2.24.12 Problem 17
Internal
problem
ID
[13576]
Book
:
Handbook
of
exact
solutions
for
ordinary
differential
equations.
By
Polyanin
and
Zaitsev.
Second
edition
Section
:
Chapter
1,
section
1.3.
Abel
Equations
of
the
Second
Kind.
subsection
1.3.3-2.
Problem
number
:
17
Date
solved
:
Sunday, January 18, 2026 at 08:44:43 PM
CAS
classification
:
[[_Abel, `2nd type`, `class B`]]
Entering first order ode abel second kind solver
\begin{align*}
y y^{\prime }&=\frac {3 y}{\left (a x +b \right )^{{1}/{3}} x^{{5}/{3}}}+\frac {3}{\left (a x +b \right )^{{2}/{3}} x^{{7}/{3}}} \\
\end{align*}
2.24.12.1 Solved using first_order_ode_abel_second_kind_solved_by_converting_to_first_kind
27.914 (sec)
This is Abel second kind ODE, it has the form
\[ \left (y+g\right )y^{\prime }= f_0(x)+f_1(x) y +f_2(x)y^{2}+f_3(x)y^{3} \]
Comparing the above to given ODE which is
\begin{align*}y y^{\prime } = \frac {3 y}{\left (a x +b \right )^{{1}/{3}} x^{{5}/{3}}}+\frac {3}{\left (a x +b \right )^{{2}/{3}} x^{{7}/{3}}}\tag {1} \end{align*}
Shows that
\begin{align*} g &= 0\\ f_0 &= \frac {3}{\left (a x +b \right )^{{2}/{3}} x^{{7}/{3}}}\\ f_1 &= \frac {3}{\left (a x +b \right )^{{1}/{3}} x^{{5}/{3}}}\\ f_2 &= 0\\ f_3 &= 0 \end{align*}
Applying transformation
\begin{align*} y&=\frac {1}{u(x)}-g \end{align*}
Results in the new ode which is Abel first kind
\begin{align*} u^{\prime }\left (x \right ) = -\frac {3 u \left (x \right )^{3}}{\left (a x +b \right )^{{2}/{3}} x^{{7}/{3}}}-\frac {3 u \left (x \right )^{2}}{\left (a x +b \right )^{{1}/{3}} x^{{5}/{3}}} \end{align*}
Which is now solved. Entering first order ode abel first kind solverThis is Abel first kind ODE, it
has the form
\[ u^{\prime }\left (x \right )= f_0(x)+f_1(x) u \left (x \right ) +f_2(x)u \left (x \right )^{2}+f_3(x)u \left (x \right )^{3} \]
Comparing the above to given ODE which is \begin{align*}u^{\prime }\left (x \right )&=-\frac {3 u \left (x \right )^{3}}{\left (a x +b \right )^{{2}/{3}} x^{{7}/{3}}}-\frac {3 u \left (x \right )^{2}}{\left (a x +b \right )^{{1}/{3}} x^{{5}/{3}}}\tag {1} \end{align*}
Therefore
\begin{align*} f_0 &= 0\\ f_1 &= 0\\ f_2 &= -\frac {3}{\left (a x +b \right )^{{1}/{3}} x^{{5}/{3}}}\\ f_3 &= -\frac {3}{\left (a x +b \right )^{{2}/{3}} x^{{7}/{3}}} \end{align*}
Hence
\begin{align*} f'_{0} &= 0\\ f'_{3} &= \frac {2 a}{\left (a x +b \right )^{{5}/{3}} x^{{7}/{3}}}+\frac {7}{\left (a x +b \right )^{{2}/{3}} x^{{10}/{3}}} \end{align*}
Since \(f_2(x)=-\frac {3}{\left (a x +b \right )^{{1}/{3}} x^{{5}/{3}}}\) is not zero, then the followingtransformation is used to remove \(f_2\). Let \(u \left (x \right ) = u(x) - \frac {f_2}{3 f_3}\) or
\begin{align*} u \left (x \right ) &= u(x) - \left ( \frac {-\frac {3}{\left (a x +b \right )^{{1}/{3}} x^{{5}/{3}}}}{-\frac {9}{\left (a x +b \right )^{{2}/{3}} x^{{7}/{3}}}} \right ) \\ &= u \left (x \right )-\frac {\left (a x +b \right )^{{1}/{3}} x^{{2}/{3}}}{3} \end{align*}
The above transformation applied to (1) gives a new ODE as
\begin{align*} u^{\prime }\left (x \right ) = \frac {x^{{2}/{3}} a}{9 \left (a x +b \right )^{{2}/{3}}}+\frac {u \left (x \right )}{x}-\frac {3 u \left (x \right )^{3}}{\left (a x +b \right )^{{2}/{3}} x^{{7}/{3}}}\tag {2} \end{align*}
The above ODE (2) can now be solved.
