2.7.3 Problem 3

Internal problem ID [13348]
Book : Handbook of exact solutions for ordinary differential equations. By Polyanin and Zaitsev. Second edition
Section : Chapter 1, section 1.2. Riccati Equation. subsection 1.2.5-1. Equations Containing Logarithmic Functions
Problem number : 3
Date solved : Wednesday, December 31, 2025 at 01:41:26 PM
CAS classification : [_Riccati]

2.7.3.1 Solved using first_order_ode_riccati

16.672 (sec)

Entering first order ode riccati solver

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

In canonical form the ODE is

\begin{align*} y' &= F(x,y)\\ &= \frac {a y^{2}+b \ln \left (x \right )^{k}+c \ln \left (x \right )^{2 k +2}}{x} \end{align*}

This is a Riccati ODE. Comparing the ODE to solve

\[ y' = \textit {the\_rhs} \]

With Riccati ODE standard form

\[ y' = f_0(x)+ f_1(x)y+f_2(x)y^{2} \]

Shows that \(f_0(x)=\frac {b \ln \left (x \right )^{k}}{x}+\frac {c \ln \left (x \right )^{2 k} \ln \left (x \right )^{2}}{x}\), \(f_1(x)=0\) and \(f_2(x)=\frac {a}{x}\). Let

\begin{align*} y &= \frac {-u'}{f_2 u} \\ &= \frac {-u'}{\frac {u a}{x}} \tag {1} \end{align*}

Using the above substitution in the given ODE results (after some simplification) in a second order ODE to solve for \(u(x)\) which is

\begin{align*} f_2 u''(x) -\left ( f_2' + f_1 f_2 \right ) u'(x) + f_2^2 f_0 u(x) &= 0 \tag {2} \end{align*}

But

\begin{align*} f_2' &=-\frac {a}{x^{2}}\\ f_1 f_2 &=0\\ f_2^2 f_0 &=\frac {a^{2} \left (\frac {b \ln \left (x \right )^{k}}{x}+\frac {c \ln \left (x \right )^{2 k} \ln \left (x \right )^{2}}{x}\right )}{x^{2}} \end{align*}

Substituting the above terms back in equation (2) gives

\[ \frac {a u^{\prime \prime }\left (x \right )}{x}+\frac {a u^{\prime }\left (x \right )}{x^{2}}+\frac {a^{2} \left (\frac {b \ln \left (x \right )^{k}}{x}+\frac {c \ln \left (x \right )^{2 k} \ln \left (x \right )^{2}}{x}\right ) u \left (x \right )}{x^{2}} = 0 \]

Unable to solve. Will ask Maple to solve this ode now.

The solution for \(u \left (x \right )\) is

\begin{equation} \tag{3} u \left (x \right ) = c_1 \,{\mathrm e}^{-\frac {i \sqrt {a}\, \sqrt {c}\, \ln \left (x \right )^{k +2}}{k +2}} \operatorname {hypergeom}\left (\left [\frac {\left (k +1\right ) \sqrt {c}+i \sqrt {a}\, b}{\sqrt {c}\, \left (4+2 k \right )}\right ], \left [\frac {k +1}{k +2}\right ], \frac {2 i \sqrt {a}\, \sqrt {c}\, \ln \left (x \right )^{k +2}}{k +2}\right )+c_2 \ln \left (x \right ) {\mathrm e}^{-\frac {i \sqrt {a}\, \sqrt {c}\, \ln \left (x \right )^{k +2}}{k +2}} \operatorname {hypergeom}\left (\left [\frac {\left (k +3\right ) \sqrt {c}+i \sqrt {a}\, b}{\sqrt {c}\, \left (4+2 k \right )}\right ], \left [\frac {k +3}{k +2}\right ], \frac {2 i \sqrt {a}\, \sqrt {c}\, \ln \left (x \right )^{k +2}}{k +2}\right ) \end{equation}

