2.29.14 Problem 74

Internal problem ID [13735]
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 : 74
Date solved : Friday, December 19, 2025 at 11:27:33 AM
CAS classification : [[_2nd_order, _with_linear_symmetries]]

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

2.29.14.1 ✓ Maple. Time used: 0.039 (sec). Leaf size: 109

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

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

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 
   -> Kummer 
      -> hyper3: Equivalence to 1F1 under a power @ Moebius 
      <- hyper3 successful: received ODE is equivalent to the 1F1 ODE 
   <- Kummer successful 
<- special function solution 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 )+\left (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 (c x +d \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 {\left (c x +d \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 x +d}{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 )+\left (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..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 (-1+r +b \right ) x^{-1+r}+\left (a_{1} \left (1+r \right ) \left (r +b \right )+a_{0} \left (a r +d \right )\right ) x^{r}+\left (\moverset {\infty }{\munderset {k =1}{\sum }}\left (a_{k +1} \left (k +1+r \right ) \left (k +r +b \right )+a_{k} \left (a k +a r +d \right )+a_{k -1} c \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 +b \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 (r +b \right )+a_{0} \left (a r +d \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 (k +r +b \right )+a k a_{k}+a r a_{k}+a_{k -1} c +a_{k} d =0 \\ \bullet & {} & \textrm {Shift index using}\hspace {3pt} k \mathrm {->}k +1 \\ {} & {} & a_{k +2} \left (k +2+r \right ) \left (k +1+r +b \right )+a \left (k +1\right ) a_{k +1}+a r a_{k +1}+a_{k} c +a_{k +1} d =0 \\ \bullet & {} & \textrm {Recursion relation that defines series solution to ODE}\hspace {3pt} \\ {} & {} & a_{k +2}=-\frac {a k a_{k +1}+a r a_{k +1}+a a_{k +1}+a_{k} c +a_{k +1} d}{\left (k +2+r \right ) \left (k +1+r +b \right )} \\ \bullet & {} & \textrm {Recursion relation for}\hspace {3pt} r =0 \\ {} & {} & a_{k +2}=-\frac {a k a_{k +1}+a a_{k +1}+a_{k} c +a_{k +1} d}{\left (k +2\right ) \left (k +1+b \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_{k +1}+a a_{k +1}+a_{k} c +a_{k +1} d}{\left (k +2\right ) \left (k +1+b \right )}, b a_{1}+d a_{0}=0\right ] \\ \bullet & {} & \textrm {Recursion relation for}\hspace {3pt} r =-b +1 \\ {} & {} & a_{k +2}=-\frac {a k a_{k +1}+a \left (-b +1\right ) a_{k +1}+a a_{k +1}+a_{k} c +a_{k +1} d}{\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_{k +1}+a \left (-b +1\right ) a_{k +1}+a a_{k +1}+a_{k} c +a_{k +1} d}{\left (k +3-b \right ) \left (k +2\right )}, a_{1} \left (-b +2\right )+a_{0} \left (a \left (-b +1\right )+d \right )=0\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}\right )+\left (\moverset {\infty }{\munderset {k =0}{\sum }}f_{k} x^{k -b +1}\right ), e_{k +2}=-\frac {a k e_{k +1}+a e_{k +1}+c e_{k}+d e_{k +1}}{\left (k +2\right ) \left (k +1+b \right )}, b e_{1}+d e_{0}=0, f_{k +2}=-\frac {a k f_{k +1}+a \left (-b +1\right ) f_{k +1}+a f_{k +1}+f_{k} c +f_{k +1} d}{\left (k +3-b \right ) \left (k +2\right )}, f_{1} \left (-b +2\right )+f_{0} \left (a \left (-b +1\right )+d \right )=0\right ] \end {array} \]

2.29.14.2 ✓ Mathematica. Time used: 0.049 (sec). Leaf size: 135

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

\begin{align*} y(x)&\to e^{-\frac {1}{2} x \left (\sqrt {a^2-4 c}+a\right )} \left (c_1 \operatorname {HypergeometricU}\left (\frac {a b+\sqrt {a^2-4 c} b-2 d}{2 \sqrt {a^2-4 c}},b,\sqrt {a^2-4 c} x\right )+c_2 L_{-\frac {a b+\sqrt {a^2-4 c} b-2 d}{2 \sqrt {a^2-4 c}}}^{b-1}\left (\sqrt {a^2-4 c} x\right )\right ) \end{align*}

2.29.14.3 ✗ Sympy

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

ValueError : Expected Expr or iterable but got None