Internal problem ID [10983]
Internal file name [OUTPUT/10240_Sunday_December_31_2023_11_10_41_AM_52504833/index.tex]
Book: Handbook of exact solutions for ordinary differential equations. By Polyanin and Zaitsev.
Second edition
Section: Chapter 2, Second-Order Differential Equations. section 2.1.2-5 Equation of form
\((a x^2+b x+c) y''+f(x)y'+g(x)y=0\)
Problem number: 160.
ODE order: 2.
ODE degree: 1.
The type(s) of ODE detected by this program : "unknown"
Maple gives the following as the ode type
[[_2nd_order, _with_linear_symmetries]]
Unable to solve or complete the solution.
\[ \boxed {\left (-x^{2}+1\right ) y^{\prime \prime }+\left (\beta -\alpha -\left (\alpha +\beta +2\right ) x \right ) y^{\prime }+n \left (n +\alpha +\beta +1\right ) y=0} \]
\[ \begin {array}{lll} & {} & \textrm {Let's solve}\hspace {3pt} \\ {} & {} & \left (-x^{2}+1\right ) \left (\frac {d}{d x}y^{\prime }\right )+\left (\left (-\beta -\alpha -2\right ) x +\beta -\alpha \right ) y^{\prime }+n \left (n +\alpha +\beta +1\right ) y=0 \\ \bullet & {} & \textrm {Highest derivative means the order of the ODE is}\hspace {3pt} 2 \\ {} & {} & \frac {d}{d x}y^{\prime } \\ \bullet & {} & \textrm {Isolate 2nd derivative}\hspace {3pt} \\ {} & {} & \frac {d}{d x}y^{\prime }=\frac {n \left (n +\alpha +\beta +1\right ) y}{x^{2}-1}-\frac {\left (x \alpha +x \beta +\alpha -\beta +2 x \right ) y^{\prime }}{x^{2}-1} \\ \bullet & {} & \textrm {Group terms with}\hspace {3pt} y\hspace {3pt}\textrm {on the lhs of the ODE and the rest on the rhs of the ODE; ODE is linear}\hspace {3pt} \\ {} & {} & \frac {d}{d x}y^{\prime }+\frac {\left (x \alpha +x \beta +\alpha -\beta +2 x \right ) y^{\prime }}{x^{2}-1}-\frac {n \left (n +\alpha +\beta +1\right ) y}{x^{2}-1}=0 \\ \square & {} & \textrm {Check to see if}\hspace {3pt} x_{0}\hspace {3pt}\textrm {is a regular singular point}\hspace {3pt} \\ {} & \circ & \textrm {Define functions}\hspace {3pt} \\ {} & {} & \left [P_{2}\left (x \right )=\frac {x \alpha +x \beta +\alpha -\beta +2 x}{x^{2}-1}, P_{3}\left (x \right )=-\frac {n \left (n +\alpha +\beta +1\right )}{x^{2}-1}\right ] \\ {} & \circ & \left (x +1\right )\cdot P_{2}\left (x \right )\textrm {is analytic at}\hspace {3pt} x =-1 \\ {} & {} & \left (\left (x +1\right )\cdot P_{2}\left (x \right )\right )\bigg | {\mstack {}{_{x \hiderel {=}-1}}}=\beta +1 \\ {} & \circ & \left (x +1\right )^{2}\cdot P_{3}\left (x \right )\textrm {is analytic at}\hspace {3pt} x =-1 \\ {} & {} & \left (\left (x +1\right )^{2}\cdot P_{3}\left (x \right )\right )\bigg | {\mstack {}{_{x \hiderel {=}-1}}}=0 \\ {} & \circ & x =-1\textrm {is a regular singular point}\hspace {3pt} \\ & {} & \textrm {Check to see if}\hspace {3pt} x_{0}\hspace {3pt}\textrm {is a regular singular point}\hspace {3pt} \\ {} & {} & x_{0}=-1 \\ \bullet & {} & \textrm {Multiply by denominators}\hspace {3pt} \\ {} & {} & \left (x^{2}-1\right ) \left (\frac {d}{d x}y^{\prime }\right )+\left (x \alpha +x \beta +\alpha -\beta +2 x \right ) y^{\prime }-n \left (n +\alpha +\beta +1\right ) y=0 \\ \bullet & {} & \textrm {Change variables using}\hspace {3pt} x =u -1\hspace {3pt}\textrm {so that the regular singular point is at}\hspace {3pt} u =0 \\ {} & {} & \left (u^{2}-2 u \right ) \left (\frac {d}{d u}\frac {d}{d u}y \left (u \right )\right )+\left (\alpha u +\beta u -2 \beta +2 u -2\right ) \left (\frac {d}{d u}y \left (u \right )\right )+\left (-\alpha n -\beta n -n^{2}-n \right ) y \left (u \right )=0 \\ \bullet & {} & \textrm {Assume series solution for}\hspace {3pt} y \left (u \right ) \\ {} & {} & y \left (u \right )=\moverset {\infty }{\munderset {k =0}{\sum }}a_{k} u^{k +r} \\ \square & {} & \textrm {Rewrite ODE with series expansions}\hspace {3pt} \\ {} & \circ & \textrm {Convert}\hspace {3pt} u^{m}\cdot \left (\frac {d}{d u}y \left (u \right )\right )\hspace {3pt}\textrm {to series expansion for}\hspace {3pt} m =0..1 \\ {} & {} & u^{m}\cdot \left (\frac {d}{d u}y \left (u \right )\right )=\moverset {\infty }{\munderset {k =0}{\sum }}a_{k} \left (k +r \right ) u^{k +r -1+m} \\ {} & \circ & \textrm {Shift index using}\hspace {3pt} k \mathrm {->}k +1-m \\ {} & {} & u^{m}\cdot \left (\frac {d}{d u}y \left (u \right )\right )=\moverset {\infty }{\munderset {k =-1+m}{\sum }}a_{k +1-m} \left (k +1-m +r \right ) u^{k +r} \\ {} & \circ & \textrm {Convert}\hspace {3pt} u^{m}\cdot \left (\frac {d}{d u}\frac {d}{d u}y \left (u \right )\right )\hspace {3pt}\textrm {to series expansion for}\hspace {3pt} m =1..2 \\ {} & {} & u^{m}\cdot \left (\frac {d}{d u}\frac {d}{d u}y \left (u \right )\right )=\moverset {\infty }{\munderset {k =0}{\sum }}a_{k} \left (k +r \right ) \left (k +r -1\right ) u^{k +r -2+m} \\ {} & \circ & \textrm {Shift index using}\hspace {3pt} k \mathrm {->}k +2-m \\ {} & {} & u^{m}\cdot \left (\frac {d}{d u}\frac {d}{d u}y \left (u \right )\right )=\moverset {\infty }{\munderset {k =-2+m}{\sum }}a_{k +2-m} \left (k +2-m +r \right ) \left (k +1-m +r \right ) u^{k +r} \\ & {} & \textrm {Rewrite ODE with series expansions}\hspace {3pt} \\ {} & {} & -2 a_{0} r \left (r +\beta \right ) u^{-1+r}+\left (\moverset {\infty }{\munderset {k =0}{\sum }}\left (-2 a_{k +1} \left (k +1+r \right ) \left (k +1+r +\beta \right )+a_{k} \left (k -n +r \right ) \left (k +\alpha +\beta +r +1+n \right )\right ) u^{k +r}\right )=0 \\ \bullet & {} & a_{0}\textrm {cannot be 0 by assumption, giving the indicial equation}\hspace {3pt} \\ {} & {} & -2 r \left (r +\beta \right )=0 \\ \bullet & {} & \textrm {Values of r that satisfy the indicial equation}\hspace {3pt} \\ {} & {} & r \in \left \{0, -\beta \right \} \\ \bullet & {} & \textrm {Each term in the series must be 0, giving the recursion relation}\hspace {3pt} \\ {} & {} & -2 a_{k +1} \left (k +1+r \right ) \left (k +1+r +\beta \right )+a_{k} \left (k -n +r \right ) \left (k +\alpha +\beta +r +1+n \right )=0 \\ \bullet & {} & \textrm {Recursion relation that defines series solution to ODE}\hspace {3pt} \\ {} & {} & a_{k +1}=\frac {a_{k} \left (k -n +r \right ) \left (k +\alpha +\beta +r +1+n \right )}{2 \left (k +1+r \right ) \left (k +1+r +\beta \right )} \\ \bullet & {} & \textrm {Recursion relation for}\hspace {3pt} r =0 \\ {} & {} & a_{k +1}=\frac {a_{k} \left (k -n \right ) \left (k +\alpha +\beta +1+n \right )}{2 \left (k +1\right ) \left (k +1+\beta \right )} \\ \bullet & {} & \textrm {Solution for}\hspace {3pt} r =0 \\ {} & {} & \left [y \left (u \right )=\moverset {\infty }{\munderset {k =0}{\sum }}a_{k} u^{k}, a_{k +1}=\frac {a_{k} \left (k -n \right ) \left (k +\alpha +\beta +1+n \right )}{2 \left (k +1\right ) \left (k +1+\beta \right )}\right ] \\ \bullet & {} & \textrm {Revert the change of