2.31.14 Problem 162

Internal problem ID [13823]
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 : 162
Date solved : Thursday, January 01, 2026 at 02:45:15 AM
CAS classification : [[_2nd_order, _with_linear_symmetries], [_2nd_order, _linear, `_with_symmetry_[0,F(x)]`]]

2.31.14.1 Solved as second order ode using Kovacic algorithm

195.256 (sec)

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

Entering kovacic solverWriting the ode as

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

Comparing (1) and (2) shows that

\begin{align*} A &= a \,x^{2}+b \\ B &= a x\tag {3} \\ C &= c \end{align*}

Applying the Liouville transformation on the dependent variable gives

\begin{align*} z(x) &= y e^{\int \frac {B}{2 A} \,dx} \end{align*}

Then (2) becomes

\begin{align*} z''(x) = r z(x)\tag {4} \end{align*}

Where \(r\) is given by

\begin{align*} r &= \frac {s}{t}\tag {5} \\ &= \frac {2 A B' - 2 B A' + B^2 - 4 A C}{4 A^2} \end{align*}

Substituting the values of \(A,B,C\) from (3) in the above and simplifying gives

\begin{align*} r &= \frac {-a^{2} x^{2}-4 x^{2} c a +2 a b -4 b c}{4 \left (a \,x^{2}+b \right )^{2}}\tag {6} \end{align*}

Comparing the above to (5) shows that

\begin{align*} s &= -a^{2} x^{2}-4 x^{2} c a +2 a b -4 b c\\ t &= 4 \left (a \,x^{2}+b \right )^{2} \end{align*}

Therefore eq. (4) becomes

\begin{align*} z''(x) &= \left ( \frac {-a^{2} x^{2}-4 x^{2} c a +2 a b -4 b c}{4 \left (a \,x^{2}+b \right )^{2}}\right ) z(x)\tag {7} \end{align*}

Equation (7) is now solved. After finding \(z(x)\) then \(y\) is found using the inverse transformation

\begin{align*} y &= z \left (x \right ) e^{-\int \frac {B}{2 A} \,dx} \end{align*}

The first step is to determine the case of Kovacic algorithm this ode belongs to. There are 3 cases depending on the order of poles of \(r\) and the order of \(r\) at \(\infty \). The following table summarizes these cases.

Case	Allowed pole order for \(r\)	Allowed value for \(\mathcal {O}(\infty )\)
1	\(\left \{ 0,1,2,4,6,8,\cdots \right \} \)	\(\left \{ \cdots ,-6,-4,-2,0,2,3,4,5,6,\cdots \right \} \)
2	Need to have at least one pole that is either order \(2\) or odd order greater than \(2\). Any other pole order is allowed as long as the above condition is satisfied. Hence the following set of pole orders are all allowed. \(\{1,2\}\),\(\{1,3\}\),\(\{2\}\),\(\{3\}\),\(\{3,4\}\),\(\{1,2,5\}\).	no condition
3	\(\left \{ 1,2\right \} \)	\(\left \{ 2,3,4,5,6,7,\cdots \right \} \)

Table 2.65: Necessary conditions for each Kovacic case

The order of \(r\) at \(\infty \) is the degree of \(t\) minus the degree of \(s\). Therefore

\begin{align*} O\left (\infty \right ) &= \text {deg}(t) - \text {deg}(s) \\ &= 4 - 2 \\ &= 2 \end{align*}

The poles of \(r\) in eq. (7) and the order of each pole are determined by solving for the roots of \(t=4 \left (a \,x^{2}+b \right )^{2}\). There is a pole at \(x=\frac {\sqrt {-a b}}{a}\) of order \(2\). There is a pole at \(x=-\frac {\sqrt {-a b}}{a}\) of order \(2\). Since there is no odd order pole larger than \(2\) and the order at \(\infty \) is \(2\) then the necessary conditions for case one are met. Since there is a pole of order \(2\) then necessary conditions for case two are met. Since pole order is not larger than \(2\) and the order at \(\infty \) is \(2\) then the necessary conditions for case three are met. Therefore

\begin{align*} L &= [1, 2, 4, 6, 12] \end{align*}

Attempting to find a solution using case \(n=1\).

Unable to find solution using case one

Attempting to find a solution using case \(n=2\).

Unable to find solution using case two.

Attempting to find a solution using \(n=4\).

