$
has solutions $v = -u'/u$ where $u$ satisfies the hypergeometric equation
$$ (x^2-x) u'' + ((a+b+1) x - c) u' + ab u = 0$$
and so one of its solutions is
$$ - ;_2F_1(a,b; c; x)^-1 dfracddx _2F_1(a,b; c; x) = -frac ab;mbox$_2$F$_1$(b+1,a+1;,c+1;,x)
c; mbox$_2$F$_1$(a,b;,c;,x)
$$
answered Apr 2 at 16:48
Robert IsraelRobert Israel
332k23222482
$endgroup$
A first-order Riccati equation "reduces" to a second-order linear equation, which might be hypergeometric. For example, the Riccati equation
$$ frac rm drm dxv left( x right) = left( v left( x
right) right) ^2-frac left( left( a+b+1 right) x-c
right) v left( x right) x^2-x+frac abx^2-x=0
$$
has solutions $v = -u'/u$ where $u$ satisfies the hypergeometric equation
$$ (x^2-x) u'' + ((a+b+1) x - c) u' + ab u = 0$$
and so one of its solutions is
$$ - ;_2F_1(a,b; c; x)^-1 dfracddx _2F_1(a,b; c; x) = -frac ab;mbox$_2$F$_1$(b+1,a+1;,c+1;,x)
c; mbox$_2$F$_1$(a,b;,c;,x)
$$
answered Apr 2 at 16:48
Robert IsraelRobert Israel
332k23222482
edited Apr 2 at 16:55
answered Apr 2 at 16:48
Robert IsraelRobert Israel
332k23222482
answered Apr 2 at 16:48
Robert IsraelRobert Israel
332k23222482
answered Apr 2 at 16:48
Robert IsraelRobert Israel
332k23222482
332k23222482
add a comment |
add a comment |
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$begingroup$
I think you need to put more conditions on your equation. I suspect your cited claim is only true for linear DE, and you are dealing with a non-linear Riccati DE or similar.
$endgroup$
– LutzL
Apr 2 at 16:45
$begingroup$
Which equation, which hypergeometric function? $exp,arcsin, ln$ are all hypergeometric functions and solutions of simple first order differential equations.
$endgroup$
– user647486
Apr 2 at 16:45
$begingroup$
Solve the linear equation -3/4 x^n ( dy(x))/( dx) - 3/20 x^(n - 1) ( dy(x))/( dx) + x^n (n - 1) - 3/20 x^(n - 2) (n - 1) y(x) - 3/4 x^(n - 1) (n - 1) y(x) = 0, such that y(0) = 0:
$endgroup$
– Ash
Apr 2 at 17:50