2.31.19 Problem 167

Internal problem ID [13828]
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 : 167
Date solved : Friday, December 19, 2025 at 02:15:49 PM
CAS classification : [[_2nd_order, _with_linear_symmetries]]

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

2.31.19.1 ✓ Maple. Time used: 0.029 (sec). Leaf size: 27

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

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

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 = 4*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} \\ {} & {} & \left (-x^{2}+1\right ) \left (\frac {d^{2}}{d x^{2}}y \left (x \right )\right )-x \left (\frac {d}{d x}y \left (x \right )\right )+\left (2 a \,x^{2}+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 (2 a \,x^{2}+b \right ) y \left (x \right )}{x^{2}-1}-\frac {x \left (\frac {d}{d x}y \left (x \right )\right )}{x^{2}-1} \\ \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 {x \left (\frac {d}{d x}y \left (x \right )\right )}{x^{2}-1}-\frac {\left (2 a \,x^{2}+b \right ) y \left (x \right )}{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}{x^{2}-1}, P_{3}\left (x \right )=-\frac {2 a \,x^{2}+b}{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}}}=\frac {1}{2} \\ {} & \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^{2}}{d x^{2}}y \left (x \right )\right )+x \left (\frac {d}{d x}y \left (x \right )\right )+\left (-2 a \,x^{2}-b \right ) y \left (x \right )=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^{2}}{d u^{2}}y \left (u \right )\right )+\left (u -1\right ) \left (\frac {d}{d u}y \left (u \right )\right )+\left (-2 a \,u^{2}+4 a u -2 a -b \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 y \left (u \right )\hspace {3pt}\textrm {to series expansion for}\hspace {3pt} m =0..2 \\ {} & {} & u^{m}\cdot y \left (u \right )=\moverset {\infty }{\munderset {k =0}{\sum }}a_{k} u^{k +r +m} \\ {} & \circ & \textrm {Shift index using}\hspace {3pt} k \mathrm {->}k -m \\ {} & {} & u^{m}\cdot y \left (u \right )=\moverset {\infty }{\munderset {k =m}{\sum }}a_{k -m} u^{k +r} \\ {} & \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^{2}}{d u^{2}}y \left (u \right )\right )\hspace {3pt}\textrm {to series expansion for}\hspace {3pt} m =1..2 \\ {} & {} & u^{m}\cdot \left (\frac {d^{2}}{d u^{2}}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^{2}}{d u^{2}}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} \\ {} & {} & -a_{0} r \left (-1+2 r \right ) u^{-1+r}+\left (-a_{1} \left (1+r \right ) \left (1+2 r \right )-a_{0} \left (-r^{2}+2 a +b \right )\right ) u^{r}+\left (-a_{2} \left (2+r \right ) \left (3+2 r \right )-a_{1} \left (-r^{2}+2 a +b -2 r -1\right )+4 a_{0} a \right ) u^{1+r}+\left (\moverset {\infty }{\munderset {k =2}{\sum }}\left (-a_{k +1} \left (k +1+r \right ) \left (2 k +1+2 r \right )-a_{k} \left (-k^{2}-2 k r -r^{2}+2 a +b \right )+4 a_{k -1} a -2 a_{k -2} a \right ) u^{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 {The coefficients of each power of}\hspace {3pt} u \hspace {3pt}\textrm {must be 0}\hspace {3pt} \\ {} & {} & \left [-a_{1} \left (1+r \right ) \left (1+2 r \right )-a_{0} \left (-r^{2}+2 a +b \right )=0, -a_{2} \left (2+r \right ) \left (3+2 r \right )-a_{1} \left (-r^{2}+2 a +b -2 r -1\right )+4 a_{0} a =0\right ] \\ \bullet & {} & \textrm {Solve for the dependent coefficient(s)}\hspace {3pt} \\ {} & {} & \left \{a_{1}=-\frac {a_{0} \left (-r^{2}+2 a +b \right )}{2 r^{2}+3 r +1}, a_{2}=\frac {a_{0} \left (r^{4}+4 a \,r^{2}-2 b \,r^{2}+2 r^{3}+4 a^{2}+4 a b +8 a r +b^{2}-2 b r +r^{2}+2 a -b \right )}{4 r^{4}+20 r^{3}+35 r^{2}+25 r +6}\right \} \\ \bullet & {} & \textrm {Each term in the series must be 0, giving the recursion relation}\hspace {3pt} \\ {} & {} & -2 \left (k +\frac {1}{2}+r \right ) \left (k +1+r \right ) a_{k +1}+\left (k^{2}+2 k r +r^{2}-2 a -b \right ) a_{k}-2 a \left (a_{k -2}-2 a_{k -1}\right )=0 \\ \bullet & {} & \textrm {Shift index using}\hspace {3pt} k \mathrm {->}k +2 \\ {} & {} & -2 \left (k +\frac {5}{2}+r \right ) \left (k +3+r \right ) a_{k +3}+\left (\left (k +2\right )^{2}+2 \left (k +2\right ) r +r^{2}-2 a -b \right ) a_{k +2}-2 a \left (a_{k}-2 a_{k +1}\right )=0 \\ \bullet & {} & \textrm {Recursion relation that defines series solution to ODE}\hspace {3pt} \\ {} & {} & a_{k +3}=-\frac {-k^{2} a_{k +2}-2 k r a_{k +2}-r^{2} a_{k +2}+2 a_{k} a -4 a a_{k +1}+2 a a_{k +2}+b a_{k +2}-4 k a_{k +2}-4 r a_{k +2}-4 a_{k +2}}{\left (2 k +5+2 r \right ) \left (k +3+r \right )} \\ \bullet & {} & \textrm {Recursion relation for}\hspace {3pt} r =0 \\ {} & {} & a_{k +3}=-\frac {-k^{2} a_{k +2}+2 a_{k} a -4 a a_{k +1}+2 a a_{k +2}+b a_{k +2}-4 k a_{k +2}-4 a_{k +2}}{\left (2 k +5\right ) \left (k +3\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 +3}=-\frac {-k^{2} a_{k +2}+2 a_{k} a -4 a a_{k +1}+2 a a_{k +2}+b a_{k +2}-4 k a_{k +2}-4 a_{k +2}}{\left (2 k +5\right ) \left (k +3\right )}, a_{1}=-a_{0} \left (2 a +b \right ), a_{2}=\frac {a_{0} \left (4 a^{2}+4 a b +b^{2}+2 a -b \right )}{6}\right ] \\ \bullet & {} & \textrm {Revert the change of variables}\hspace {3pt} u =x +1 \\ {} & {} & \left [y \left (x \right )=\moverset {\infty }{\munderset {k =0}{\sum }}a_{k} \left (x +1\right )^{k}, a_{k +3}=-\frac {-k^{2} a_{k +2}+2 a_{k} a -4 a a_{k +1}+2 a a_{k +2}+b a_{k +2}-4 k a_{k +2}-4 a_{k +2}}{\left (2 k +5\right ) \left (k +3\right )}, a_{1}=-a_{0} \left (2 a +b \right ), a_{2}=\frac {a_{0} \left (4 a^{2}+4 a b +b^{2}+2 a -b \right )}{6}\right ] \\ \bullet & {} & \textrm {Recursion relation for}\hspace {3pt} r =\frac {1}{2} \\ {} & {} & a_{k +3}=-\frac {-k^{2} a_{k +2}+2 a_{k} a -4 a a_{k +1}+2 a a_{k +2}+b a_{k +2}-5 k a_{k +2}-\frac {25}{4} a_{k +2}}{\left (2 k +6\right ) \left (k +\frac {7}{2}\right )} \\ \bullet & {} & \textrm {Solution for}\hspace {3pt} r =\frac {1}{2} \\ {} & {} & \left [y \left (u \right )=\moverset {\infty }{\munderset {k =0}{\sum }}a_{k} u^{k +\frac {1}{2}}, a_{k +3}=-\frac {-k^{2} a_{k +2}+2 a_{k} a -4 a a_{k +1}+2 a a_{k +2}+b a_{k +2}-5 k a_{k +2}-\frac {25}{4} a_{k +2}}{\left (2 k +6\right ) \left (k +\frac {7}{2}\right )}, a_{1}=-\frac {a_{0} \left (-\frac {1}{4}+2 a +b \right )}{3}, a_{2}=\frac {a_{0} \left (4 a^{2}+4 a b +b^{2}+7 a -\frac {5}{2} b +\frac {9}{16}\right )}{30}\right ] \\ \bullet & {} & \textrm {Revert the change of variables}\hspace {3pt} u =x +1 \\ {} & {} & \left [y \left (x \right )=\moverset {\infty }{\munderset {k =0}{\sum }}a_{k} \left (x +1\right )^{k +\frac {1}{2}}, a_{k +3}=-\frac {-k^{2} a_{k +2}+2 a_{k} a -4 a a_{k +1}+2 a a_{k +2}+b a_{k +2}-5 k a_{k +2}-\frac {25}{4} a_{k +2}}{\left (2 k +6\right ) \left (k +\frac {7}{2}\right )}, a_{1}=-\frac {a_{0} \left (-\frac {1}{4}+2 a +b \right )}{3}, a_{2}=\frac {a_{0} \left (4 a^{2}+4 a b +b^{2}+7 a -\frac {5}{2} b +\frac {9}{16}\right )}{30}\right ] \\ \bullet & {} & \textrm {Combine solutions and rename parameters}\hspace {3pt} \\ {} & {} & \left [y \left (x \right )=\left (\moverset {\infty }{\munderset {k =0}{\sum }}c_{k} \left (x +1\right )^{k}\right )+\left (\moverset {\infty }{\munderset {k =0}{\sum }}d_{k} \left (x +1\right )^{k +\frac {1}{2}}\right ), c_{k +3}=-\frac {-k^{2} c_{k +2}+2 a c_{k}-4 a c_{k +1}+2 a c_{k +2}+b c_{k +2}-4 k c_{k +2}-4 c_{k +2}}{\left (2 k +5\right ) \left (k +3\right )}, c_{1}=-c_{0} \left (2 a +b \right ), c_{2}=\frac {c_{0} \left (4 a^{2}+4 a b +b^{2}+2 a -b \right )}{6}, d_{k +3}=-\frac {-k^{2} d_{k +2}+2 d_{k} a -4 a d_{k +1}+2 a d_{k +2}+b d_{k +2}-5 k d_{k +2}-\frac {25}{4} d_{k +2}}{\left (2 k +6\right ) \left (k +\frac {7}{2}\right )}, d_{1}=-\frac {d_{0} \left (-\frac {1}{4}+2 a +b \right )}{3}, d_{2}=\frac {d_{0} \left (4 a^{2}+4 a b +b^{2}+7 a -\frac {5}{2} b +\frac {9}{16}\right )}{30}\right ] \end {array} \]

2.31.19.2 ✓ Mathematica. Time used: 0.02 (sec). Leaf size: 34

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

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

2.31.19.3 ✗ Sympy

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

False