Looking at poles of order 2. The partial fractions decomposition of \(r\) is

\[ r = -\frac {3}{16 \left (x -\sqrt {-\frac {b}{a}}\right )^{2}}-\frac {3}{16 \left (x +\sqrt {-\frac {b}{a}}\right )^{2}}+\frac {-a b +8 b c}{16 a^{2} \left (-\frac {b}{a}\right )^{{3}/{2}} \left (x -\sqrt {-\frac {b}{a}}\right )}-\frac {-a b +8 b c}{16 a^{2} \left (-\frac {b}{a}\right )^{{3}/{2}} \left (x +\sqrt {-\frac {b}{a}}\right )} \]

For the pole at \(x=\frac {\sqrt {-a b}}{a}\) let \(b\) be the coefficient of \(\frac {1}{ \left (x -\frac {\sqrt {-a b}}{a}\right )^{2}}\) in the partial fractions decomposition of \(r\) given above. This shows that \(b=0\). Hence

\begin{align*} E_c &= \left \{ 6+\frac {12 k}{n}\sqrt {1+4 b}|k=0,\pm 1, \pm 2, \dots , \pm \frac {n}{2} \right \} \cap \mathbb {Z} \end{align*}

Where \(n\) for case \(3\) is \(4,6\) or \(12\). For the current case \(n=4\). Hence the above becomes

\begin{align*} E_c &= \{0, 3, 6, 9, 12\} \end{align*}

For the pole at \(x=-\frac {\sqrt {-a b}}{a}\) let \(b\) be the coefficient of \(\frac {1}{ \left (x +\frac {\sqrt {-a b}}{a}\right )^{2}}\) in the partial fractions decomposition of \(r\) given above. This shows that \(b=0\). Hence

\begin{align*} E_c &= \left \{ 6+\frac {12 k}{n}\sqrt {1+4 b}|k=0,\pm 1, \pm 2, \dots , \pm \frac {n}{2} \right \} \cap \mathbb {Z} \end{align*}

Where \(n\) for case \(3\) is \(4,6\) or \(12\). For the current case \(n=4\). Hence the above becomes

\begin{align*} E_c &= \{0, 3, 6, 9, 12\} \end{align*}

Let

\begin{align*} E_\infty &= \left \{ 6+\frac {12 k}{n}\sqrt {1+4 b}|k=0,\pm 1, \pm 2, \dots , \pm \frac {n}{2} \right \} \cap \mathbb {Z} \tag {B1} \end{align*}

Where \(b\) is the coefficient of \(\frac {1}{x^{2}}\) in the Laurent series for \(r\) at \(\infty \) given by

\[ r \approx \frac {-a^{2}-4 a c}{4 a^{2} x^{2}}+\frac {-\frac {\left (-a^{2}-4 a c \right ) b}{2 a^{3}}+\frac {2 a b -4 b c}{4 a^{2}}}{x^{4}}+\frac {\frac {3 \left (-a^{2}-4 a c \right ) b^{2}}{4 a^{4}}-\frac {\left (2 a b -4 b c \right ) b}{2 a^{3}}}{x^{6}}+\frac {-\frac {\left (-a^{2}-4 a c \right ) b^{3}}{a^{5}}+\frac {3 \left (2 a b -4 b c \right ) b^{2}}{4 a^{4}}}{x^{8}}+\frac {\frac {5 \left (-a^{2}-4 a c \right ) b^{4}}{4 a^{6}}-\frac {\left (2 a b -4 b c \right ) b^{3}}{a^{5}}}{x^{10}}+\frac {-\frac {3 \left (-a^{2}-4 a c \right ) b^{5}}{2 a^{7}}+\frac {5 \left (2 a b -4 b c \right ) b^{4}}{4 a^{6}}}{x^{12}} + \cdots \]

The above shows that

\[ b = -{\frac {1}{4}} \]

The value of \(n\) in eq. (B1) for case \(3\) is \(4,6\) or \(2\).For the current case \(n=4\). eq. (B1) simplifies to the following, after removing any duplicate and non integer entries in the set.

\begin{align*} E_\infty &= \{6\} \end{align*}

The following table summarizes the results found so far for poles and for the order of \(r\) at \(\infty \) for case 3 of Kovacic algorithm using \(n=4\).

pole \(c\) location	pole order	set \(\{E_c\}\)
\(\frac {\sqrt {-a b}}{a}\)	\(2\)	\(\{0, 3, 6, 9, 12\}\)
\(-\frac {\sqrt {-a b}}{a}\)	\(2\)	\(\{0, 3, 6, 9, 12\}\)

