2.7.6 Problem 7
Internal
problem
ID
[13351]
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
:
7
Date
solved
:
Wednesday, December 31, 2025 at 01:47:28 PM
CAS
classification
:
[_Riccati]
2.7.6.1 Solved using first_order_ode_riccati
9.753 (sec)
Entering first order ode riccati solver
\begin{align*}
y^{\prime } x^{2}&=y^{2} x^{2}+\ln \left (x \right )^{2} a +b \ln \left (x \right )+c \\
\end{align*}
In canonical form the ODE is \begin{align*} y' &= F(x,y)\\ &= \frac {y^{2} x^{2}+\ln \left (x \right )^{2} a +b \ln \left (x \right )+c}{x^{2}} \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 \(f_0(x)=\frac {\ln \left (x \right )^{2} a}{x^{2}}+\frac {b \ln \left (x \right )}{x^{2}}+\frac {c}{x^{2}}\), \(f_1(x)=0\) and \(f_2(x)=1\). Let \begin{align*} y &= \frac {-u'}{f_2 u} \\ &= \frac {-u'}{u} \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' &=0\\ f_1 f_2 &=0\\ f_2^2 f_0 &=\frac {\ln \left (x \right )^{2} a}{x^{2}}+\frac {b \ln \left (x \right )}{x^{2}}+\frac {c}{x^{2}} \end{align*}
Substituting the above terms back in equation (2) gives
\[
u^{\prime \prime }\left (x \right )+\left (\frac {\ln \left (x \right )^{2} a}{x^{2}}+\frac {b \ln \left (x \right )}{x^{2}}+\frac {c}{x^{2}}\right ) u \left (x \right ) = 0
\]
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 \,x^{\frac {-i b +\sqrt {a}}{2 \sqrt {a}}} {\mathrm e}^{-\frac {i \ln \left (x \right )^{2} \sqrt {a}}{2}} \operatorname {hypergeom}\left (\left [\frac {4 a^{{3}/{2}}+i \left (4 c -1\right ) a -i b^{2}}{16 a^{{3}/{2}}}\right ], \left [\frac {1}{2}\right ], \frac {i \left (2 a \ln \left (x \right )+b \right )^{2}}{4 a^{{3}/{2}}}\right )+c_2 \,x^{\frac {-i b +\sqrt {a}}{2 \sqrt {a}}} \left (2 a \ln \left (x \right )+b \right ) {\mathrm e}^{-\frac {i \ln \left (x \right )^{2} \sqrt {a}}{2}} \operatorname {hypergeom}\left (\left [\frac {12 a^{{3}/{2}}+i \left (4 c -1\right ) a -i b^{2}}{16 a^{{3}/{2}}}\right ], \left [\frac {3}{2}\right ], \frac {i \left (2 a \ln \left (x \right )+b \right )^{2}}{4 a^{{3}/{2}}}\right )
\end{equation}
Taking derivative gives \begin{equation}
\tag{4} u^{\prime }\left (x \right ) = \frac {c_1 \,x^{\frac {-i b +\sqrt {a}}{2 \sqrt {a}}} \left (-i b +\sqrt {a}\right ) {\mathrm e}^{-\frac {i \ln \left (x \right )^{2} \sqrt {a}}{2}} \operatorname {hypergeom}\left (\left [\frac {4 a^{{3}/{2}}+i \left (4 c -1\right ) a -i b^{2}}{16 a^{{3}/{2}}}\right ], \left [\frac {1}{2}\right ], \frac {i \left (2 a \ln \left (x \right )+b \right )^{2}}{4 a^{{3}/{2}}}\right )}{2 \sqrt {a}\, x}-\frac {i c_1 \,x^{\frac {-i b +\sqrt {a}}{2 \sqrt {a}}} \ln \left (x \right ) \sqrt {a}\, {\mathrm e}^{-\frac {i \ln \left (x \right )^{2} \sqrt {a}}{2}} \operatorname {hypergeom}\left (\left [\frac {4 a^{{3}/{2}}+i \left (4 c -1\right ) a -i b^{2}}{16 a^{{3}/{2}}}\right ], \left [\frac {1}{2}\right ], \frac {i \left (2 a \ln \left (x \right )+b \right )^{2}}{4 a^{{3}/{2}}}\right )}{x}+\frac {i c_1 \,x^{\frac {-i b +\sqrt {a}}{2 \sqrt {a}}} {\mathrm e}^{-\frac {i \ln \left (x \right )^{2} \sqrt {a}}{2}} \left (4 a^{{3}/{2}}+i \left (4 c -1\right ) a -i b^{2}\right ) \operatorname {hypergeom}\left (\left [\frac {4 