Taking derivative gives

\begin{equation} \tag{4} u^{\prime }\left (x \right ) = -\frac {i c_1 \sqrt {a}\, \sqrt {c}\, \ln \left (x \right )^{k +2} {\mathrm e}^{-\frac {i \sqrt {a}\, \sqrt {c}\, \ln \left (x \right )^{k +2}}{k +2}} \operatorname {hypergeom}\left (\left [\frac {\left (k +1\right ) \sqrt {c}+i \sqrt {a}\, b}{\sqrt {c}\, \left (4+2 k \right )}\right ], \left [\frac {k +1}{k +2}\right ], \frac {2 i \sqrt {a}\, \sqrt {c}\, \ln \left (x \right )^{k +2}}{k +2}\right )}{x \ln \left (x \right )}+\frac {2 i c_1 \,{\mathrm e}^{-\frac {i \sqrt {a}\, \sqrt {c}\, \ln \left (x \right )^{k +2}}{k +2}} \left (\left (k +1\right ) \sqrt {c}+i \sqrt {a}\, b \right ) \left (k +2\right ) \operatorname {hypergeom}\left (\left [\frac {\left (k +1\right ) \sqrt {c}+i \sqrt {a}\, b}{\sqrt {c}\, \left (4+2 k \right )}+1\right ], \left [\frac {k +1}{k +2}+1\right ], \frac {2 i \sqrt {a}\, \sqrt {c}\, \ln \left (x \right )^{k +2}}{k +2}\right ) \sqrt {a}\, \ln \left (x \right )^{k +2}}{\left (4+2 k \right ) \left (k +1\right ) x \ln \left (x \right )}+\frac {c_2 \,{\mathrm e}^{-\frac {i \sqrt {a}\, \sqrt {c}\, \ln \left (x \right )^{k +2}}{k +2}} \operatorname {hypergeom}\left (\left [\frac {\left (k +3\right ) \sqrt {c}+i \sqrt {a}\, b}{\sqrt {c}\, \left (4+2 k \right )}\right ], \left [\frac {k +3}{k +2}\right ], \frac {2 i \sqrt {a}\, \sqrt {c}\, \ln \left (x \right )^{k +2}}{k +2}\right )}{x}-\frac {i c_2 \sqrt {a}\, \sqrt {c}\, \ln \left (x \right )^{k +2} {\mathrm e}^{-\frac {i \sqrt {a}\, \sqrt {c}\, \ln \left (x \right )^{k +2}}{k +2}} \operatorname {hypergeom}\left (\left [\frac {\left (k +3\right ) \sqrt {c}+i \sqrt {a}\, b}{\sqrt {c}\, \left (4+2 k \right )}\right ], \left [\frac {k +3}{k +2}\right ], \frac {2 i \sqrt {a}\, \sqrt {c}\, \ln \left (x \right )^{k +2}}{k +2}\right )}{x}+\frac {2 i c_2 \,{\mathrm e}^{-\frac {i \sqrt {a}\, \sqrt {c}\, \ln \left (x \right )^{k +2}}{k +2}} \left (\left (k +3\right ) \sqrt {c}+i \sqrt {a}\, b \right ) \left (k +2\right ) \operatorname {hypergeom}\left (\left [\frac {\left (k +3\right ) \sqrt {c}+i \sqrt {a}\, b}{\sqrt {c}\, \left (4+2 k \right )}+1\right ], \left [\frac {k +3}{k +2}+1\right ], \frac {2 i \sqrt {a}\, \sqrt {c}\, \ln \left (x \right )^{k +2}}{k +2}\right ) \sqrt {a}\, \ln \left (x \right )^{k +2}}{\left (4+2 k \right ) \left (k +3\right ) x} \end{equation}

