2.29.46 Problem 106

Internal problem ID [13767]
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 : 106
Date solved : Friday, December 19, 2025 at 12:08:36 PM
CAS classification : [[_2nd_order, _with_linear_symmetries]]

\begin{align*} \left (a_{1} x +a_{0} \right ) y^{\prime \prime }+\left (b_{1} x +b_{0} \right ) y^{\prime }-m b_{1} y&=0 \\ \end{align*}

2.29.46.1 ✓ Maple. Time used: 0.008 (sec). Leaf size: 101

ode:=(a__1*x+a__0)*diff(diff(y(x),x),x)+(b__1*x+b__0)*diff(y(x),x)-m*b__1*y(x) = 0; 
dsolve(ode,y(x), singsol=all);
 

\[ y = {\mathrm e}^{-\frac {b_{1} x}{a_{1}}} \left (a_{1} x +a_{0} \right )^{\frac {a_{0} b_{1} +a_{1}^{2}-a_{1} b_{0}}{a_{1}^{2}}} \left (\operatorname {KummerU}\left (m +1, \frac {a_{0} b_{1} +2 a_{1}^{2}-a_{1} b_{0}}{a_{1}^{2}}, \frac {b_{1} \left (a_{1} x +a_{0} \right )}{a_{1}^{2}}\right ) c_2 +\operatorname {KummerM}\left (m +1, \frac {a_{0} b_{1} +2 a_{1}^{2}-a_{1} b_{0}}{a_{1}^{2}}, \frac {b_{1} \left (a_{1} x +a_{0} \right )}{a_{1}^{2}}\right ) c_1 \right ) \]