Order of \(r\) at \(\infty \)	set \(\{E_\infty \}\)
\(2\)	\(\{6\}\)

Now that \(E_c\) sets for all poles are found and \(E_\infty \) set is found, the next step is to determine a non negative integer \(d\) using the following

\begin{align*} d &= \frac {n}{12} \left ( e_\infty - \sum _{c \in \Gamma } e_c \right ) \end{align*}

Where in the above \(e_c\) is a distinct element from each corresponding \(E_c\). This means all possible tuples \(\{e_{c_1},e_{c_2},\dots ,e_{c_n}\}\) are tried in the sum above, where \(e_{c_i}\) is one element of each \(E_c\) found earlier. Using the following family \(\{e_1,e_2,\dots ,e_\infty \}\) given by

\[ e_1=3,\hspace {3pt} e_2=3,\hspace {3pt} e_\infty =6 \]

Gives a non negative integer \(d\) (the degree of the polynomial \(p(x)\)), which is generated using

\begin{align*} d &= \frac {n}{12} \left ( e_\infty - \sum _{c \in \Gamma } e_c \right )\\ &= \frac {4}{12} \left ( 6 - \left (3+\left (3\right )\right )\right )\\ &= 0 \end{align*}

The following rational function is

\begin{align*} \theta &= \frac {n}{12} \sum _{c \in \Gamma } \frac {e_c}{x-c} \\ &= \frac {4}{12} \left (\frac {3}{\left (x-\left (\frac {\sqrt {-a b}}{a}\right )\right )}+\frac {3}{\left (x-\left (-\frac {\sqrt {-a b}}{a}\right )\right )}\right ) \\ &= \frac {2 a x}{a \,x^{2}+b} \end{align*}

And

\begin{align*} S &= \prod _{c\in \Gamma } (x-c) \\ &= \left (x -\frac {\sqrt {-a b}}{a}\right ) \left (x +\frac {\sqrt {-a b}}{a}\right ) \end{align*}

The polynomial \(p(x)\) is now determined. Since the degree of the polynomial is \(d=0\), then let

\[ p(x) = 1 \]

The following set of equations are set up in order to determine the coefficients \(a_i\) (if any) of the above polynomial

\begin{align*} P_n &= - p(x) \\ &= - 1 \\ P_{i-1} &= - S p'_{i} + ((n-i)S' -S\theta ) P_i - (n-1)(i+1) S^2 r P_{i+1} \qquad i=n,n-1,\dots , 0 \tag {1A} \end{align*}

The coefficients \(a_i\) are solved for from

\[ P_{-1} = 0 \tag {2A} \]

By using method of undetermined coefficients. Carrying the above computation in eq. (1A) gives the following sequence of polynomials \(P_i\) (noting that \(n=4\) and \(r=\frac {-a^{2} x^{2}-4 x^{2} c a +2 a b -4 b c}{4 \left (a \,x^{2}+b \right )^{2}}\)).

\begin{align*} P_{4} &= - p = -1 \\ P_{3} &= 2 x \\ P_{2} &= \frac {-3 a^{2} x^{2}-4 x^{2} c a -4 b c}{a^{2}} \\ P_{1} &= \frac {3 x \left (a \left (a +4 c \right ) x^{2}+4 b c \right )}{a^{2}} \\ P_{0} &= -\frac {3 \left (a^{2} x^{2}+4 x^{2} c a +4 b c \right )^{2}}{2 a^{4}} \\ P_{-1} &= 0 \\ \end{align*}

Because \(P_{-1} = 0\) then \(z=e^{\int \omega }\) is a solution. \(\omega \) is found by finding a solution to the equation generated by the following sum

\begin{align*} \sum _{i=0}^{n} S^i \frac {P_i}{(n-i)!} \omega ^i &= 0 \\ \sum _{i=0}^{4} S^i \frac {P_i}{(4-i)!} \omega ^i &= 0 \end{align*}

Where the \(P_i\) are the polynomials found earlier. Computing the above sum gives

\begin{equation} \tag{3A} \frac {1}{16 a^{4}}\left (-\left (4 a^{2} x^{4} \omega ^{2}-4 a^{2} x^{3} \omega +8 a b \,\omega ^{2} x^{2}+a^{2} x^{2}-4 a b \omega x +4 x^{2} c a +4 b^{2} \omega ^{2}+4 b c \right )^{2}\right ) =0 \end{equation}