Substituting equations (3,4) into (1) results in

\begin{align*} y &= \frac {-u'}{f_2 u} \\ y &= \frac {-u'}{\frac {u a}{x}} \\ y &= -\frac {\left (-\frac {i c_1 \sqrt {a}\, \sqrt {c}\, \ln \left (x \right )^{k +2} {\mathrm e}^{-\frac {i \sqrt {a}\, \sqrt {c}\, \ln \left (x \right )^{k +2}}{k +2}} \operatorname {hypergeom}\left (\left [\frac {\left (k +1\right ) \sqrt {c}+i \sqrt {a}\, b}{\sqrt {c}\, \left (4+2 k \right )}\right ], \left [\frac {k +1}{k +2}\right ], \frac {2 i \sqrt {a}\, \sqrt {c}\, \ln \left (x \right )^{k +2}}{k +2}\right )}{x \ln \left (x \right )}+\frac {2 i c_1 \,{\mathrm e}^{-\frac {i \sqrt {a}\, \sqrt {c}\, \ln \left (x \right )^{k +2}}{k +2}} \left (\left (k +1\right ) \sqrt {c}+i \sqrt {a}\, b \right ) \left (k +2\right ) \operatorname {hypergeom}\left (\left [\frac {\left (k +1\right ) \sqrt {c}+i \sqrt {a}\, b}{\sqrt {c}\, \left (4+2 k \right )}+1\right ], \left [\frac {k +1}{k +2}+1\right ], \frac {2 i \sqrt {a}\, \sqrt {c}\, \ln \left (x \right )^{k +2}}{k +2}\right ) \sqrt {a}\, \ln \left (x \right )^{k +2}}{\left (4+2 k \right ) \left (k +1\right ) x \ln \left (x \right )}+\frac {c_2 \,{\mathrm e}^{-\frac {i \sqrt {a}\, \sqrt {c}\, \ln \left (x \right )^{k +2}}{k +2}} \operatorname {hypergeom}\left (\left [\frac {\left (k +3\right ) \sqrt {c}+i \sqrt {a}\, b}{\sqrt {c}\, \left (4+2 k \right )}\right ], \left [\frac {k +3}{k +2}\right ], \frac {2 i \sqrt {a}\, \sqrt {c}\, \ln \left (x \right )^{k +2}}{k +2}\right )}{x}-\frac {i c_2 \sqrt {a}\, \sqrt {c}\, \ln \left (x \right )^{k +2} {\mathrm e}^{-\frac {i \sqrt {a}\, \sqrt {c}\, \ln \left (x \right )^{k +2}}{k +2}} \operatorname {hypergeom}\left (\left [\frac {\left (k +3\right ) \sqrt {c}+i \sqrt {a}\, b}{\sqrt {c}\, \left (4+2 k \right )}\right ], \left [\frac {k +3}{k +2}\right ], \frac {2 i \sqrt {a}\, \sqrt {c}\, \ln \left (x \right )^{k +2}}{k +2}\right )}{x}+\frac {2 i c_2 \,{\mathrm e}^{-\frac {i \sqrt {a}\, \sqrt {c}\, \ln \left (x \right )^{k +2}}{k +2}} \left (\left (k +3\right ) \sqrt {c}+i \sqrt {a}\, b \right ) \left (k +2\right ) \operatorname {hypergeom}\left (\left [\frac {\left (k +3\right ) \sqrt {c}+i \sqrt {a}\, b}{\sqrt {c}\, \left (4+2 k \right )}+1\right ], \left [\frac {k +3}{k +2}+1\right ], \frac {2 i \sqrt {a}\, \sqrt {c}\, \ln \left (x \right )^{k +2}}{k +2}\right ) \sqrt {a}\, \ln \left (x \right )^{k +2}}{\left (4+2 k \right ) \left (k +3\right ) x}\right ) x}{a \left (c_1 \,{\mathrm e}^{-\frac {i \sqrt {a}\, \sqrt {c}\, \ln \left (x \right )^{k +2}}{k +2}} \operatorname {hypergeom}\left (\left [\frac {\left (k +1\right ) \sqrt {c}+i \sqrt {a}\, b}{\sqrt {c}\, \left (4+2 k \right )}\right ], \left [\frac {k +1}{k +2}\right ], \frac {2 i \sqrt {a}\, \sqrt {c}\, \ln \left (x \right )^{k +2}}{k +2}\right )+c_2 \ln \left (x \right ) {\mathrm e}^{-\frac {i \sqrt {a}\, \sqrt {c}\, \ln \left (x \right )^{k +2}}{k +2}} \operatorname {hypergeom}\left (\left [\frac {\left (k +3\right ) \sqrt {c}+i \sqrt {a}\, b}{\sqrt {c}\, \left (4+2 k \right )}\right ], \left [\frac {k +3}{k +2}\right ], \frac {2 i \sqrt {a}\, \sqrt {c}\, \ln \left (x \right )^{k +2}}{k +2}\right )\right )} \\ \end{align*}