Entering first order ode homog type D2 solverApplying change of variables \(u \left (x \right ) = u \left (x \right ) x\), then the ode becomes
\begin{align*} u^{\prime }\left (x \right ) x +u \left (x \right ) = \frac {x^{{2}/{3}} a}{9 \left (a x +b \right )^{{2}/{3}}}+u \left (x \right )-\frac {3 x^{{2}/{3}} u \left (x \right )^{3}}{\left (a x +b \right )^{{2}/{3}}} \end{align*}
Which is now solved The ode
\begin{equation}
u^{\prime }\left (x \right ) = -\frac {27 u \left (x \right )^{3}-a}{9 x^{{1}/{3}} \left (a x +b \right )^{{2}/{3}}}
\end{equation}
is separable as it can be written as \begin{align*} u^{\prime }\left (x \right )&= -\frac {27 u \left (x \right )^{3}-a}{9 x^{{1}/{3}} \left (a x +b \right )^{{2}/{3}}}\\ &= f(x) g(u) \end{align*}
Where
\begin{align*} f(x) &= \frac {1}{\left (a x +b \right )^{{2}/{3}} x^{{1}/{3}}}\\ g(u) &= \frac {a}{9}-3 u^{3} \end{align*}
Integrating gives
\begin{align*}
\int { \frac {1}{g(u)} \,du} &= \int { f(x) \,dx} \\
\int { \frac {1}{\frac {a}{9}-3 u^{3}}\,du} &= \int { \frac {1}{\left (a x +b \right )^{{2}/{3}} x^{{1}/{3}}} \,dx} \\
\end{align*}
\[
\frac {2 \sqrt {3}\, \arctan \left (\frac {\sqrt {3}\, \left (6 u \left (x \right )+a^{{1}/{3}}\right )}{3 a^{{1}/{3}}}\right )-2 \ln \left (3 u \left (x \right )-a^{{1}/{3}}\right )+\ln \left (a^{{2}/{3}}+3 a^{{1}/{3}} u \left (x \right )+9 u \left (x \right )^{2}\right )}{2 a^{{2}/{3}}}=\int \frac {1}{\left (a x +b \right )^{{2}/{3}} x^{{1}/{3}}}d x +c_1
\]
We now need to find the singular solutions, these are found by finding
for what values \(g(u)\) is zero, since we had to divide by this above. Solving \(g(u)=0\) or \[
\frac {a}{9}-3 u^{3}=0
\]
for \(u \left (x \right )\) gives
\begin{align*} u \left (x \right )&=\frac {a^{{1}/{3}}}{3}\\ u \left (x \right )&=-\frac {a^{{1}/{3}}}{6}-\frac {i \sqrt {3}\, a^{{1}/{3}}}{6} \end{align*}
Now we go over each such singular solution and check if it verifies the ode itself and any initial
conditions given. If it does not then the singular solution will not be used.
