2.33.8 Problem 218

Internal problem ID [13878]
Book : Handbook of exact solutions for ordinary differential equations. By Polyanin and Zaitsev. Second edition
Section : Chapter 2, Second-Order Differential Equations. section 2.1.2-7
Problem number : 218
Date solved : Friday, December 19, 2025 at 07:10:18 PM
CAS classification : [[_2nd_order, _with_linear_symmetries]]

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

2.33.8.1 ✓ Maple. Time used: 0.015 (sec). Leaf size: 260

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

\[ y = \sqrt {x}\, \left (x -1\right )^{-\frac {\sqrt {a -4 d -4 c -4 b}-\sqrt {a}}{2 \sqrt {a}}} \left (\operatorname {hypergeom}\left (\left [-\frac {\sqrt {a -4 d -4 c -4 b}-\sqrt {a}+\sqrt {a -4 d}+\sqrt {a -4 b}}{2 \sqrt {a}}, -\frac {\sqrt {a -4 d -4 c -4 b}-\sqrt {a}+\sqrt {a -4 d}-\sqrt {a -4 b}}{2 \sqrt {a}}\right ], \left [1-\frac {\sqrt {a -4 d}}{\sqrt {a}}\right ], x\right ) x^{-\frac {\sqrt {a -4 d}}{2 \sqrt {a}}} c_2 +\operatorname {hypergeom}\left (\left [\frac {-\sqrt {a -4 d -4 c -4 b}+\sqrt {a}+\sqrt {a -4 d}+\sqrt {a -4 b}}{2 \sqrt {a}}, -\frac {\sqrt {a -4 d -4 c -4 b}-\sqrt {a}-\sqrt {a -4 d}+\sqrt {a -4 b}}{2 \sqrt {a}}\right ], \left [1+\frac {\sqrt {a -4 d}}{\sqrt {a}}\right ], x\right ) x^{\frac {\sqrt {a -4 d}}{2 \sqrt {a}}} c_1 \right ) \]