Doing change of constants, the above solution becomes

\[ y = -\frac {\left (-\frac {i \sqrt {a}\, \sqrt {c}\, \ln \left (x \right )^{k +2} {\mathrm e}^{-\frac {i \sqrt {a}\, \sqrt {c}\, \ln \left (x \right )^{k +2}}{k +2}} \operatorname {hypergeom}\left (\left [\frac {\left (k +1\right ) \sqrt {c}+i \sqrt {a}\, b}{\sqrt {c}\, \left (4+2 k \right )}\right ], \left [\frac {k +1}{k +2}\right ], \frac {2 i \sqrt {a}\, \sqrt {c}\, \ln \left (x \right )^{k +2}}{k +2}\right )}{x \ln \left (x \right )}+\frac {2 i {\mathrm e}^{-\frac {i \sqrt {a}\, \sqrt {c}\, \ln \left (x \right )^{k +2}}{k +2}} \left (\left (k +1\right ) \sqrt {c}+i \sqrt {a}\, b \right ) \left (k +2\right ) \operatorname {hypergeom}\left (\left [\frac {\left (k +1\right ) \sqrt {c}+i \sqrt {a}\, b}{\sqrt {c}\, \left (4+2 k \right )}+1\right ], \left [\frac {k +1}{k +2}+1\right ], \frac {2 i \sqrt {a}\, \sqrt {c}\, \ln \left (x \right )^{k +2}}{k +2}\right ) \sqrt {a}\, \ln \left (x \right )^{k +2}}{\left (4+2 k \right ) \left (k +1\right ) x \ln \left (x \right )}+\frac {c_3 \,{\mathrm e}^{-\frac {i \sqrt {a}\, \sqrt {c}\, \ln \left (x \right )^{k +2}}{k +2}} \operatorname {hypergeom}\left (\left [\frac {\left (k +3\right ) \sqrt {c}+i \sqrt {a}\, b}{\sqrt {c}\, \left (4+2 k \right )}\right ], \left [\frac {k +3}{k +2}\right ], \frac {2 i \sqrt {a}\, \sqrt {c}\, \ln \left (x \right )^{k +2}}{k +2}\right )}{x}-\frac {i c_3 \sqrt {a}\, \sqrt {c}\, \ln \left (x \right )^{k +2} {\mathrm e}^{-\frac {i \sqrt {a}\, \sqrt {c}\, \ln \left (x \right )^{k +2}}{k +2}} \operatorname {hypergeom}\left (\left [\frac {\left (k +3\right ) \sqrt {c}+i \sqrt {a}\, b}{\sqrt {c}\, \left (4+2 k \right )}\right ], \left [\frac {k +3}{k +2}\right ], \frac {2 i \sqrt {a}\, \sqrt {c}\, \ln \left (x \right )^{k +2}}{k +2}\right )}{x}+\frac {2 i c_3 \,{\mathrm e}^{-\frac {i \sqrt {a}\, \sqrt {c}\, \ln \left (x \right )^{k +2}}{k +2}} \left (\left (k +3\right ) \sqrt {c}+i \sqrt {a}\, b \right ) \left (k +2\right ) \operatorname {hypergeom}\left (\left [\frac {\left (k +3\right ) \sqrt {c}+i \sqrt {a}\, b}{\sqrt {c}\, \left (4+2 k \right )}+1\right ], \left [\frac {k +3}{k +2}+1\right ], \frac {2 i \sqrt {a}\, \sqrt {c}\, \ln \left (x \right )^{k +2}}{k +2}\right ) \sqrt {a}\, \ln \left (x \right )^{k +2}}{\left (4+2 k \right ) \left (k +3\right ) x}\right ) x}{a \left ({\mathrm e}^{-\frac {i \sqrt {a}\, \sqrt {c}\, \ln \left (x \right )^{k +2}}{k +2}} \operatorname {hypergeom}\left (\left [\frac {\left (k +1\right ) \sqrt {c}+i \sqrt {a}\, b}{\sqrt {c}\, \left (4+2 k \right )}\right ], \left [\frac {k +1}{k +2}\right ], \frac {2 i \sqrt {a}\, \sqrt {c}\, \ln \left (x \right )^{k +2}}{k +2}\right )+c_3 \ln \left (x \right ) {\mathrm e}^{-\frac {i \sqrt {a}\, \sqrt {c}\, \ln \left (x \right )^{k +2}}{k +2}} \operatorname {hypergeom}\left (\left [\frac {\left (k +3\right ) \sqrt {c}+i \sqrt {a}\, b}{\sqrt {c}\, \left (4+2 k \right )}\right ], \left [\frac {k +3}{k +2}\right ], \frac {2 i \sqrt {a}\, \sqrt {c}\, \ln \left (x \right )^{k +2}}{k +2}\right )\right )} \]