a^{{3}/{2}}+i \left (4 c -1\right ) a -i b^{2}}{16 a^{{3}/{2}}}+1\right ], \left [\frac {3}{2}\right ], \frac {i \left (2 a \ln \left (x \right )+b \right )^{2}}{4 a^{{3}/{2}}}\right ) \left (2 a \ln \left (x \right )+b \right )}{8 a^{2} x}+\frac {c_2 \,x^{\frac {-i b +\sqrt {a}}{2 \sqrt {a}}} \left (-i b +\sqrt {a}\right ) \left (2 a \ln \left (x \right )+b \right ) {\mathrm e}^{-\frac {i \ln \left (x \right )^{2} \sqrt {a}}{2}} \operatorname {hypergeom}\left (\left [\frac {12 a^{{3}/{2}}+i \left (4 c -1\right ) a -i b^{2}}{16 a^{{3}/{2}}}\right ], \left [\frac {3}{2}\right ], \frac {i \left (2 a \ln \left (x \right )+b \right )^{2}}{4 a^{{3}/{2}}}\right )}{2 \sqrt {a}\, x}+\frac {2 c_2 \,x^{\frac {-i b +\sqrt {a}}{2 \sqrt {a}}} a \,{\mathrm e}^{-\frac {i \ln \left (x \right )^{2} \sqrt {a}}{2}} \operatorname {hypergeom}\left (\left [\frac {12 a^{{3}/{2}}+i \left (4 c -1\right ) a -i b^{2}}{16 a^{{3}/{2}}}\right ], \left [\frac {3}{2}\right ], \frac {i \left (2 a \ln \left (x \right )+b \right )^{2}}{4 a^{{3}/{2}}}\right )}{x}-\frac {i c_2 \,x^{\frac {-i b +\sqrt {a}}{2 \sqrt {a}}} \left (2 a \ln \left (x \right )+b \right ) \ln \left (x \right ) \sqrt {a}\, {\mathrm e}^{-\frac {i \ln \left (x \right )^{2} \sqrt {a}}{2}} \operatorname {hypergeom}\left (\left [\frac {12 a^{{3}/{2}}+i \left (4 c -1\right ) a -i b^{2}}{16 a^{{3}/{2}}}\right ], \left [\frac {3}{2}\right ], \frac {i \left (2 a \ln \left (x \right )+b \right )^{2}}{4 a^{{3}/{2}}}\right )}{x}+\frac {i c_2 \,x^{\frac {-i b +\sqrt {a}}{2 \sqrt {a}}} \left (2 a \ln \left (x \right )+b \right )^{2} {\mathrm e}^{-\frac {i \ln \left (x \right )^{2} \sqrt {a}}{2}} \left (12 a^{{3}/{2}}+i \left (4 c -1\right ) a -i b^{2}\right ) \operatorname {hypergeom}\left (\left [\frac {12 a^{{3}/{2}}+i \left (4 c -1\right ) a -i b^{2}}{16 a^{{3}/{2}}}+1\right ], \left [\frac {5}{2}\right ], \frac {i \left (2 a \ln \left (x \right )+b \right )^{2}}{4 a^{{3}/{2}}}\right )}{24 a^{2} x}
\end{equation}
Substituting equations (3,4) into (1) results in \begin{align*}
y &= \frac {-u'}{f_2 u} \\
y &= \frac {-u'}{u} \\
y &= \text {Expression too large to display} \\
\end{align*}
Doing change of constants, the above solution becomes \[
\text {Expression too large to display}
\]
Simplifying the above gives
\begin{align*}
y &= \frac {\left (-\frac {b \ln \left (x \right ) \left (b \ln \left (x \right )-4 c +1\right ) a^{{5}/{2}}}{12}-\frac {b^{2} \left (b \ln \left (x \right )-c +\frac {1}{4}\right ) a^{{3}/{2}}}{12}+\frac {\ln \left (x \right )^{2} \left (c -\frac {1}{4}\right ) a^{{7}/{2}}}{3}-\frac {\sqrt {a}\, b^{4}}{48}-i a^{2} \left (a \ln \left (x \right )+\frac {b}{2}\right )^{2}\right ) c_3 \operatorname {hypergeom}\left (\left [\frac {28 a^{{3}/{2}}+i \left (4 c -1\right ) a -i b^{2}}{16 a^{{3}/{2}}}\right ], \left [\frac {5}{2}\right ], \frac {i \left (2 a \ln \left (x \right )+b \right )^{2}}{4 a^{{3}/{2}}}\right )+\left (-\frac {a^{{5}/{2}} b}{4}+\left (-\frac {\ln \left (x \right )}{2}-1\right ) a^{{7}/{2}}+i a^{2} \left (a \ln \left (x \right )+\frac {b}{2}\right )^{2}\right ) c_3 \operatorname {hypergeom}\left (\left [\frac {12 a^{{3}/{2}}+i \left (4 c -1\right ) a -i b^{2}}{16 a^{{3}/{2}}}\right ], \left [\frac {3}{2}\right ], \frac {i \left (2 a \ln \left (x \right )+b \right )^{2}}{4 a^{{3}/{2}}}\right )+\frac {\left (\ln \left (x \right ) \left (c -\frac {1}{4}\right ) a^{{5}/{2}}-\frac {b \left (b \ln \left (x \right )-2 c +\frac {1}{2}\right ) a^{{3}/{2}}}{4}-\frac {\sqrt {a}\, b^{3}}{8}-i a^{2} \left (a \ln \left (x \right )+\frac {b}{2}\right )\right ) \operatorname {hypergeom}\left (\left [\frac {20 a^{{3}/{2}}+i \left (4 c -1\right ) a -i b^{2}}{16 a^{{3}/{2}}}\right ], \left [\frac {3}{2}\right ], \frac {i \left (2 a \ln \left (x \right )+b \right )^{2}}{4 a^{{3}/{2}}}\right )}{2}+\frac {\left (i \ln \left (x \right ) a^{3}+\frac {i a^{2} b}{2}-\frac {a^{{5}/{2}}}{2}\right ) \operatorname {hypergeom}\left (\left [\frac {4 a^{{3}/{2}}+i \left (4 c -1\right ) a -i b^{2}}{16 a^{{3}/{2}}}\right ], \left [\frac {1}{2}\right ], \frac {i \left (2 a \ln \left (x \right )+b \right )^{2}}{4 a^{{3}/{2}}}\right )}{2}}{a^{{5}/{2}} x \left (c_3 \left (a \ln \left (x \right )+\frac {b}{2}\right ) \operatorname {hypergeom}\left (\left [\frac {12 a^{{3}/{2}}+i \left (4 c -1\right ) a -i b^{2}}{16 a^{{3}/{2}}}\right ], \left [\frac {3}{2}\right ], \frac {i \left (2 a \ln \left (x \right )+b \right )^{2}}{4 a^{{3}/{2}}}\right )+\frac {\operatorname {hypergeom}\left (\left [\frac {4 a^{{3}/{2}}+i \left (4 c -1\right ) a -i b^{2}}{16 a^{{3}/{2}}}\right ], \left [\frac {1}{2}\right ], \frac {i \left (2 a \ln \left (x \right )+b \right )^{2}}{4 a^{{3}/{2}}}\right )}{2}\right )} \\
\end{align*}
Summary of solutions found
\begin{align*}
y &= \frac {\left (-\frac {b \ln \left (x \right ) \left (b \ln \left (x \right )-4 c +1\right ) a^{{5}/{2}}}{12}-\frac {b^{2} \left (b \ln \left (x \right )-c +\frac {1}{4}\right ) a^{{3}/{2}}}{12}+\frac {\ln \left (x \right )^{2} \left (c -\frac {1}{4}\right ) a^{{7}/{2}}}{3}-\frac {\sqrt {a}\, b^{4}}{48}-i a^{2} \left (a \ln \left (x \right )+\frac {b}{2}\right )^{2}\right ) c_3 \operatorname {hypergeom}\left (\left [\frac {28 a^{{3}/{2}}+i \left (4 c -1\right ) a -i b^{2}}{16 a^{{3}/{2}}}\right ], \left [\frac {5}{2}\right ], \frac {i \left (2 a \ln \left (x \right )+b \right )^{2}}{4 a^{{3}/{2}}}\right )+\left (-\frac {a^{{5}/{2}} b}{4}+\left (-\frac {\ln \left (x \right )}{2}-1\right ) a^{{7}/{2}}+i a^{2} \left (a \ln \left (x \right )+\frac {b}{2}\right )^{2}\right ) c_3 \operatorname {hypergeom}\left (\left [\frac {12 a^{{3}/{2}}+i \left (4 c -1\right ) a -i b^{2}}{16 a^{{3}/{2}}}\right ], \left [\frac {3}{2}\right ], \frac {i \left (2 a \ln \left (x \right )+b \right )^{2}}{4 a^{{3}/{2}}}\right )+\frac {\left (\ln \left (x \right ) \left (c -\frac {1}{4}\right ) a^{{5}/{2}}-\frac {b \left (b \ln \left (x \right )-2 c +\frac {1}{2}\right ) a^{{3}/{2}}}{4}-\frac {\sqrt {a}\, b^{3}}{8}-i a^{2} \left (a \ln \left (x \right )+\frac {b}{2}\right )\right ) \operatorname {hypergeom}\left (\left [\frac {20 a^{{3}/{2}}+i \left (4 c -1\right ) a -i b^{2}}{16 a^{{3}/{2}}}\right ], \left [\frac {3}{2}\right ], \frac {i \left (2 a \ln \left (x \right )+b \right )^{2}}{4 a^{{3}/{2}}}\right )}{2}+\frac {\left (i \ln \left (x \right ) a^{3}+\frac {i a^{2} b}{2}-\frac {a^{{5}/{2}}}{2}\right ) \operatorname {hypergeom}\left (\left [\frac {4 a^{{3}/{2}}+i \left (4 c -1\right ) a -i b^{2}}{16 a^{{3}/{2}}}\right ], \left [\frac {1}{2}\right ], \frac {i \left (2 a \ln \left (x \right )+b \right )^{2}}{4 a^{{3}/{2}}}\right )}{2}}{a^{{5}/{2}} x \left (c_3 \left (a \ln \left (x \right )+\frac {b}{2}\right ) \operatorname {hypergeom}\left (\left [\frac {12 a^{{3}/{2}}+i \left (4 c -1\right ) a -i b^{2}}{16 a^{{3}/{2}}}\right ], \left [\frac {3}{2}\right ], \frac {i \left (2 a \ln \left (x \right )+b \right )^{2}}{4 a^{{3}/{2}}}\right )+\frac {\operatorname {hypergeom}\left (\left [\frac {4 a^{{3}/{2}}+i \left (4 c -1\right ) a -i b^{2}}{16 a^{{3}/{2}}}\right ], \left [\frac {1}{2}\right ], \frac {i \left (2 a \ln \left (x \right )+b \right )^{2}}{4 a^{{3}/{2}}}\right )}{2}\right )} \\
\end{align*}
2.7.6.2 ✓ Maple. Time used: 0.052 (sec). Leaf size: 477
ode:=x^2*diff(y(x),x) = x^2*y(x)^2+a*ln(x)^2+b*ln(x)+c;
dsolve(ode,y(x), singsol=all);
\begin{align*} \text {Solution too large to show}\end{align*}
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 Special
trying Riccati sub-methods:
trying Riccati_symmetries
trying Riccati to 2nd Order
-> Calling odsolve with the ODE, diff(diff(y(x),x),x) = -(a*ln(x)^2+b*ln(x)+
c)/x^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 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
-> hypergeometric
-> heuristic approach
-> hyper3: Equivalence to 2F1, 1F1 or 0F1 under a power @ Moebi\
us
<- hyper3 successful: indirect Equivalence to 0F1 under ``^ @ Mo\
ebius`` is resolved
<- hypergeometric successful
<- special function solution successful
Change of variables used:
[x = exp(t)]
Linear ODE actually solved:
(a*t^2+b*t+c)*u(t)-diff(u(t),t)+diff(diff(u(t),t),t) = 0
<- change of variables successful
<- Riccati to 2nd Order successful
Maple step by step
\[ \begin {array}{lll} & {} & \textrm {Let's solve}\hspace {3pt} \\ {} & {} & x^{2} \left (\frac {d}{d x}y \left (x \right )\right )=x^{2} y \left (x \right )^{2}+a \ln \left (x \right )^{2}+b \ln \left (x \right )+c \\ \bullet & {} & \textrm {Highest derivative means the order of the ODE is}\hspace {3pt} 1 \\ {} & {} & \frac {d}{d x}y \left (x \right ) \\ \bullet & {} & \textrm {Solve for the highest derivative}\hspace {3pt} \\ {} & {} & \frac {d}{d x}y \left (x \right )=\frac {x^{2} y \left (x \right )^{2}+a \ln \left (x \right )^{2}+b \ln \left (x \right )+c}{x^{2}} \end {array} \]
2.7.6.3 ✓ Mathematica. Time used: 0.388 (sec). Leaf size: 868
ode=x^2*D[y[x],x]==x^2*y[x]^2+a*(Log[x])^2+b*Log[x]+c;
ic={};
DSolve[{ode,ic},y[x],x,IncludeSingularSolutions->True]
\begin{align*} \text {Solution too large to show}\end{align*}
2.7.6.4 ✗ Sympy
from sympy import *
x = symbols("x")
a = symbols("a")
b = symbols("b")
c = symbols("c")
y = Function("y")
ode = Eq(-a*log(x)**2 - b*log(x) - c - x**2*y(x)**2 + x**2*Derivative(y(x), x),0)
ics = {}
dsolve(ode,func=y(x),ics=ics)
NotImplementedError : The given ODE Derivative(y(x), x) - (a*log(x)**2 + b*log(x) + c + x**2*y(x)**2)/x**2 cannot be solved by the factorable group method