variables}\hspace {3pt} u =x +1 \\ {} & {} & \left [y=\moverset {\infty }{\munderset {k =0}{\sum }}a_{k} \left (x +1\right )^{k}, a_{k +1}=\frac {a_{k} \left (k -n \right ) \left (k +\alpha +\beta +1+n \right )}{2 \left (k +1\right ) \left (k +1+\beta \right )}\right ] \\ \bullet & {} & \textrm {Recursion relation for}\hspace {3pt} r =-\beta \\ {} & {} & a_{k +1}=\frac {a_{k} \left (k -n -\beta \right ) \left (k +\alpha +1+n \right )}{2 \left (k +1-\beta \right ) \left (k +1\right )} \\ \bullet & {} & \textrm {Solution for}\hspace {3pt} r =-\beta \\ {} & {} & \left [y \left (u \right )=\moverset {\infty }{\munderset {k =0}{\sum }}a_{k} u^{k -\beta }, a_{k +1}=\frac {a_{k} \left (k -n -\beta \right ) \left (k +\alpha +1+n \right )}{2 \left (k +1-\beta \right ) \left (k +1\right )}\right ] \\ \bullet & {} & \textrm {Revert the change of variables}\hspace {3pt} u =x +1 \\ {} & {} & \left [y=\moverset {\infty }{\munderset {k =0}{\sum }}a_{k} \left (x +1\right )^{k -\beta }, a_{k +1}=\frac {a_{k} \left (k -n -\beta \right ) \left (k +\alpha +1+n \right )}{2 \left (k +1-\beta \right ) \left (k +1\right )}\right ] \\ \bullet & {} & \textrm {Combine solutions and rename parameters}\hspace {3pt} \\ {} & {} & \left [y=\left (\moverset {\infty }{\munderset {k =0}{\sum }}a_{k} \left (x +1\right )^{k}\right )+\left (\moverset {\infty }{\munderset {k =0}{\sum }}b_{k} \left (x +1\right )^{k -\beta }\right ), a_{k +1}=\frac {a_{k} \left (k -n \right ) \left (k +\alpha +\beta +1+n \right )}{2 \left (k +1\right ) \left (k +1+\beta \right )}, b_{k +1}=\frac {b_{k} \left (k -n -\beta \right ) \left (k +\alpha +1+n \right )}{2 \left (k +1-\beta \right ) \left (k +1\right )}\right ] \end {array} \]
Maple trace
`Methods for second order ODEs: --- Trying classification methods --- trying a quadrature checking if the LODE has constant coefficients checking if the LODE is of Euler type trying a symmetry of the form [xi=0, eta=F(x)] checking if the LODE is missing y -> Trying a Liouvillian solution using Kovacics algorithm <- No Liouvillian solutions exists -> Trying a solution in terms of special functions: -> Bessel -> elliptic -> Legendre -> Kummer -> hyper3: Equivalence to 1F1 under a power @ Moebius -> hypergeometric -> heuristic approach <- heuristic approach successful <- hypergeometric successful <- special function solution successful`
✓ Solution by Maple
Time used: 0.094 (sec). Leaf size: 61
dsolve((1-x^2)*diff(y(x),x$2)+(beta-alpha-(alpha+beta+2)*x)*diff(y(x),x)+n*(n+alpha+beta+1)*y(x)=0,y(x), singsol=all)
\[ y \left (x \right ) = c_{1} \operatorname {hypergeom}\left (\left [-n , n +\alpha +\beta +1\right ], \left [\beta +1\right ], \frac {1}{2}+\frac {x}{2}\right )+c_{2} \left (\frac {1}{2}+\frac {x}{2}\right )^{-\beta } \operatorname {hypergeom}\left (\left [-n -\beta , n +\alpha +1\right ], \left [1-\beta \right ], \frac {1}{2}+\frac {x}{2}\right ) \]
✓ Solution by Mathematica
Time used: 0.28 (sec). Leaf size: 69
DSolve[(1-x^2)*y''[x]+(\[Beta]-\[Alpha]-(\[Alpha]+\[Beta]+2)*x)*y'[x]+n*(n+\[Alpha]+\[Beta]+1)*y[x]==0,y[x],x,IncludeSingularSolutions -> True]
\[ y(x)\to 2^{\alpha } c_2 (x-1)^{-\alpha } \operatorname {Hypergeometric2F1}\left (-n-\alpha ,n+\beta +1,1-\alpha ,\frac {1-x}{2}\right )+c_1 \operatorname {Hypergeometric2F1}\left (-n,n+\alpha +\beta +1,\alpha +1,\frac {1-x}{2}\right ) \]