The solution \(\omega \) of eq. 3A is found as

\[ \omega =\frac {1}{2 a \,x^{2}+2 b}\left (a x -2 \sqrt {-\left (a \,x^{2}+b \right ) c}\right ) \]

This \(\omega \) is used to find a solution to \(z''=r z\).

\[ z_1(x) = e^{ \int \omega \,dx}\tag {5A} \]

Doing the integration gives in eq. (4A) gives

\begin{align*} z_1(x) &= e^{ \int \omega \,dx} \\ &= {\mathrm e}^{\int \frac {a x -2 \sqrt {-\left (a \,x^{2}+b \right ) c}}{2 a \,x^{2}+2 b}d x} \\ &= \left (a \,x^{2}+b \right )^{{1}/{4}} {\mathrm e}^{\frac {c \arctan \left (\frac {\sqrt {a c}\, x}{\sqrt {-\left (a \,x^{2}+b \right ) c}}\right )}{\sqrt {a c}}} \\ \end{align*}

Which simplifies to

\[ z_1(x) = \left (a \,x^{2}+b \right )^{{1}/{4}} {\mathrm e}^{\frac {c \arctan \left (x \sqrt {-\frac {a}{a \,x^{2}+b}}\right )}{\sqrt {a c}}} \]

The first solution to the original ode in \(y\) is found from

\begin{align*} y_1 &= z_1 e^{ \int -\frac {1}{2} \frac {B}{A} \,dx} \\ &= z_1 e^{ -\int \frac {1}{2} \frac {a x}{a \,x^{2}+b} \,dx} \\ &= z_1 e^{-\frac {\ln \left (a \,x^{2}+b \right )}{4}} \\ &= z_1 \left (\frac {1}{\left (a \,x^{2}+b \right )^{{1}/{4}}}\right ) \\ \end{align*}

Which simplifies to

\[ y_1 = {\mathrm e}^{\frac {c \arctan \left (x \sqrt {-\frac {a}{a \,x^{2}+b}}\right )}{\sqrt {a c}}} \]

The second solution \(y_2\) to the original ode is found using reduction of order

\[ y_2 = y_1 \int \frac { e^{\int -\frac {B}{A} \,dx}}{y_1^2} \,dx \]

Substituting gives

\begin{align*} y_2 &= y_1 \int \frac { e^{\int -\frac {a x}{a \,x^{2}+b} \,dx}}{\left (y_1\right )^2} \,dx \\ &= y_1 \int \frac { e^{-\frac {\ln \left (a \,x^{2}+b \right )}{2}}}{\left (y_1\right )^2} \,dx \\ &= y_1 \left (\int \frac {{\mathrm e}^{-\frac {2 c \arctan \left (x \sqrt {-\frac {a}{a \,x^{2}+b}}\right )}{\sqrt {a c}}}}{\sqrt {a \,x^{2}+b}}d x\right ) \\ \end{align*}

Therefore the solution is

\begin{align*} y &= c_1 y_1 + c_2 y_2 \\ &= c_1 \left ({\mathrm e}^{\frac {c \arctan \left (x \sqrt {-\frac {a}{a \,x^{2}+b}}\right )}{\sqrt {a c}}}\right ) + c_2 \left ({\mathrm e}^{\frac {c \arctan \left (x \sqrt {-\frac {a}{a \,x^{2}+b}}\right )}{\sqrt {a c}}}\left (\int \frac {{\mathrm e}^{-\frac {2 c \arctan \left (x \sqrt {-\frac {a}{a \,x^{2}+b}}\right )}{\sqrt {a c}}}}{\sqrt {a \,x^{2}+b}}d x\right )\right ) \\ \end{align*}

2.31.14.2 ✓ Maple. Time used: 0.002 (sec). Leaf size: 59

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

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

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)] 
<- linear_1 successful
 

2.31.14.3 ✓ Mathematica. Time used: 0.044 (sec). Leaf size: 62

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

\begin{align*} y(x)&\to c_1 \cos \left (\frac {\sqrt {c} \text {arcsinh}\left (\frac {\sqrt {a} x}{\sqrt {b}}\right )}{\sqrt {a}}\right )+c_2 \sin \left (\frac {\sqrt {c} \text {arcsinh}\left (\frac {\sqrt {a} x}{\sqrt {b}}\right )}{\sqrt {a}}\right ) \end{align*}

2.31.14.4 ✗ Sympy

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

False