Therefore the solutions found are
\begin{align*}
\frac {2 \sqrt {3}\, \arctan \left (\frac {\sqrt {3}\, \left (6 u \left (x \right )+a^{{1}/{3}}\right )}{3 a^{{1}/{3}}}\right )-2 \ln \left (3 u \left (x \right )-a^{{1}/{3}}\right )+\ln \left (a^{{2}/{3}}+3 a^{{1}/{3}} u \left (x \right )+9 u \left (x \right )^{2}\right )}{2 a^{{2}/{3}}} &= \int \frac {1}{\left (a x +b \right )^{{2}/{3}} x^{{1}/{3}}}d x +c_1 \\
u \left (x \right ) &= \frac {a^{{1}/{3}}}{3} \\
u \left (x \right ) &= -\frac {a^{{1}/{3}}}{6}-\frac {i \sqrt {3}\, a^{{1}/{3}}}{6} \\
\end{align*}
Converting \(\frac {2 \sqrt {3}\, \arctan \left (\frac {\sqrt {3}\, \left (6 u \left (x \right )+a^{{1}/{3}}\right )}{3 a^{{1}/{3}}}\right )-2 \ln \left (3 u \left (x \right )-a^{{1}/{3}}\right )+\ln \left (a^{{2}/{3}}+3 a^{{1}/{3}} u \left (x \right )+9 u \left (x \right )^{2}\right )}{2 a^{{2}/{3}}} = \int \frac {1}{\left (a x +b \right )^{{2}/{3}} x^{{1}/{3}}}d x +c_1\) back to \(u \left (x \right )\) gives \begin{align*} \frac {2 \sqrt {3}\, \arctan \left (\frac {\sqrt {3}\, \left (\frac {6 u \left (x \right )}{x}+a^{{1}/{3}}\right )}{3 a^{{1}/{3}}}\right )-2 \ln \left (\frac {3 u \left (x \right )}{x}-a^{{1}/{3}}\right )+\ln \left (a^{{2}/{3}}+\frac {3 a^{{1}/{3}} u \left (x \right )}{x}+\frac {9 u \left (x \right )^{2}}{x^{2}}\right )}{2 a^{{2}/{3}}} = \int \frac {1}{\left (a x +b \right )^{{2}/{3}} x^{{1}/{3}}}d x +c_1 \end{align*}
Converting \(u \left (x \right ) = \frac {a^{{1}/{3}}}{3}\) back to \(u \left (x \right )\) gives
\begin{align*} u \left (x \right ) = \frac {x \,a^{{1}/{3}}}{3} \end{align*}
Converting \(u \left (x \right ) = -\frac {a^{{1}/{3}}}{6}-\frac {i \sqrt {3}\, a^{{1}/{3}}}{6}\) back to \(u \left (x \right )\) gives
\begin{align*} u \left (x \right ) = x \left (-\frac {a^{{1}/{3}}}{6}-\frac {i \sqrt {3}\, a^{{1}/{3}}}{6}\right ) \end{align*}
Substituting \(u=u \left (x \right )+\frac {\left (a x +b \right )^{{1}/{3}} x^{{2}/{3}}}{3}\) in the above solution gives
\begin{align*} \frac {2 \sqrt {3}\, \arctan \left (\frac {\sqrt {3}\, \left (\frac {6 u \left (x \right )+2 \left (a x +b \right )^{{1}/{3}} x^{{2}/{3}}}{x}+a^{{1}/{3}}\right )}{3 a^{{1}/{3}}}\right )-2 \ln \left (\frac {3 u \left (x \right )+\left (a x +b \right )^{{1}/{3}} x^{{2}/{3}}}{x}-a^{{1}/{3}}\right )+\ln \left (a^{{2}/{3}}+\frac {3 a^{{1}/{3}} \left (u \left (x \right )+\frac {\left (a x +b \right )^{{1}/{3}} x^{{2}/{3}}}{3}\right )}{x}+\frac {9 \left (u \left (x \right )+\frac {\left (a x +b \right )^{{1}/{3}} x^{{2}/{3}}}{3}\right )^{2}}{x^{2}}\right )}{2 a^{{2}/{3}}} = \int \frac {1}{\left (a x +b \right )^{{2}/{3}} x^{{1}/{3}}}d x +c_1 \end{align*}
Now we transform the solution \(u \left (x \right ) = x \left (-\frac {a^{{1}/{3}}}{6}-\frac {i \sqrt {3}\, a^{{1}/{3}}}{6}\right )\) to \(u \left (x \right )\) using \(u \left (x \right ) = u(x) - \frac {f_2}{3 f_3}\), which gives
