2.31.23 Problem 171

Internal problem ID [13832]
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
Problem number : 171
Date solved : Friday, December 19, 2025 at 02:30:30 PM
CAS classification : [_Jacobi]

\begin{align*} x \left (x -1\right ) y^{\prime \prime }+\left (\left (\alpha +\beta +1\right ) x -\gamma \right ) y^{\prime }+\alpha \beta y&=0 \\ \end{align*}

2.31.23.1 ✓ Maple. Time used: 0.010 (sec). Leaf size: 44

ode:=x*(x-1)*diff(diff(y(x),x),x)+((alpha+beta+1)*x-gamma)*diff(y(x),x)+alpha*beta*y(x) = 0; 
dsolve(ode,y(x), singsol=all);
 

\[ y = c_1 \operatorname {hypergeom}\left (\left [\alpha , \beta \right ], \left [\gamma \right ], x\right )+c_2 \,x^{1-\gamma } \operatorname {hypergeom}\left (\left [\beta +1-\gamma , \alpha +1-\gamma \right ], \left [2-\gamma \right ], x\right ) \]

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
 

Maple step by step

\[ \begin {array}{lll} & {} & \textrm {Let's solve}\hspace {3pt} \\ {} & {} & x \left (x -1\right ) \left (\frac {d^{2}}{d x^{2}}y \left (x \right )\right )+\left (\left (\alpha +\beta +1\right ) x -\gamma \right ) \left (\frac {d}{d x}y \left (x \right )\right )+\alpha \beta y \left (x \right )=0 \\ \bullet & {} & \textrm {Highest derivative means the order of the ODE is}\hspace {3pt} 2 \\ {} & {} & \frac {d^{2}}{d x^{2}}y \left (x \right ) \\ \bullet & {} & \textrm {Isolate 2nd derivative}\hspace {3pt} \\ {} & {} & \frac {d^{2}}{d x^{2}}y \left (x \right )=-\frac {\alpha \beta y \left (x \right )}{x \left (x -1\right )}-\frac {\left (x \alpha +x \beta -\gamma +x \right ) \left (\frac {d}{d x}y \left (x \right )\right )}{x \left (x -1\right )} \\ \bullet & {} & \textrm {Group terms with}\hspace {3pt} y \left (x \right )\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^{2}}{d x^{2}}y \left (x \right )+\frac {\left (x \alpha +x \beta -\gamma +x \right ) \left (\frac {d}{d x}y \left (x \right )\right )}{x \left (x -1\right )}+\frac {\alpha \beta y \left (x \right )}{x \left (x -1\right )}=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 -\gamma +x}{x \left (x -1\right )}, P_{3}\left (x \right )=\frac {\alpha \beta }{x \left (x -1\right )}\right ] \\ {} & \circ & x \cdot P_{2}\left (x \right )\textrm {is analytic at}\hspace {3pt} x =0 \\ {} & {} & \left (x \cdot P_{2}\left (x \right )\right )\bigg | {\mstack {}{_{x \hiderel {=}0}}}=\gamma \\ {} & \circ & x^{2}\cdot P_{3}\left (x \right )\textrm {is analytic at}\hspace {3pt} x =0 \\ {} & {} & \left (x^{2}\cdot P_{3}\left (x \right )\right )\bigg | {\mstack {}{_{x \hiderel {=}0}}}=0 \\ {} & \circ & x =0\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}=0 \\ \bullet & {} & \textrm {Multiply by denominators}\hspace {3pt} \\ {} & {} & x \left (x -1\right ) \left (\frac {d^{2}}{d x^{2}}y \left (x \right )\right )+\left (x \alpha +x \beta -\gamma +x \right ) \left (\frac {d}{d x}y \left (x \right )\right )+\alpha \beta y \left (x \right )=0 \\ \bullet & {} & \textrm {Assume series solution for}\hspace {3pt} y \left (x \right ) \\ {} & {} & y \left (x \right )=\moverset {\infty }{\munderset {k =0}{\sum }}a_{k} x^{k +r} \\ \square & {} & \textrm {Rewrite ODE with series expansions}\hspace {3pt} \\ {} & \circ & \textrm {Convert}\hspace {3pt} x^{m}\cdot \left (\frac {d}{d x}y \left (x \right )\right )\hspace {3pt}\textrm {to series expansion for}\hspace {3pt} m =0..1 \\ {} & {} & x^{m}\cdot \left (\frac {d}{d x}y \left (x \right )\right )=\moverset {\infty }{\munderset {k =0}{\sum }}a_{k} \left (k +r \right ) x^{k +r -1+m} \\ {} & \circ & \textrm {Shift index using}\hspace {3pt} k \mathrm {->}k +1-m \\ {} & {} & x^{m}\cdot \left (\frac {d}{d x}y \left (x \right )\right )=\moverset {\infty }{\munderset {k =-1+m}{\sum }}a_{k +1-m} \left (k +1-m +r \right ) x^{k +r} \\ {} & \circ & \textrm {Convert}\hspace {3pt} x^{m}\cdot \left (\frac {d^{2}}{d x^{2}}y \left (x \right )\right )\hspace {3pt}\textrm {to series expansion for}\hspace {3pt} m =1..2 \\ {} & {} & x^{m}\cdot \left (\frac {d^{2}}{d x^{2}}y \left (x \right )\right )=\moverset {\infty }{\munderset {k =0}{\sum }}a_{k} \left (k +r \right ) \left (k +r -1\right ) x^{k +r -2+m} \\ {} & \circ & \textrm {Shift index using}\hspace {3pt} k \mathrm {->}k +2-m \\ {} & {} & x^{m}\cdot \left (\frac {d^{2}}{d x^{2}}y \left (x \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 ) x^{k +r} \\ & {} & \textrm {Rewrite ODE with series expansions}\hspace {3pt} \\ {} & {} & -a_{0} r \left (-1+r +\gamma \right ) x^{-1+r}+\left (\moverset {\infty }{\munderset {k =0}{\sum }}\left (-a_{k +1} \left (k +1+r \right ) \left (k +r +\gamma \right )+a_{k} \left (\beta +k +r \right ) \left (\alpha +k +r \right )\right ) x^{k +r}\right )=0 \\ \bullet & {} & a_{0}\textrm {cannot be 0 by assumption, giving the indicial equation}\hspace {3pt} \\ {} & {} & -r \left (-1+r +\gamma \right )=0 \\ \bullet & {} & \textrm {Values of r that satisfy the indicial equation}\hspace {3pt} \\ {} & {} & r \in \left \{0, 1-\gamma \right \} \\ \bullet & {} & \textrm {Each term in the series must be 0, giving the recursion relation}\hspace {3pt} \\ {} & {} & -a_{k +1} \left (k +1+r \right ) \left (k +r +\gamma \right )+a_{k} \left (\beta +k +r \right ) \left (\alpha +k +r \right )=0 \\ \bullet & {} & \textrm {Recursion relation that defines series solution to ODE}\hspace {3pt} \\ {} & {} & a_{k +1}=\frac {a_{k} \left (\beta +k +r \right ) \left (\alpha +k +r \right )}{\left (k +1+r \right ) \left (k +r +\gamma \right )} \\ \bullet & {} & \textrm {Recursion relation for}\hspace {3pt} r =0 \\ {} & {} & a_{k +1}=\frac {a_{k} \left (\beta +k \right ) \left (\alpha +k \right )}{\left (k +1\right ) \left (k +\gamma \right )} \\ \bullet & {} & \textrm {Solution for}\hspace {3pt} r =0 \\ {} & {} & \left [y \left (x \right )=\moverset {\infty }{\munderset {k =0}{\sum }}a_{k} x^{k}, a_{k +1}=\frac {a_{k} \left (\beta +k \right ) \left (\alpha +k \right )}{\left (k +1\right ) \left (k +\gamma \right )}\right ] \\ \bullet & {} & \textrm {Recursion relation for}\hspace {3pt} r =1-\gamma \\ {} & {} & a_{k +1}=\frac {a_{k} \left (\beta +k +1-\gamma \right ) \left (\alpha +k +1-\gamma \right )}{\left (k +2-\gamma \right ) \left (k +1\right )} \\ \bullet & {} & \textrm {Solution for}\hspace {3pt} r =1-\gamma \\ {} & {} & \left [y \left (x \right )=\moverset {\infty }{\munderset {k =0}{\sum }}a_{k} x^{k +1-\gamma }, a_{k +1}=\frac {a_{k} \left (\beta +k +1-\gamma \right ) \left (\alpha +k +1-\gamma \right )}{\left (k +2-\gamma \right ) \left (k +1\right )}\right ] \\ \bullet & {} & \textrm {Combine solutions and rename parameters}\hspace {3pt} \\ {} & {} & \left [y \left (x \right )=\left (\moverset {\infty }{\munderset {k =0}{\sum }}a_{k} x^{k}\right )+\left (\moverset {\infty }{\munderset {k =0}{\sum }}b_{k} x^{k +1-\gamma }\right ), a_{k +1}=\frac {a_{k} \left (\beta +k \right ) \left (\alpha +k \right )}{\left (k +1\right ) \left (k +\gamma \right )}, b_{k +1}=\frac {b_{k} \left (\beta +k +1-\gamma \right ) \left (\alpha +k +1-\gamma \right )}{\left (k +2-\gamma \right ) \left (k +1\right )}\right ] \end {array} \]