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 
   -> Whittaker 
      -> 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} \\ {} & {} & a \,x^{2} \left (x -1\right )^{2} \left (\frac {d^{2}}{d x^{2}}y \left (x \right )\right )+\left (b \,x^{2}+c x +d \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 (b \,x^{2}+c x +d \right ) y \left (x \right )}{a \,x^{2} \left (x -1\right )^{2}} \\ \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 (b \,x^{2}+c x +d \right ) y \left (x \right )}{a \,x^{2} \left (x -1\right )^{2}}=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 )=0, P_{3}\left (x \right )=\frac {b \,x^{2}+c x +d}{a \,x^{2} \left (x -1\right )^{2}}\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}}}=0 \\ {} & \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}}}=\frac {d}{a} \\ {} & \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} \\ {} & {} & a \,x^{2} \left (x -1\right )^{2} \left (\frac {d^{2}}{d x^{2}}y \left (x \right )\right )+\left (b \,x^{2}+c x +d \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..2 \\ {} & {} & 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^{2}}{d x^{2}}y \left (x \right )\right )\hspace {3pt}\textrm {to series expansion for}\hspace {3pt} m =2..4 \\ {} & {} & 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} \left (a \,r^{2}-a r +d \right ) x^{r}+\left (\left (a \,r^{2}+a r +d \right ) a_{1}-a_{0} \left (2 a \,r^{2}-2 a r -c \right )\right ) x^{1+r}+\left (\moverset {\infty }{\munderset {k =2}{\sum }}\left (a_{k} \left (a \,k^{2}+2 a k r +a \,r^{2}-a k -a r +d \right )-a_{k -1} \left (2 a \left (k -1\right )^{2}+4 a \left (k -1\right ) r +2 a \,r^{2}-2 a \left (k -1\right )-2 a r -c \right )+a_{k -2} \left (a \left (k -2\right )^{2}+2 a \left (k -2\right ) r +a \,r^{2}-a \left (k -2\right )-a r +b \right )\right ) x^{k +r}\right )=0 \\ \bullet & {} & a_{0}\textrm {cannot be 0 by assumption, giving the indicial equation}\hspace {3pt} \\ {} & {} & a \,r^{2}-a r +d =0 \\ \bullet & {} & \textrm {Values of r that satisfy the indicial equation}\hspace {3pt} \\ {} & {} & r \in \left \{\frac {a +\sqrt {a^{2}-4 a d}}{2 a}, -\frac {-a +\sqrt {a^{2}-4 a d}}{2 a}\right \} \\ \bullet & {} & \textrm {Each term must be 0}\hspace {3pt} \\ {} & {} & \left (a \,r^{2}+a r +d \right ) a_{1}-a_{0} \left (2 a \,r^{2}-2 a r -c \right )=0 \\ \bullet & {} & \textrm {Solve for the dependent coefficient(s)}\hspace {3pt} \\ {} & {} & a_{1}=\frac {a_{0} \left (2 a \,r^{2}-2 a r -c \right )}{a \,r^{2}+a r +d} \\ \bullet & {} & \textrm {Each term in the series must be 0, giving the recursion relation}\hspace {3pt} \\ {} & {} & \left (\left (a_{k}+a_{k -2}-2 a_{k -1}\right ) k^{2}+\left (\left (2 a_{k}+2 a_{k -2}-4 a_{k -1}\right ) r -a_{k}-5 a_{k -2}+6 a_{k -1}\right ) k +\left (a_{k}+a_{k -2}-2 a_{k -1}\right ) r^{2}+\left (-a_{k}-5 a_{k -2}+6 a_{k -1}\right ) r +6 a_{k -2}-4 a_{k -1}\right ) a +b a_{k -2}+c a_{k -1}+a_{k} d =0 \\ \bullet & {} & \textrm {Shift index using}\hspace {3pt} k \mathrm {->}k +2 \\ {} & {} & \left (\left (a_{k +2}+a_{k}-2 a_{k +1}\right ) \left (k +2\right )^{2}+\left (\left (2 a_{k +2}+2 a_{k}-4 a_{k +1}\right ) r -a_{k +2}-5 a_{k}+6 a_{k +1}\right ) \left (k +2\right )+\left (a_{k +2}+a_{k}-2 a_{k +1}\right ) r^{2}+\left (-a_{k +2}-5 a_{k}+6 a_{k +1}\right ) r +6 a_{k}-4 a_{k +1}\right ) a +a_{k} b +c a_{k +1}+a_{k +2} d =0 \\ \bullet & {} & \textrm {Recursion relation that defines series solution to ODE}\hspace {3pt} \\ {} & {} & a_{k +2}=-\frac {a \,k^{2} a_{k}-2 a \,k^{2} a_{k +1}+2 a k r a_{k}-4 a k r a_{k +1}+a \,r^{2} a_{k}-2 a \,r^{2} a_{k +1}-a k a_{k}-2 a k a_{k +1}-a r a_{k}-2 a r a_{k +1}+a_{k} b +c a_{k +1}}{a \,k^{2}+2 a k r +a \,r^{2}+3 a k +3 a r +2 a +d} \\ \bullet & {} & \textrm {Recursion relation for}\hspace {3pt} r =\frac {a +\sqrt {a^{2}-4 a d}}{2 a} \\ {} & {} & a_{k +2}=-\frac {a \,k^{2} a_{k}-2 a \,k^{2} a_{k +1}+k \left (a +\sqrt {a^{2}-4 a d}\right ) a_{k}-2 k \left (a +\sqrt {a^{2}-4 a d}\right ) a_{k +1}+\frac {\left (a +\sqrt {a^{2}-4 a d}\right )^{2} a_{k}}{4 a}-\frac {\left (a +\sqrt {a^{2}-4 a d}\right )^{2} a_{k +1}}{2 a}-a k a_{k}-2 a k a_{k +1}-\frac {\left (a +\sqrt {a^{2}-4 a d}\right ) a_{k}}{2}-\left (a +\sqrt {a^{2}-4 a d}\right ) a_{k +1}+a_{k} b +c a_{k +1}}{a \,k^{2}+k \left (a +\sqrt {a^{2}-4 a d}\right )+\frac {\left (a +\sqrt {a^{2}-4 a d}\right )^{2}}{4 a}+3 a k +\frac {7 a}{2}+\frac {3 \sqrt {a^{2}-4 a d}}{2}+d} \\ \bullet & {} & \textrm {Solution for}\hspace {3pt} r =\frac {a +\sqrt {a^{2}-4 a d}}{2 a} \\ {} & {} & \left [y \left (x \right )=\moverset {\infty }{\munderset {k =0}{\sum }}a_{k} x^{k +\frac {a +\sqrt {a^{2}-4 a d}}{2 a}}, a_{k +2}=-\frac {a \,k^{2} a_{k}-2 a \,k^{2} a_{k +1}+k \left (a +\sqrt {a^{2}-4 a d}\right ) a_{k}-2 k \left (a +\sqrt {a^{2}-4 a d}\right ) a_{k +1}+\frac {\left (a +\sqrt {a^{2}-4 a d}\right )^{2} a_{k}}{4 a}-\frac {\left (a +\sqrt {a^{2}-4 a d}\right )^{2} a_{k +1}}{2 a}-a k a_{k}-2 a k a_{k +1}-\frac {\left (a +\sqrt {a^{2}-4 a d}\right ) a_{k}}{2}-\left (a +\sqrt {a^{2}-4 a d}\right ) a_{k +1}+a_{k} b +c a_{k +1}}{a \,k^{2}+k \left (a +\sqrt {a^{2}-4 a d}\right )+\frac {\left (a +\sqrt {a^{2}-4 a d}\right )^{2}}{4 a}+3 a k +\frac {7 a}{2}+\frac {3 \sqrt {a^{2}-4 a d}}{2}+d}, a_{1}=\frac {a_{0} \left (\frac {\left (a +\sqrt {a^{2}-4 a d}\right )^{2}}{2 a}-a -\sqrt {a^{2}-4 a d}-c \right )}{\frac {\left (a +\sqrt {a^{2}-4 a d}\right )^{2}}{4 a}+\frac {a}{2}+\frac {\sqrt {a^{2}-4 a d}}{2}+d}\right ] \\ \bullet & {} & \textrm {Recursion relation for}\hspace {3pt} r =-\frac {-a +\sqrt {a^{2}-4 a d}}{2 a} \\ {} & {} & a_{k +2}=-\frac {a \,k^{2} a_{k}-2 a \,k^{2} a_{k +1}-k \left (-a +\sqrt {a^{2}-4 a d}\right ) a_{k}+2 k \left (-a +\sqrt {a^{2}-4 a d}\right ) a_{k +1}+\frac {\left (-a +\sqrt {a^{2}-4 a d}\right )^{2} a_{k}}{4 a}-\frac {\left (-a +\sqrt {a^{2}-4 a d}\right )^{2} a_{k +1}}{2 a}-a k a_{k}-2 a k a_{k +1}+\frac {\left (-a +\sqrt {a^{2}-4 a d}\right ) a_{k}}{2}+\left (-a +\sqrt {a^{2}-4 a d}\right ) a_{k +1}+a_{k} b +c a_{k +1}}{a \,k^{2}-k \left (-a +\sqrt {a^{2}-4 a d}\right )+\frac {\left (-a +\sqrt {a^{2}-4 a d}\right )^{2}}{4 a}+3 a k +\frac {7 a}{2}-\frac {3 \sqrt {a^{2}-4 a d}}{2}+d} \\ \bullet & {} & \textrm {Solution for}\hspace {3pt} r =-\frac {-a +\sqrt {a^{2}-4 a d}}{2 a} \\ {} & {} & \left [y \left (x \right )=\moverset {\infty }{\munderset {k =0}{\sum }}a_{k} x^{k -\frac {-a +\sqrt {a^{2}-4 a d}}{2 a}}, a_{k +2}=-\frac {a \,k^{2} a_{k}-2 a \,k^{2} a_{k +1}-k \left (-a +\sqrt {a^{2}-4 a d}\right ) a_{k}+2 k \left (-a +\sqrt {a^{2}-4 a d}\right ) a_{k +1}+\frac {\left (-a +\sqrt {a^{2}-4 a d}\right )^{2} a_{k}}{4 a}-\frac {\left (-a +\sqrt {a^{2}-4 a d}\right )^{2} a_{k +1}}{2 a}-a k a_{k}-2 a k a_{k +1}+\frac {\left (-a +\sqrt {a^{2}-4 a d}\right ) a_{k}}{2}+\left (-a +\sqrt {a^{2}-4 a d}\right ) a_{k +1}+a_{k} b +c a_{k +1}}{a \,k^{2}-k \left (-a +\sqrt {a^{2}-4 a d}\right )+\frac {\left (-a +\sqrt {a^{2}-4 a d}\right )^{2}}{4 a}+3 a k +\frac {7 a}{2}-\frac {3 \sqrt {a^{2}-4 a d}}{2}+d}, a_{1}=\frac {a_{0} \left (\frac {\left (-a +\sqrt {a^{2}-4 a d}\right )^{2}}{2 a}-a +\sqrt {a^{2}-4 a d}-c \right )}{\frac {\left (-a +\sqrt {a^{2}-4 a d}\right )^{2}}{4 a}+\frac {a}{2}-\frac {\sqrt {a^{2}-4 a d}}{2}+d}\right ] \\ \bullet & {} & \textrm {Combine solutions and rename parameters}\hspace {3pt} \\ {} & {} & \left [y \left (x \right )=\left (\moverset {\infty }{\munderset {k =0}{\sum }}e_{k} x^{k +\frac {a +\sqrt {a^{2}-4 a d}}{2 a}}\right )+\left (\moverset {\infty }{\munderset {k =0}{\sum }}f_{k} x^{k -\frac {-a +\sqrt {a^{2}-4 a d}}{2 a}}\right ), e_{k +2}=-\frac {a \,k^{2} e_{k}-2 a \,k^{2} e_{k +1}+k \left (a +\sqrt {a^{2}-4 a d}\right ) e_{k}-2 k \left (a +\sqrt {a^{2}-4 a d}\right ) e_{k +1}+\frac {\left (a +\sqrt {a^{2}-4 a d}\right )^{2} e_{k}}{4 a}-\frac {\left (a +\sqrt {a^{2}-4 a d}\right )^{2} e_{k +1}}{2 a}-a k e_{k}-2 a k e_{k +1}-\frac {\left (a +\sqrt {a^{2}-4 a d}\right ) e_{k}}{2}-\left (a +\sqrt {a^{2}-4 a d}\right ) e_{k +1}+e_{k} b +c e_{k +1}}{a \,k^{2}+k \left (a +\sqrt {a^{2}-4 a d}\right )+\frac {\left (a +\sqrt {a^{2}-4 a d}\right )^{2}}{4 a}+3 a k +\frac {7 a}{2}+\frac {3 \sqrt {a^{2}-4 a d}}{2}+d}, e_{1}=\frac {e_{0} \left (\frac {\left (a +\sqrt {a^{2}-4 a d}\right )^{2}}{2 a}-a -\sqrt {a^{2}-4 a d}-c \right )}{\frac {\left (a +\sqrt {a^{2}-4 a d}\right )^{2}}{4 a}+\frac {a}{2}+\frac {\sqrt {a^{2}-4 a d}}{2}+d}, f_{k +2}=-\frac {a \,k^{2} f_{k}-2 a \,k^{2} f_{k +1}-k \left (-a +\sqrt {a^{2}-4 a d}\right ) f_{k}+2 k \left (-a +\sqrt {a^{2}-4 a d}\right ) f_{k +1}+\frac {\left (-a +\sqrt {a^{2}-4 a d}\right )^{2} f_{k}}{4 a}-\frac {\left (-a +\sqrt {a^{2}-4 a d}\right )^{2} f_{k +1}}{2 a}-a k f_{k}-2 a k f_{k +1}+\frac {\left (-a +\sqrt {a^{2}-4 a d}\right ) f_{k}}{2}+\left (-a +\sqrt {a^{2}-4 a d}\right ) f_{k +1}+f_{k} b +c f_{k +1}}{a \,k^{2}-k \left (-a +\sqrt {a^{2}-4 a d}\right )+\frac {\left (-a +\sqrt {a^{2}-4 a d}\right )^{2}}{4 a}+3 a k +\frac {7 a}{2}-\frac {3 \sqrt {a^{2}-4 a d}}{2}+d}, f_{1}=\frac {f_{0} \left (\frac {\left (-a +\sqrt {a^{2}-4 a d}\right )^{2}}{2 a}-a +\sqrt {a^{2}-4 a d}-c \right )}{\frac {\left (-a +\sqrt {a^{2}-4 a d}\right )^{2}}{4 a}+\frac {a}{2}-\frac {\sqrt {a^{2}-4 a d}}{2}+d}\right ] \end {array} \]

2.33.8.2 ✓ Mathematica. Time used: 165.204 (sec). Leaf size: 413606

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

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2.33.8.3 ✗ Sympy

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

NotImplementedError : The given ODE a*x**2*(x - 1)**2*Derivative(y(x), (x, 2)) + (b*x**2 + c*x + d)*y(x) cannot be solved by the hypergeometric method