Simplifying the above gives

\begin{align*} y &= \frac {\left (-i \left (k +1\right ) \sqrt {c}\, \sqrt {a}+a b \right ) \ln \left (x \right )^{k +1} \left (k +3\right ) \operatorname {hypergeom}\left (\left [\frac {\left (3 k +5\right ) \sqrt {c}+i \sqrt {a}\, b}{\sqrt {c}\, \left (4+2 k \right )}\right ], \left [\frac {3+2 k}{k +2}\right ], \frac {2 i \sqrt {a}\, \sqrt {c}\, \ln \left (x \right )^{k +2}}{k +2}\right )+\left (\left (-i \left (k +3\right ) \sqrt {c}\, \sqrt {a}+a b \right ) \ln \left (x \right )^{k +2} c_3 \operatorname {hypergeom}\left (\left [\frac {\left (3 k +7\right ) \sqrt {c}+i \sqrt {a}\, b}{\sqrt {c}\, \left (4+2 k \right )}\right ], \left [\frac {2 k +5}{k +2}\right ], \frac {2 i \sqrt {a}\, \sqrt {c}\, \ln \left (x \right )^{k +2}}{k +2}\right )+\left (c_3 \left (i \sqrt {a}\, \sqrt {c}\, \ln \left (x \right )^{k +2}-1\right ) \operatorname {hypergeom}\left (\left [\frac {\left (k +3\right ) \sqrt {c}+i \sqrt {a}\, b}{\sqrt {c}\, \left (4+2 k \right )}\right ], \left [\frac {k +3}{k +2}\right ], \frac {2 i \sqrt {a}\, \sqrt {c}\, \ln \left (x \right )^{k +2}}{k +2}\right )+i \ln \left (x \right )^{k +1} \sqrt {a}\, \sqrt {c}\, \operatorname {hypergeom}\left (\left [\frac {\left (k +1\right ) \sqrt {c}+i \sqrt {a}\, b}{\sqrt {c}\, \left (4+2 k \right )}\right ], \left [\frac {k +1}{k +2}\right ], \frac {2 i \sqrt {a}\, \sqrt {c}\, \ln \left (x \right )^{k +2}}{k +2}\right )\right ) \left (k +3\right )\right ) \left (k +1\right )}{\left (c_3 \operatorname {hypergeom}\left (\left [\frac {\left (k +3\right ) \sqrt {c}+i \sqrt {a}\, b}{\sqrt {c}\, \left (4+2 k \right )}\right ], \left [\frac {k +3}{k +2}\right ], \frac {2 i \sqrt {a}\, \sqrt {c}\, \ln \left (x \right )^{k +2}}{k +2}\right ) \ln \left (x \right )+\operatorname {hypergeom}\left (\left [\frac {\left (k +1\right ) \sqrt {c}+i \sqrt {a}\, b}{\sqrt {c}\, \left (4+2 k \right )}\right ], \left [\frac {k +1}{k +2}\right ], \frac {2 i \sqrt {a}\, \sqrt {c}\, \ln \left (x \right )^{k +2}}{k +2}\right )\right ) a \left (k +3\right ) \left (k +1\right )} \\ \end{align*}