\begin{align*} u \left (x \right ) &= x \left (-\frac {a^{{1}/{3}}}{6}-\frac {i \sqrt {3}\, a^{{1}/{3}}}{6}\right ) - \left (\frac {\left (a x +b \right )^{{1}/{3}} x^{{2}/{3}}}{3}\right ) \\ &= x \left (-\frac {a^{{1}/{3}}}{6}-\frac {i \sqrt {3}\, a^{{1}/{3}}}{6}\right )-\frac {\left (a x +b \right )^{{1}/{3}} x^{{2}/{3}}}{3}\\ &= -\frac {\left (a x +b \right )^{{1}/{3}} x^{{2}/{3}}}{3}-\frac {a^{{1}/{3}} x \left (1+i \sqrt {3}\right )}{6} \end{align*}
Now we transform the solution \(u \left (x \right ) = \frac {x \,a^{{1}/{3}}}{3}\) to \(u \left (x \right )\) using \(u \left (x \right ) = u(x) - \frac {f_2}{3 f_3}\), which gives
\begin{align*} u \left (x \right ) &= \frac {x \,a^{{1}/{3}}}{3} - \left (\frac {\left (a x +b \right )^{{1}/{3}} x^{{2}/{3}}}{3}\right ) \\ &= \frac {x \,a^{{1}/{3}}}{3}-\frac {\left (a x +b \right )^{{1}/{3}} x^{{2}/{3}}}{3}\\ &= \frac {x \,a^{{1}/{3}}}{3}-\frac {\left (a x +b \right )^{{1}/{3}} x^{{2}/{3}}}{3} \end{align*}
Substituting \(u \left (x \right )=\frac {1}{y}\) in the above solution gives
\[
\frac {2 \sqrt {3}\, \arctan \left (\frac {\sqrt {3}\, \left (\frac {\frac {6}{y}+2 \left (a x +b \right )^{{1}/{3}} x^{{2}/{3}}}{x}+a^{{1}/{3}}\right )}{3 a^{{1}/{3}}}\right )-2 \ln \left (\frac {\frac {3}{y}+\left (a x +b \right )^{{1}/{3}} x^{{2}/{3}}}{x}-a^{{1}/{3}}\right )+\ln \left (a^{{2}/{3}}+\frac {3 a^{{1}/{3}} \left (\frac {1}{y}+\frac {\left (a x +b \right )^{{1}/{3}} x^{{2}/{3}}}{3}\right )}{x}+\frac {9 \left (\frac {1}{y}+\frac {\left (a x +b \right )^{{1}/{3}} x^{{2}/{3}}}{3}\right )^{2}}{x^{2}}\right )}{2 a^{{2}/{3}}} = \int \frac {1}{\left (a x +b \right )^{{2}/{3}} x^{{1}/{3}}}d x +c_1
\]
Now we transform the solution \(u \left (x \right ) = \frac {x \,a^{{1}/{3}}}{3}-\frac {\left (a x +b \right )^{{1}/{3}} x^{{2}/{3}}}{3}\) to \(y\) using \(u \left (x \right )=\frac {1}{y}\) which gives \[
y = -\frac {3}{\left (a x +b \right )^{{1}/{3}} x^{{2}/{3}}-x \,a^{{1}/{3}}}
\]
Now we transform the solution \(u \left (x \right ) = -\frac {\left (a x +b \right )^{{1}/{3}} x^{{2}/{3}}}{3}-\frac {a^{{1}/{3}} x \left (1+i \sqrt {3}\right )}{6}\) to \(y\) using \(u \left (x \right )=\frac {1}{y}\) which gives \[
y = -\frac {6}{2 \left (a x +b \right )^{{1}/{3}} x^{{2}/{3}}+a^{{1}/{3}} x \left (1+i \sqrt {3}\right )}
\]
Summary of solutions found
\begin{align*}
\frac {2 \sqrt {3}\, \arctan \left (\frac {\sqrt {3}\, \left (\frac {\frac {6}{y}+2 \left (a x +b \right )^{{1}/{3}} x^{{2}/{3}}}{x}+a^{{1}/{3}}\right )}{3 a^{{1}/{3}}}\right )-2 \ln \left (\frac {\frac {3}{y}+\left (a x +b \right )^{{1}/{3}} x^{{2}/{3}}}{x}-a^{{1}/{3}}\right )+\ln \left (a^{{2}/{3}}+\frac {3 a^{{1}/{3}} \left (\frac {1}{y}+\frac {\left (a x +b \right )^{{1}/{3}} x^{{2}/{3}}}{3}\right )}{x}+\frac {9 \left (\frac {1}{y}+\frac {\left (a x +b \right )^{{1}/{3}} x^{{2}/{3}}}{3}\right )^{2}}{x^{2}}\right )}{2 a^{{2}/{3}}} &= \int \frac {1}{\left (a x +b \right )^{{2}/{3}} x^{{1}/{3}}}d x +c_1 \\
y &= -\frac {3}{\left (a x +b \right )^{{1}/{3}} x^{{2}/{3}}-x \,a^{{1}/{3}}} \\
y &= -\frac {6}{2 \left (a x +b \right )^{{1}/{3}} x^{{2}/{3}}+a^{{1}/{3}} x \left (1+i \sqrt {3}\right )} \\
\end{align*}
