The candidate_points command determines candidate points for which power series solutions with d'Alembertian, hypergeometric, rational, or polynomial coefficients of the given linear ordinary differential equation exist.

If ode is an expression, then it is equated to zero.

The command returns an error message if the differential equation ode does not satisfy the following conditions.

ode must be linear in var

ode must have polynomial coefficients in $x$ over the rational number field which can be extended by one or more parameters.

ode must either be homogeneous or have a right hand side that is rational in $x$

If opt=all, the output is a list of three elements:

a set of hypergeometric points, which may include the symbol 'any_ordinary_point'

a set of rational points;

a set of polynomial points.

Otherwise, the output is the set of the required points.

Note that the computation of candidate points for power series solutions with d'Alembertian coefficients is currently considerably more expensive computationally than for the other three types of coefficients.

$\mathrm{with}\left(\mathrm{Slode}\right)\:$

$\mathrm{ode}≔\left(3x^2-6x+3\right)\mathrm{diff}\left(\mathrm{diff}\left(y\left(x\right)\,x\right)\,x\right)+\left(12x-12\right)\mathrm{diff}\left(y\left(x\right)\,x\right)+6y\left(x\right)$

$\mathrm{ode}≔\left(3x^2-6x+3\right)\left(\frac{{\ⅆ}^2}{\ⅆx^2}y\left(x\right)\right)+\left(12x-12\right)\left(\frac{\ⅆ}{\ⅆx}y\left(x\right)\right)+6y\left(x\right)$

$\mathrm{candidate\_points}\left(\mathrm{ode}\,y\left(x\right)\,'\mathrm{type}'='\mathrm{polynomial}'\right)$

$\left\{0\right\}$

$\mathrm{candidate\_points}\left(\mathrm{ode}\,y\left(x\right)\,'\mathrm{type}'='\mathrm{rational}'\right)$

$\left\{1\right\}$

$\mathrm{candidate\_points}\left(\mathrm{ode}\,y\left(x\right)\,'\mathrm{type}'='\mathrm{hypergeometric}'\right)$

$\left\{1\,\mathrm{any\_ordinary\_point}\right\}$

$\mathrm{candidate\_points}\left(\mathrm{ode}\,y\left(x\right)\,'\mathrm{type}'='\mathrm{all}'\right)$

$\left[\left\{1\,\mathrm{any\_ordinary\_point}\right\}\,\left\{1\right\}\,\left\{0\right\}\right]$

$\mathrm{candidate\_points}\left(\mathrm{ode}\,y\left(x\right)\,'\mathrm{type}'='\mathrm{dAlembertian}'\right)$

$\left\{1\,\mathrm{any\_ordinary\_point}\right\}$

$\mathrm{ode1}≔60y\left(x\right)+2x\left(x-30\right)\mathrm{diff}\left(y\left(x\right)\,x\right)-x^2\left(2x-27\right)\mathrm{diff}\left(y\left(x\right)\,x\,x\right)+x^3\left(4x-27\right)\mathrm{diff}\left(y\left(x\right)\,x\,x\,x\right)=-\frac{2x^2\left(-5-330x+60x^4-1137x^2+32x^3\right)}{{\left(x-1\right)}^6}$

$\mathrm{ode1}≔60y\left(x\right)+2x\left(x-30\right)\left(\frac{\ⅆ}{\ⅆx}y\left(x\right)\right)-x^2\left(2x-27\right)\left(\frac{{\ⅆ}^2}{\ⅆx^2}y\left(x\right)\right)+x^3\left(4x-27\right)\left(\frac{{\ⅆ}^3}{\ⅆx^3}y\left(x\right)\right)=-\frac{2x^2\left(60x^4+32x^3-1137x^2-330x-5\right)}{{\left(x-1\right)}^6}$

$\mathrm{candidate\_points}\left(\mathrm{ode1}\,y\left(x\right)\right)$

$\left[\left\{0\,1\,\frac{27}{4}\,\mathrm{any\_ordinary\_point}\,\mathrm{RootOf}\left(60{\mathrm{\_Z}}^4+32{\mathrm{\_Z}}^3-1137{\mathrm{\_Z}}^2-330\mathrm{\_Z}-5\right)\right\}\,\left\{\mathrm{-1}\,0\,1\,\frac{23}{4}\,\frac{27}{4}\,\mathrm{RootOf}\left(49{\mathrm{\_Z}}^4-287{\mathrm{\_Z}}^3-1418{\mathrm{\_Z}}^2-714\mathrm{\_Z}-45\right)-1\,\mathrm{RootOf}\left(49{\mathrm{\_Z}}^4-287{\mathrm{\_Z}}^3-1418{\mathrm{\_Z}}^2-714\mathrm{\_Z}-45\right)\,\mathrm{RootOf}\left(60{\mathrm{\_Z}}^4+32{\mathrm{\_Z}}^3-1137{\mathrm{\_Z}}^2-330\mathrm{\_Z}-5\right)\right\}\,\left\{\mathrm{-1}\,0\,\frac{23}{4}\,\mathrm{RootOf}\left(60{\mathrm{\_Z}}^4+32{\mathrm{\_Z}}^3-1137{\mathrm{\_Z}}^2-330\mathrm{\_Z}-5\right)-1\right\}\right]$

$\mathrm{ode2}≔\left(x-1\right)\mathrm{diff}\left(y\left(x\right)\,x\right)-\left(x-2\right)y\left(x\right)$

$\mathrm{ode2}≔\left(x-1\right)\left(\frac{\ⅆ}{\ⅆx}y\left(x\right)\right)-\left(x-2\right)y\left(x\right)$

$\mathrm{candidate\_points}\left(\mathrm{ode2}\,y\left(x\right)\,'\mathrm{type}'='\mathrm{hypergeometric}'\right)$

$\left\{1\right\}$

$\mathrm{candidate\_points}\left(\mathrm{ode2}\,y\left(x\right)\,'\mathrm{type}'='\mathrm{dAlembertian}'\right)$

$\left\{1\,\mathrm{any\_ordinary\_point}\right\}$