2.29.11 Problem 71

Internal problem ID [13732]
Book : Handbook of exact solutions for ordinary differential equations. By Polyanin and Zaitsev. Second edition
Section : Chapter 2, Second-Order Differential Equations. section 2.1.2-3
Problem number : 71
Date solved : Friday, December 19, 2025 at 11:23:58 AM
CAS classification : [[_2nd_order, _with_linear_symmetries]]

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

2.29.11.1 ✓ Maple. Time used: 0.004 (sec). Leaf size: 40

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

\[ y = c_1 \,{\mathrm e}^{-x c}+c_2 \,x^{-\frac {b}{2}} \operatorname {WhittakerM}\left (-\frac {b}{2}, -\frac {b}{2}+\frac {1}{2}, \left (-2 c +a \right ) x \right ) {\mathrm e}^{-\frac {a x}{2}} \]

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 
   A Liouvillian solution exists 
   Reducible group (found an exponential solution) 
   Group is reducible, not completely reducible 
<- Kovacics algorithm successful
 

Maple step by step

\[ \begin {array}{lll} & {} & \textrm {Let's solve}\hspace {3pt} \\ {} & {} & x \left (\frac {d^{2}}{d x^{2}}y \left (x \right )\right )+\left (a x +b \right ) \left (\frac {d}{d x}y \left (x \right )\right )+c \left (\left (a -c \right ) 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 {c \left (a x -x c +b \right ) y \left (x \right )}{x}-\frac {\left (a x +b \right ) \left (\frac {d}{d x}y \left (x \right )\right )}{x} \\ \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 (a x +b \right ) \left (\frac {d}{d x}y \left (x \right )\right )}{x}+\frac {c \left (a x -x c +b \right ) y \left (x \right )}{x}=0 \\ \square & {} & \textrm {Check to see if}\hspace {3pt} x_{0}=0\hspace {3pt}\textrm {is a regular singular point}\hspace {3pt} \\ {} & \circ & \textrm {Define functions}\hspace {3pt} \\ {} & {} & \left [P_{2}\left (x \right )=\frac {a x +b}{x}, P_{3}\left (x \right )=\frac {c \left (a x -x c +b \right )}{x}\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}}}=b \\ {} & \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}=0\hspace {3pt}\textrm {is a regular singular point}\hspace {3pt} \\ {} & {} & x_{0}=0 \\ \bullet & {} & \textrm {Multiply by denominators}\hspace {3pt} \\ {} & {} & x \left (\frac {d^{2}}{d x^{2}}y \left (x \right )\right )+\left (a x +b \right ) \left (\frac {d}{d x}y \left (x \right )\right )+c \left (a x -x c +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 \cdot \left (\frac {d^{2}}{d x^{2}}y \left (x \right )\right )\hspace {3pt}\textrm {to series expansion}\hspace {3pt} \\ {} & {} & x \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 -1} \\ {} & \circ & \textrm {Shift index using}\hspace {3pt} k \mathrm {->}k +1 \\ {} & {} & x \cdot \left (\frac {d^{2}}{d x^{2}}y \left (x \right )\right )=\moverset {\infty }{\munderset {k =-1}{\sum }}a_{k +1} \left (k +1+r \right ) \left (k +r \right ) x^{k +r} \\ & {} & \textrm {Rewrite ODE with series expansions}\hspace {3pt} \\ {} & {} & a_{0} r \left (b -1+r \right ) x^{-1+r}+\left (a_{1} \left (1+r \right ) \left (b +r \right )+a_{0} \left (a r +c b \right )\right ) x^{r}+\left (\moverset {\infty }{\munderset {k =1}{\sum }}\left (a_{k +1} \left (k +1+r \right ) \left (b +k +r \right )+a_{k} \left (a k +a r +c b \right )+c a_{k -1} \left (a -c \right )\right ) x^{k +r}\right )=0 \\ \bullet & {} & a_{0}\textrm {cannot be 0 by assumption, giving the indicial equation}\hspace {3pt} \\ {} & {} & r \left (b -1+r \right )=0 \\ \bullet & {} & \textrm {Values of r that satisfy the indicial equation}\hspace {3pt} \\ {} & {} & r \in \left \{0, -b +1\right \} \\ \bullet & {} & \textrm {Each term must be 0}\hspace {3pt} \\ {} & {} & a_{1} \left (1+r \right ) \left (b +r \right )+a_{0} \left (a r +c b \right )=0 \\ \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 (b +k +r \right )+a k a_{k}+a r a_{k}+c \left (a_{k} b +a_{k -1} \left (a -c \right )\right )=0 \\ \bullet & {} & \textrm {Shift index using}\hspace {3pt} k \mathrm {->}k +1 \\ {} & {} & a_{k +2} \left (k +2+r \right ) \left (b +k +1+r \right )+a \left (k +1\right ) a_{k +1}+a r a_{k +1}+c \left (b a_{k +1}+a_{k} \left (a -c \right )\right )=0 \\ \bullet & {} & \textrm {Recursion relation that defines series solution to ODE}\hspace {3pt} \\ {} & {} & a_{k +2}=-\frac {a_{k} a c +a k a_{k +1}+a r a_{k +1}+b c a_{k +1}-a_{k} c^{2}+a a_{k +1}}{\left (k +2+r \right ) \left (b +k +1+r \right )} \\ \bullet & {} & \textrm {Recursion relation for}\hspace {3pt} r =0 \\ {} & {} & a_{k +2}=-\frac {a_{k} a c +a k a_{k +1}+b c a_{k +1}-a_{k} c^{2}+a a_{k +1}}{\left (k +2\right ) \left (b +k +1\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 {a_{k} a c +a k a_{k +1}+b c a_{k +1}-a_{k} c^{2}+a a_{k +1}}{\left (k +2\right ) \left (b +k +1\right )}, b c a_{0}+b a_{1}=0\right ] \\ \bullet & {} & \textrm {Recursion relation for}\hspace {3pt} r =-b +1 \\ {} & {} & a_{k +2}=-\frac {a_{k} a c +a k a_{k +1}+a \left (-b +1\right ) a_{k +1}+b c a_{k +1}-a_{k} c^{2}+a a_{k +1}}{\left (k +3-b \right ) \left (k +2\right )} \\ \bullet & {} & \textrm {Solution for}\hspace {3pt} r =-b +1 \\ {} & {} & \left [y \left (x \right )=\moverset {\infty }{\munderset {k =0}{\sum }}a_{k} x^{k -b +1}, a_{k +2}=-\frac {a_{k} a c +a k a_{k +1}+a \left (-b +1\right ) a_{k +1}+b c a_{k +1}-a_{k} c^{2}+a a_{k +1}}{\left (k +3-b \right ) \left (k +2\right )}, a_{1} \left (-b +2\right )+a_{0} \left (a \left (-b +1\right )+c b \right )=0\right ] \\ \bullet & {} & \textrm {Combine solutions and rename parameters}\hspace {3pt} \\ {} & {} & \left [y \left (x \right )=\left (\moverset {\infty }{\munderset {k =0}{\sum }}d_{k} x^{k}\right )+\left (\moverset {\infty }{\munderset {k =0}{\sum }}e_{k} x^{k -b +1}\right ), d_{k +2}=-\frac {a c d_{k}+a k d_{k +1}+b c d_{k +1}-c^{2} d_{k}+a d_{k +1}}{\left (k +2\right ) \left (b +k +1\right )}, b c d_{0}+b d_{1}=0, e_{k +2}=-\frac {e_{k} a c +a k e_{k +1}+a \left (-b +1\right ) e_{k +1}+b c e_{k +1}-e_{k} c^{2}+a e_{k +1}}{\left (k +3-b \right ) \left (k +2\right )}, e_{1} \left (-b +2\right )+e_{0} \left (a \left (-b +1\right )+c b \right )=0\right ] \end {array} \]

2.29.11.2 ✓ Mathematica. Time used: 0.213 (sec). Leaf size: 50

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

\begin{align*} y(x)&\to e^{-c x} \left (c_1-c_2 x^{1-b} (x (a-2 c))^{b-1} \Gamma (1-b,(a-2 c) x)\right ) \end{align*}

2.29.11.3 ✗ Sympy

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

ValueError : Expected Expr or iterable but got None