2.24.12.2 ✓ Maple. Time used: 0.004 (sec). Leaf size: 144
ode:=y(x)*diff(y(x),x) = 3/(a*x+b)^(1/3)/x^(5/3)*y(x)+3/(a*x+b)^(2/3)/x^(7/3);
dsolve(ode,y(x), singsol=all);
\[
y = \frac {6 \sqrt {3}\, \left (a x +b \right )^{{5}/{3}}}{x^{{1}/{3}} \left (3 \left (\tan \left (\operatorname {RootOf}\left (\sqrt {3}\, \ln \left (\frac {\tan \left (\textit {\_Z} \right )^{2}+1}{\left (\sqrt {3}-\tan \left (\textit {\_Z} \right )\right )^{2}}\right )+6 \sqrt {3}\, c_1 -2 \sqrt {3}\, \int \frac {\left (x^{2} a \left (a x +b \right )^{4}\right )^{{2}/{3}}}{\left (a x +b \right )^{{10}/{3}} x^{{5}/{3}}}d x +6 \textit {\_Z} \right )\right )-\frac {\sqrt {3}}{3}\right ) \left (a x +b \right )^{{1}/{3}} \left (x^{2} a \left (a x +b \right )^{4}\right )^{{1}/{3}}-2 \sqrt {3}\, \left (x^{{7}/{3}} a^{2}+2 x^{{4}/{3}} a b +b^{2} x^{{1}/{3}}\right )\right )}
\]
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
trying Abel
<- Abel successful
Maple step by step
\[ \begin {array}{lll} & {} & \textrm {Let's solve}\hspace {3pt} \\ {} & {} & y \left (x \right ) \left (\frac {d}{d x}y \left (x \right )\right )=\frac {3 y \left (x \right )}{\left (a x +b \right )^{{1}/{3}} x^{{5}/{3}}}+\frac {3}{\left (a x +b \right )^{{2}/{3}} x^{{7}/{3}}} \\ \bullet & {} & \textrm {Highest derivative means the order of the ODE is}\hspace {3pt} 1 \\ {} & {} & \frac {d}{d x}y \left (x \right ) \\ \bullet & {} & \textrm {Solve for the highest derivative}\hspace {3pt} \\ {} & {} & \frac {d}{d x}y \left (x \right )=\frac {\frac {3 y \left (x \right )}{\left (a x +b \right )^{{1}/{3}} x^{{5}/{3}}}+\frac {3}{\left (a x +b \right )^{{2}/{3}} x^{{7}/{3}}}}{y \left (x \right )} \end {array} \]
2.24.12.3 ✓ Mathematica. Time used: 1.221 (sec). Leaf size: 304
ode=y[x]*D[y[x],x]==3*(a*x+b)^(-1/3)*x^(-5/3)*y[x]+3*(a*x+b)^(-2/3)*x^(-7/3);
ic={};
DSolve[{ode,ic},y[x],x,IncludeSingularSolutions->True]
\[
\text {Solve}\left [\frac {1}{6} \left (\frac {\sqrt [3]{a} x \left (\log \left (a^{2/3} x^{2/3}+\sqrt [3]{a} \sqrt [3]{x} \sqrt [3]{a x+b}+(a x+b)^{2/3}\right )+2 \sqrt {3} \arctan \left (\frac {\sqrt {3} \sqrt [3]{a x+b}}{\sqrt [3]{a x+b}+2 \sqrt [3]{a} \sqrt [3]{x}}\right )-2 \log \left (\sqrt [3]{a x+b}-\sqrt [3]{a} \sqrt [3]{x}\right )\right )}{\sqrt [3]{a x^3}}+2 \sqrt {3} \arctan \left (\frac {-\frac {2 \left (x^{2/3} y(x) \sqrt [3]{a x+b}+3\right )}{\sqrt [3]{a x^3} y(x)}-1}{\sqrt {3}}\right )+2 \log \left (\frac {-x^{2/3} y(x) \sqrt [3]{a x+b}-3}{\sqrt [3]{a x^3} y(x)}+1\right )-\log \left (\frac {\left (x^{2/3} y(x) \sqrt [3]{a x+b}+3\right )^2}{\left (a x^3\right )^{2/3} y(x)^2}+\frac {x^{2/3} y(x) \sqrt [3]{a x+b}+3}{\sqrt [3]{a x^3} y(x)}+1\right )\right )=c_1,y(x)\right ]
\]
2.24.12.4 ✗ Sympy
from sympy import *
x = symbols("x")
a = symbols("a")
b = symbols("b")
y = Function("y")
ode = Eq(y(x)*Derivative(y(x), x) - 3*y(x)/(x**(5/3)*(a*x + b)**(1/3)) - 3/(x**(7/3)*(a*x + b)**(2/3)),0)
ics = {}
dsolve(ode,func=y(x),ics=ics)
Timed Out
Python version: 3.12.3 (main, Aug 14 2025, 17:47:21) [GCC 13.3.0]
Sympy version 1.14.0