Summary of solutions found

\begin{align*} y &= \frac {\left (-i \left (k +1\right ) \sqrt {c}\, \sqrt {a}+a b \right ) \ln \left (x \right )^{k +1} \left (k +3\right ) \operatorname {hypergeom}\left (\left [\frac {\left (3 k +5\right ) \sqrt {c}+i \sqrt {a}\, b}{\sqrt {c}\, \left (4+2 k \right )}\right ], \left [\frac {3+2 k}{k +2}\right ], \frac {2 i \sqrt {a}\, \sqrt {c}\, \ln \left (x \right )^{k +2}}{k +2}\right )+\left (\left (-i \left (k +3\right ) \sqrt {c}\, \sqrt {a}+a b \right ) \ln \left (x \right )^{k +2} c_3 \operatorname {hypergeom}\left (\left [\frac {\left (3 k +7\right ) \sqrt {c}+i \sqrt {a}\, b}{\sqrt {c}\, \left (4+2 k \right )}\right ], \left [\frac {2 k +5}{k +2}\right ], \frac {2 i \sqrt {a}\, \sqrt {c}\, \ln \left (x \right )^{k +2}}{k +2}\right )+\left (c_3 \left (i \sqrt {a}\, \sqrt {c}\, \ln \left (x \right )^{k +2}-1\right ) \operatorname {hypergeom}\left (\left [\frac {\left (k +3\right ) \sqrt {c}+i \sqrt {a}\, b}{\sqrt {c}\, \left (4+2 k \right )}\right ], \left [\frac {k +3}{k +2}\right ], \frac {2 i \sqrt {a}\, \sqrt {c}\, \ln \left (x \right )^{k +2}}{k +2}\right )+i \ln \left (x \right )^{k +1} \sqrt {a}\, \sqrt {c}\, \operatorname {hypergeom}\left (\left [\frac {\left (k +1\right ) \sqrt {c}+i \sqrt {a}\, b}{\sqrt {c}\, \left (4+2 k \right )}\right ], \left [\frac {k +1}{k +2}\right ], \frac {2 i \sqrt {a}\, \sqrt {c}\, \ln \left (x \right )^{k +2}}{k +2}\right )\right ) \left (k +3\right )\right ) \left (k +1\right )}{\left (c_3 \operatorname {hypergeom}\left (\left [\frac {\left (k +3\right ) \sqrt {c}+i \sqrt {a}\, b}{\sqrt {c}\, \left (4+2 k \right )}\right ], \left [\frac {k +3}{k +2}\right ], \frac {2 i \sqrt {a}\, \sqrt {c}\, \ln \left (x \right )^{k +2}}{k +2}\right ) \ln \left (x \right )+\operatorname {hypergeom}\left (\left [\frac {\left (k +1\right ) \sqrt {c}+i \sqrt {a}\, b}{\sqrt {c}\, \left (4+2 k \right )}\right ], \left [\frac {k +1}{k +2}\right ], \frac {2 i \sqrt {a}\, \sqrt {c}\, \ln \left (x \right )^{k +2}}{k +2}\right )\right ) a \left (k +3\right ) \left (k +1\right )} \\ \end{align*}

2.7.3.2 ✓ Maple. Time used: 0.003 (sec). Leaf size: 480

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

\[ y = \frac {-\ln \left (x \right )^{k +1} \left (k +3\right ) \left (i \sqrt {c}\, \left (k +1\right ) \sqrt {a}-a b \right ) \operatorname {hypergeom}\left (\left [\frac {\left (3 k +5\right ) \sqrt {c}+i \sqrt {a}\, b}{\sqrt {c}\, \left (2 k +4\right )}\right ], \left [\frac {3+2 k}{k +2}\right ], \frac {2 i \sqrt {a}\, \sqrt {c}\, \ln \left (x \right )^{k +2}}{k +2}\right )+\left (-c_1 \left (i \sqrt {c}\, \left (k +3\right ) \sqrt {a}-a b \right ) \ln \left (x \right )^{k +2} \operatorname {hypergeom}\left (\left [\frac {\left (3 k +7\right ) \sqrt {c}+i \sqrt {a}\, b}{\sqrt {c}\, \left (2 k +4\right )}\right ], \left [\frac {2 k +5}{k +2}\right ], \frac {2 i \sqrt {a}\, \sqrt {c}\, \ln \left (x \right )^{k +2}}{k +2}\right )+\left (\left (i \sqrt {a}\, \sqrt {c}\, \ln \left (x \right )^{k +2}-1\right ) c_1 \operatorname {hypergeom}\left (\left [\frac {\left (k +3\right ) \sqrt {c}+i \sqrt {a}\, b}{\sqrt {c}\, \left (2 k +4\right )}\right ], \left [\frac {k +3}{k +2}\right ], \frac {2 i \sqrt {a}\, \sqrt {c}\, \ln \left (x \right )^{k +2}}{k +2}\right )+i \ln \left (x \right )^{k +1} \sqrt {a}\, \sqrt {c}\, \operatorname {hypergeom}\left (\left [\frac {\left (k +1\right ) \sqrt {c}+i \sqrt {a}\, b}{\sqrt {c}\, \left (2 k +4\right )}\right ], \left [\frac {k +1}{k +2}\right ], \frac {2 i \sqrt {a}\, \sqrt {c}\, \ln \left (x \right )^{k +2}}{k +2}\right )\right ) \left (k +3\right )\right ) \left (k +1\right )}{\left (\ln \left (x \right ) \operatorname {hypergeom}\left (\left [\frac {\left (k +3\right ) \sqrt {c}+i \sqrt {a}\, b}{\sqrt {c}\, \left (2 k +4\right )}\right ], \left [\frac {k +3}{k +2}\right ], \frac {2 i \sqrt {a}\, \sqrt {c}\, \ln \left (x \right )^{k +2}}{k +2}\right ) c_1 +\operatorname {hypergeom}\left (\left [\frac {\left (k +1\right ) \sqrt {c}+i \sqrt {a}\, b}{\sqrt {c}\, \left (2 k +4\right )}\right ], \left [\frac {k +1}{k +2}\right ], \frac {2 i \sqrt {a}\, \sqrt {c}\, \ln \left (x \right )^{k +2}}{k +2}\right )\right ) a \left (k +3\right ) \left (k +1\right )} \]

