2.31.25 Problem 173

Internal problem ID [13834]
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 : 173
Date solved : Friday, December 19, 2025 at 02:40:17 PM
CAS classification : [_Jacobi]

\begin{align*} 2 x \left (x -1\right ) y^{\prime \prime }+\left (2 x -1\right ) y^{\prime }+\left (x a +b \right ) y&=0 \\ \end{align*}

2.31.25.1 ✓ Maple. Time used: 0.027 (sec). Leaf size: 39

ode:=2*x*(x-1)*diff(diff(y(x),x),x)+(2*x-1)*diff(y(x),x)+(a*x+b)*y(x) = 0; 
dsolve(ode,y(x), singsol=all);
 

\[ y = c_1 \operatorname {MathieuC}\left (-2 b -a , \frac {a}{2}, \arccos \left (\sqrt {x}\right )\right )+c_2 \operatorname {MathieuS}\left (-2 b -a , \frac {a}{2}, \arccos \left (\sqrt {x}\right )\right ) \]

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 
      -> hyper3: Equivalence to 2F1, 1F1 or 0F1 under a power @ Moebius 
   -> Mathieu 
      -> Equivalence to the rational form of Mathieu ODE under a power @ Moebi\ 
us 
      Equivalence transformation and function parameters: {x = t}, {kappa = -8*\ 
b-4, mu = 8*a} 
      <- Equivalence to the rational form of Mathieu ODE successful 
   <- Mathieu successful 
<- special function solution successful
 