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} \\ {} & {} & \left (a_{1} x +a_{0} \right ) \left (\frac {d^{2}}{d x^{2}}y \left (x \right )\right )+\left (b_{1} x +b_{0} \right ) \left (\frac {d}{d x}y \left (x \right )\right )-m b_{1} 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 {m b_{1} y \left (x \right )}{a_{1} x +a_{0}}-\frac {\left (b_{1} x +b_{0} \right ) \left (\frac {d}{d x}y \left (x \right )\right )}{a_{1} x +a_{0}} \\ \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_{1} x +b_{0} \right ) \left (\frac {d}{d x}y \left (x \right )\right )}{a_{1} x +a_{0}}-\frac {m b_{1} y \left (x \right )}{a_{1} x +a_{0}}=0 \\ \square & {} & \textrm {Check to see if}\hspace {3pt} x_{0}=-\frac {a_{0}}{a_{1}}\hspace {3pt}\textrm {is a regular singular point}\hspace {3pt} \\ {} & \circ & \textrm {Define functions}\hspace {3pt} \\ {} & {} & \left [P_{2}\left (x \right )=\frac {b_{1} x +b_{0}}{a_{1} x +a_{0}}, P_{3}\left (x \right )=-\frac {m b_{1}}{a_{1} x +a_{0}}\right ] \\ {} & \circ & \left (x +\frac {a_{0}}{a_{1}}\right )\cdot P_{2}\left (x \right )\textrm {is analytic at}\hspace {3pt} x =-\frac {a_{0}}{a_{1}} \\ {} & {} & \left (\left (x +\frac {a_{0}}{a_{1}}\right )\cdot P_{2}\left (x \right )\right )\bigg | {\mstack {}{_{x \hiderel {=}-\frac {a_{0}}{a_{1}}}}}=\frac {-\frac {b_{1} a_{0}}{a_{1}}+b_{0}}{a_{1}} \\ {} & \circ & \left (x +\frac {a_{0}}{a_{1}}\right )^{2}\cdot P_{3}\left (x \right )\textrm {is analytic at}\hspace {3pt} x =-\frac {a_{0}}{a_{1}} \\ {} & {} & \left (\left (x +\frac {a_{0}}{a_{1}}\right )^{2}\cdot P_{3}\left (x \right )\right )\bigg | {\mstack {}{_{x \hiderel {=}-\frac {a_{0}}{a_{1}}}}}=0 \\ {} & \circ & x =-\frac {a_{0}}{a_{1}}\textrm {is a regular singular point}\hspace {3pt} \\ & {} & \textrm {Check to see if}\hspace {3pt} x_{0}=-\frac {a_{0}}{a_{1}}\hspace {3pt}\textrm {is a regular singular point}\hspace {3pt} \\ {} & {} & x_{0}=-\frac {a_{0}}{a_{1}} \\ \bullet & {} & \textrm {Multiply by denominators}\hspace {3pt} \\ {} & {} & \left (a_{1} x +a_{0} \right ) \left (\frac {d^{2}}{d x^{2}}y \left (x \right )\right )+\left (b_{1} x +b_{0} \right ) \left (\frac {d}{d x}y \left (x \right )\right )-m b_{1} y \left (x \right )=0 \\ \bullet & {} & \textrm {Change variables using}\hspace {3pt} x =u -\frac {a_{0}}{a_{1}}\hspace {3pt}\textrm {so that the regular singular point is at}\hspace {3pt} u =0 \\ {} & {} & a_{1} u \left (\frac {d^{2}}{d u^{2}}y \left (u \right )\right )+\left (b_{1} u -\frac {b_{1} a_{0}}{a_{1}}+b_{0} \right ) \left (\frac {d}{d u}y \left (u \right )\right )-m b_{1} 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 \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 \cdot \left (\frac {d^{2}}{d u^{2}}y \left (u \right )\right )\hspace {3pt}\textrm {to series expansion}\hspace {3pt} \\ {} & {} & u \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 -1} \\ {} & \circ & \textrm {Shift index using}\hspace {3pt} k \mathrm {->}k +1 \\ {} & {} & u \cdot \left (\frac {d^{2}}{d u^{2}}y \left (u \right )\right )=\moverset {\infty }{\munderset {k =-1}{\sum }}a_{k +1} \left (k +1+r \right ) \left (k +r \right ) u^{k +r} \\ & {} & \textrm {Rewrite ODE with series expansions}\hspace {3pt} \\ {} & {} & -\frac {a_{0} r \left (-a_{1}^{2} r +a_{0} b_{1} +a_{1}^{2}-a_{1} b_{0} \right ) u^{-1+r}}{a_{1}}+\left (\moverset {\infty }{\munderset {k =0}{\sum }}\left (-\frac {a_{k +1} \left (k +1+r \right ) \left (-a_{1}^{2} \left (k +1\right )-a_{1}^{2} r +a_{0} b_{1} +a_{1}^{2}-a_{1} b_{0} \right )}{a_{1}}+a_{k} b_{1} \left (k +r -m \right )\right ) u^{k +r}\right )=0 \\ \bullet & {} & a_{0}\textrm {cannot be 0 by assumption, giving the indicial equation}\hspace {3pt} \\ {} & {} & -\frac {r \left (-a_{1}^{2} r +a_{0} b_{1} +a_{1}^{2}-a_{1} b_{0} \right )}{a_{1}}=0 \\ \bullet & {} & \textrm {Values of r that satisfy the indicial equation}\hspace {3pt} \\ {} & {} & r \in \left \{0, \frac {a_{0} b_{1} +a_{1}^{2}-a_{1} b_{0}}{a_{1}^{2}}\right \} \\ \bullet & {} & \textrm {Each term in the series must be 0, giving the recursion relation}\hspace {3pt} \\ {} & {} & \frac {a_{k +1} \left (k +1+r \right ) \left (k +r \right ) a_{1}^{2}+\left (b_{0} \left (k +1+r \right ) a_{k +1}+a_{k} b_{1} \left (k +r -m \right )\right ) a_{1} -a_{0} b_{1} a_{k +1} \left (k +1+r \right )}{a_{1}}=0 \\ \bullet & {} & \textrm {Recursion relation that defines series solution to ODE}\hspace {3pt} \\ {} & {} & a_{k +1}=\frac {a_{k} b_{1} \left (k +r -m \right ) a_{1}}{\left (k +1+r \right ) \left (-a_{1}^{2} k -a_{1}^{2} r +a_{0} b_{1} -a_{1} b_{0} \right )} \\ \bullet & {} & \textrm {Recursion relation for}\hspace {3pt} r =0 \\ {} & {} & a_{k +1}=\frac {a_{k} b_{1} \left (k -m \right ) a_{1}}{\left (k +1\right ) \left (-a_{1}^{2} k +a_{0} b_{1} -a_{1} b_{0} \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 +1}=\frac {a_{k} b_{1} \left (k -m \right ) a_{1}}{\left (k +1\right ) \left (-a_{1}^{2} k +a_{0} b_{1} -a_{1} b_{0} \right )}\right ] \\ \bullet & {} & \textrm {Revert the change of variables}\hspace {3pt} u =x +\frac {a_{0}}{a_{1}} \\ {} & {} & \left [y \left (x \right )=\moverset {\infty }{\munderset {k =0}{\sum }}a_{k} \left (x +\frac {a_{0}}{a_{1}}\right )^{k}, a_{k +1}=\frac {a_{k} b_{1} \left (k -m \right ) a_{1}}{\left (k +1\right ) \left (-a_{1}^{2} k +a_{0} b_{1} -a_{1} b_{0} \right )}\right ] \\ \bullet & {} & \textrm {Recursion relation for}\hspace {3pt} r =\frac {a_{0} b_{1} +a_{1}^{2}-a_{1} b_{0}}{a_{1}^{2}} \\ {} & {} & a_{k +1}=\frac {a_{k} b_{1} \left (k +\frac {a_{0} b_{1} +a_{1}^{2}-a_{1} b_{0}}{a_{1}^{2}}-m \right ) a_{1}}{\left (k +1+\frac {a_{0} b_{1} +a_{1}^{2}-a_{1} b_{0}}{a_{1}^{2}}\right ) \left (-a_{1}^{2} k -a_{1}^{2}\right )} \\ \bullet & {} & \textrm {Solution for}\hspace {3pt} r =\frac {a_{0} b_{1} +a_{1}^{2}-a_{1} b_{0}}{a_{1}^{2}} \\ {} & {} & \left [y \left (u \right )=\moverset {\infty }{\munderset {k =0}{\sum }}a_{k} u^{k +\frac {a_{0} b_{1} +a_{1}^{2}-a_{1} b_{0}}{a_{1}^{2}}}, a_{k +1}=\frac {a_{k} b_{1} \left (k +\frac {a_{0} b_{1} +a_{1}^{2}-a_{1} b_{0}}{a_{1}^{2}}-m \right ) a_{1}}{\left (k +1+\frac {a_{0} b_{1} +a_{1}^{2}-a_{1} b_{0}}{a_{1}^{2}}\right ) \left (-a_{1}^{2} k -a_{1}^{2}\right )}\right ] \\ \bullet & {} & \textrm {Revert the change of variables}\hspace {3pt} u =x +\frac {a_{0}}{a_{1}} \\ {} & {} & \left [y \left (x \right )=\moverset {\infty }{\munderset {k =0}{\sum }}a_{k} \left (x +\frac {a_{0}}{a_{1}}\right )^{k +\frac {a_{0} b_{1} +a_{1}^{2}-a_{1} b_{0}}{a_{1}^{2}}}, a_{k +1}=\frac {a_{k} b_{1} \left (k +\frac {a_{0} b_{1} +a_{1}^{2}-a_{1} b_{0}}{a_{1}^{2}}-m \right ) a_{1}}{\left (k +1+\frac {a_{0} b_{1} +a_{1}^{2}-a_{1} b_{0}}{a_{1}^{2}}\right ) \left (-a_{1}^{2} k -a_{1}^{2}\right )}\right ] \\ \bullet & {} & \textrm {Combine solutions and rename parameters}\hspace {3pt} \\ {} & {} & \left [y \left (x \right )=\left (\moverset {\infty }{\munderset {k =0}{\sum }}a_{k} \left (x +\frac {a_{0}}{a_{1}}\right )^{k}\right )+\left (\moverset {\infty }{\munderset {k =0}{\sum }}b_{k} \left (x +\frac {a_{0}}{a_{1}}\right )^{k +\frac {a_{0} b_{1} +a_{1}^{2}-a_{1} b_{0}}{a_{1}^{2}}}\right ), a_{k +1}=\frac {a_{k} b_{1} \left (k -m \right ) a_{1}}{\left (k +1\right ) \left (-a_{1}^{2} k +a_{0} b_{1} -a_{1} b_{0} \right )}, b_{k +1}=\frac {b_{k} b_{1} \left (k +\frac {a_{0} b_{1} +a_{1}^{2}-a_{1} b_{0}}{a_{1}^{2}}-m \right ) a_{1}}{\left (k +1+\frac {a_{0} b_{1} +a_{1}^{2}-a_{1} b_{0}}{a_{1}^{2}}\right ) \left (-a_{1}^{2} k -a_{1}^{2}\right )}\right ] \end {array} \]