Maple trace

Methods for first order ODEs: 
--- Trying classification methods --- 
trying a quadrature 
trying 1st order linear 
trying Bernoulli 
trying separable 
trying inverse linear 
trying homogeneous types: 
trying Chini 
differential order: 1; looking for linear symmetries 
trying exact 
Looking for potential symmetries 
trying Riccati 
trying Riccati sub-methods: 
   trying Riccati_symmetries 
   trying Riccati to 2nd Order 
   -> Calling odsolve with the ODE, diff(diff(y(x),x),x) = -1/x*diff(y(x),x)-a/ 
x^2*(b*ln(x)^k+c*ln(x)^(2*k+2))*y(x), y(x) 
      *** Sublevel 2 *** 
      Methods for second order ODEs: 
      --- Trying classification methods --- 
      trying a symmetry of the form [xi=0, eta=F(x)] 
      checking if the LODE is missing y 
      -> Heun: Equivalence to the GHE or one of its 4 confluent cases under a \ 
power @ Moebius 
      -> trying a solution of the form r0(x) * Y + r1(x) * Y where Y = exp(int\ 
(r(x), dx)) * 2F1([a1, a2], [b1], f) 
      -> Trying changes of variables to rationalize or make the ODE simpler 
         trying a symmetry of the form [xi=0, eta=F(x)] 
         checking if the LODE is missing y 
         -> Trying an equivalence, under non-integer power transformations, 
            to LODEs admitting Liouvillian solutions. 
            -> 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 
               <- hyper3 successful: received ODE is equivalent to the 1F1 ODE 
            <- Whittaker successful 
         <- special function solution successful 
         Change of variables used: 
            [x = exp(t)] 
         Linear ODE actually solved: 
            a*(b*t^k+c*t^(2*k+2))*u(t)+diff(diff(u(t),t),t) = 0 
      <- change of variables successful 
   <- Riccati to 2nd Order successful
 

Maple step by step

\[ \begin {array}{lll} & {} & \textrm {Let's solve}\hspace {3pt} \\ {} & {} & x \left (\frac {d}{d x}y \left (x \right )\right )=a y \left (x \right )^{2}+b \ln \left (x \right )^{k}+c \ln \left (x \right )^{2 k +2} \\ \bullet & {} & \textrm {Highest derivative means the order of the ODE is}\hspace {3pt} 1 \\ {} & {} & \frac {d}{d x}y \left (x \right ) \\ \bullet & {} & \textrm {Solve for the highest derivative}\hspace {3pt} \\ {} & {} & \frac {d}{d x}y \left (x \right )=\frac {a y \left (x \right )^{2}+b \ln \left (x \right )^{k}+c \ln \left (x \right )^{2 k +2}}{x} \end {array} \]