Maple step by step

\[ \begin {array}{lll} & {} & \textrm {Let's solve}\hspace {3pt} \\ {} & {} & 2 x \left (x -1\right ) \left (\frac {d^{2}}{d x^{2}}y \left (x \right )\right )+\left (2 x -1\right ) \left (\frac {d}{d x}y \left (x \right )\right )+\left (a x +b \right ) 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 {\left (a x +b \right ) y \left (x \right )}{2 x \left (x -1\right )}-\frac {\left (2 x -1\right ) \left (\frac {d}{d x}y \left (x \right )\right )}{2 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 (2 x -1\right ) \left (\frac {d}{d x}y \left (x \right )\right )}{2 x \left (x -1\right )}+\frac {\left (a x +b \right ) y \left (x \right )}{2 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 {2 x -1}{2 x \left (x -1\right )}, P_{3}\left (x \right )=\frac {a x +b}{2 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}}}=\frac {1}{2} \\ {} & \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} \\ {} & {} & 2 x \left (x -1\right ) \left (\frac {d^{2}}{d x^{2}}y \left (x \right )\right )+\left (2 x -1\right ) \left (\frac {d}{d x}y \left (x \right )\right )+\left (a x +b \right ) 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 y \left (x \right )\hspace {3pt}\textrm {to series expansion for}\hspace {3pt} m =0..1 \\ {} & {} & x^{m}\cdot y \left (x \right )=\moverset {\infty }{\munderset {k =0}{\sum }}a_{k} x^{k +r +m} \\ {} & \circ & \textrm {Shift index using}\hspace {3pt} k \mathrm {->}k -m \\ {} & {} & x^{m}\cdot y \left (x \right )=\moverset {\infty }{\munderset {k =m}{\sum }}a_{k -m} x^{k +r} \\ {} & \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+2 r \right ) x^{-1+r}+\left (-a_{1} \left (1+r \right ) \left (1+2 r \right )+a_{0} \left (2 r^{2}+b \right )\right ) x^{r}+\left (\moverset {\infty }{\munderset {k =1}{\sum }}\left (-a_{k +1} \left (k +1+r \right ) \left (2 k +1+2 r \right )+a_{k} \left (2 k^{2}+4 k r +2 r^{2}+b \right )+a_{k -1} a \right ) x^{k +r}\right )=0 \\ \bullet & {} & a_{0}\textrm {cannot be 0 by assumption, giving the indicial equation}\hspace {3pt} \\ {} & {} & -r \left (-1+2 r \right )=0 \\ \bullet & {} & \textrm {Values of r that satisfy the indicial equation}\hspace {3pt} \\ {} & {} & r \in \left \{0, \frac {1}{2}\right \} \\ \bullet & {} & \textrm {Each term must be 0}\hspace {3pt} \\ {} & {} & -a_{1} \left (1+r \right ) \left (1+2 r \right )+a_{0} \left (2 r^{2}+b \right )=0 \\ \bullet & {} & \textrm {Each term in the series must be 0, giving the recursion relation}\hspace {3pt} \\ {} & {} & -2 \left (k +1+r \right ) \left (k +\frac {1}{2}+r \right ) a_{k +1}+2 k^{2} a_{k}+4 k r a_{k}+2 r^{2} a_{k}+a_{k -1} a +a_{k} b =0 \\ \bullet & {} & \textrm {Shift index using}\hspace {3pt} k \mathrm {->}k +1 \\ {} & {} & -2 \left (k +2+r \right ) \left (k +\frac {3}{2}+r \right ) a_{k +2}+2 \left (k +1\right )^{2} a_{k +1}+4 \left (k +1\right ) r a_{k +1}+2 r^{2} a_{k +1}+a_{k} a +a_{k +1} b =0 \\ \bullet & {} & \textrm {Recursion relation that defines series solution to ODE}\hspace {3pt} \\ {} & {} & a_{k +2}=\frac {2 k^{2} a_{k +1}+4 k r a_{k +1}+2 r^{2} a_{k +1}+a_{k} a +a_{k +1} b +4 k a_{k +1}+4 r a_{k +1}+2 a_{k +1}}{\left (k +2+r \right ) \left (2 k +3+2 r \right )} \\ \bullet & {} & \textrm {Recursion relation for}\hspace {3pt} r =0 \\ {} & {} & a_{k +2}=\frac {2 k^{2} a_{k +1}+a_{k} a +a_{k +1} b +4 k a_{k +1}+2 a_{k +1}}{\left (k +2\right ) \left (2 k +3\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 +2}=\frac {2 k^{2} a_{k +1}+a_{k} a +a_{k +1} b +4 k a_{k +1}+2 a_{k +1}}{\left (k +2\right ) \left (2 k +3\right )}, b a_{0}-a_{1}=0\right ] \\ \bullet & {} & \textrm {Recursion relation for}\hspace {3pt} r =\frac {1}{2} \\ {} & {} & a_{k +2}=\frac {2 k^{2} a_{k +1}+a_{k} a +a_{k +1} b +6 k a_{k +1}+\frac {9}{2} a_{k +1}}{\left (k +\frac {5}{2}\right ) \left (2 k +4\right )} \\ \bullet & {} & \textrm {Solution for}\hspace {3pt} r =\frac {1}{2} \\ {} & {} & \left [y \left (x \right )=\moverset {\infty }{\munderset {k =0}{\sum }}a_{k} x^{k +\frac {1}{2}}, a_{k +2}=\frac {2 k^{2} a_{k +1}+a_{k} a +a_{k +1} b +6 k a_{k +1}+\frac {9}{2} a_{k +1}}{\left (k +\frac {5}{2}\right ) \left (2 k +4\right )}, -3 a_{1}+a_{0} \left (b +\frac {1}{2}\right )=0\right ] \\ \bullet & {} & \textrm {Combine solutions and rename parameters}\hspace {3pt} \\ {} & {} & \left [y \left (x \right )=\left (\moverset {\infty }{\munderset {k =0}{\sum }}c_{k} x^{k}\right )+\left (\moverset {\infty }{\munderset {k =0}{\sum }}d_{k} x^{k +\frac {1}{2}}\right ), c_{k +2}=\frac {2 k^{2} c_{k +1}+a c_{k}+b c_{k +1}+4 k c_{k +1}+2 c_{k +1}}{\left (k +2\right ) \left (2 k +3\right )}, b c_{0}-c_{1}=0, d_{k +2}=\frac {2 k^{2} d_{k +1}+d_{k} a +d_{k +1} b +6 k d_{k +1}+\frac {9}{2} d_{k +1}}{\left (k +\frac {5}{2}\right ) \left (2 k +4\right )}, -3 d_{1}+d_{0} \left (b +\frac {1}{2}\right )=0\right ] \end {array} \]

2.31.25.2 ✓ Mathematica. Time used: 0.121 (sec). Leaf size: 50

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

\begin{align*} y(x)&\to c_1 \text {MathieuC}\left [-a-2 b,\frac {a}{2},\arccos \left (\sqrt {x}\right )\right ]+c_2 \text {MathieuS}\left [-a-2 b,\frac {a}{2},\arccos \left (\sqrt {x}\right )\right ] \end{align*}

2.31.25.3 ✗ Sympy

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

False