2.29.46.2 ✓ Mathematica. Time used: 0.113 (sec). Leaf size: 102

ode=(a1*x+a0)*D[y[x],{x,2}]+(b1*x+b0)*D[y[x],x]-m*b1*y[x]==0; 
ic={}; 
DSolve[{ode,ic},y[x],x,IncludeSingularSolutions->True]
 

\begin{align*} y(x)&\to e^{-\frac {\text {b1} x}{\text {a1}}} (\text {a0}+\text {a1} x)^{\frac {\text {a0} \text {b1}+\text {a1}^2-\text {a1} \text {b0}}{\text {a1}^2}} \left (c_1 \operatorname {HypergeometricU}\left (m+1,-\frac {\text {b0}}{\text {a1}}+\frac {\text {a0} \text {b1}}{\text {a1}^2}+2,\frac {\text {b1} (\text {a0}+\text {a1} x)}{\text {a1}^2}\right )+c_2 L_{-m-1}^{\frac {\text {a1}^2-\text {b0} \text {a1}+\text {a0} \text {b1}}{\text {a1}^2}}\left (\frac {\text {b1} (\text {a0}+\text {a1} x)}{\text {a1}^2}\right )\right ) \end{align*}

2.29.46.3 ✗ Sympy

from sympy import * 
x = symbols("x") 
a__0 = symbols("a__0") 
a__1 = symbols("a__1") 
b__0 = symbols("b__0") 
b__1 = symbols("b__1") 
m = symbols("m") 
y = Function("y") 
ode = Eq(-b__1*m*y(x) + (a__0 + a__1*x)*Derivative(y(x), (x, 2)) + (b__0 + b__1*x)*Derivative(y(x), x),0) 
ics = {} 
dsolve(ode,func=y(x),ics=ics)
 

False