2.31.23.2 ✓ Mathematica. Time used: 0.15 (sec). Leaf size: 49

ode=x*(x-1)*D[y[x],{x,2}]+((\[Alpha]+\[Beta]+1)*x-\[Gamma])*D[y[x],x]+\[Alpha]*\[Beta]*y[x]==0; 
ic={}; 
DSolve[{ode,ic},y[x],x,IncludeSingularSolutions->True]
 

\begin{align*} y(x)&\to c_1 \operatorname {Hypergeometric2F1}(\alpha ,\beta ,\gamma ,x)-(-1)^{-\gamma } c_2 x^{1-\gamma } \operatorname {Hypergeometric2F1}(\alpha -\gamma +1,\beta -\gamma +1,2-\gamma ,x) \end{align*}

2.31.23.3 ✗ Sympy

from sympy import * 
x = symbols("x") 
Alpha = symbols("Alpha") 
BETA = symbols("BETA") 
Gamma = symbols("Gamma") 
y = Function("y") 
ode = Eq(Alpha*BETA*y(x) + x*(x - 1)*Derivative(y(x), (x, 2)) + (-Gamma + x*(Alpha + BETA + 1))*Derivative(y(x), x),0) 
ics = {} 
dsolve(ode,func=y(x),ics=ics)
 

ValueError : Expected Expr or iterable but got None