2.7.3.3 ✓ Mathematica. Time used: 1.183 (sec). Leaf size: 806

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

\begin{align*} y(x)&\to -\frac {\log ^{k+1}(x) \left (\sqrt {c} c_1 (k+2) \sqrt {-(k+2)^2} \operatorname {HypergeometricU}\left (\frac {1}{2} \left (\frac {\sqrt {a} b}{\sqrt {c} \sqrt {-(k+2)^2}}+\frac {k+1}{k+2}\right ),\frac {k+1}{k+2},\frac {2 \sqrt {a} \sqrt {c} \log ^{k+2}(x)}{\sqrt {-(k+2)^2}}\right )+c_1 \left (\sqrt {a} b (k+2)+\sqrt {c} \sqrt {-(k+2)^2} (k+1)\right ) \operatorname {HypergeometricU}\left (\frac {1}{2} \left (\frac {\sqrt {a} b}{\sqrt {c} \sqrt {-(k+2)^2}}+\frac {3 k+5}{k+2}\right ),\frac {2 k+3}{k+2},\frac {2 \sqrt {a} \sqrt {c} \log ^{k+2}(x)}{\sqrt {-(k+2)^2}}\right )+\sqrt {c} (k+2) \sqrt {-(k+2)^2} \left (L_{-\frac {\sqrt {a} b}{2 \sqrt {c} \sqrt {-(k+2)^2}}-\frac {k+1}{2 k+4}}^{-\frac {1}{k+2}}\left (\frac {2 \sqrt {a} \sqrt {c} \log ^{k+2}(x)}{\sqrt {-(k+2)^2}}\right )+2 L_{-\frac {\sqrt {a} b}{2 \sqrt {c} \sqrt {-(k+2)^2}}-\frac {3 k+5}{2 k+4}}^{\frac {k+1}{k+2}}\left (\frac {2 \sqrt {a} \sqrt {c} \log ^{k+2}(x)}{\sqrt {-(k+2)^2}}\right )\right )\right )}{\sqrt {a} (k+2)^2 \left (L_{-\frac {\sqrt {a} b}{2 \sqrt {c} \sqrt {-(k+2)^2}}-\frac {k+1}{2 k+4}}^{-\frac {1}{k+2}}\left (\frac {2 \sqrt {a} \sqrt {c} \log ^{k+2}(x)}{\sqrt {-(k+2)^2}}\right )+c_1 \operatorname {HypergeometricU}\left (\frac {1}{2} \left (\frac {\sqrt {a} b}{\sqrt {c} \sqrt {-(k+2)^2}}+\frac {k+1}{k+2}\right ),\frac {k+1}{k+2},\frac {2 \sqrt {a} \sqrt {c} \log ^{k+2}(x)}{\sqrt {-(k+2)^2}}\right )\right )}\\ y(x)&\to \frac {\log ^{k+1}(x) \left (-\frac {\left (\sqrt {a} b (k+2)+\sqrt {c} \sqrt {-(k+2)^2} (k+1)\right ) \operatorname {HypergeometricU}\left (\frac {1}{2} \left (\frac {\sqrt {a} b}{\sqrt {c} \sqrt {-(k+2)^2}}+\frac {k+1}{k+2}+2\right ),\frac {2 k+3}{k+2},\frac {2 \sqrt {a} \sqrt {c} \log ^{k+2}(x)}{\sqrt {-(k+2)^2}}\right )}{\operatorname {HypergeometricU}\left (\frac {1}{2} \left (\frac {\sqrt {a} b}{\sqrt {c} \sqrt {-(k+2)^2}}+\frac {k+1}{k+2}\right ),\frac {k+1}{k+2},\frac {2 \sqrt {a} \sqrt {c} \log ^{k+2}(x)}{\sqrt {-(k+2)^2}}\right )}-\sqrt {c} \sqrt {-(k+2)^2} (k+2)\right )}{\sqrt {a} (k+2)^2} \end{align*}

2.7.3.4 ✗ Sympy

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

NotImplementedError : The given ODE Derivative(y(x), x) - (a*y(x)**2 + b*log(x)**k + c*log(x)**(2*k + 2))/x cannot be solved by the factorable group method