2.25.14 Problem 28

Internal problem ID [13629]
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 1.3.4-2.
Problem number : 28
Date solved : Wednesday, December 31, 2025 at 10:05:51 PM
CAS classification : [[_1st_order, _with_linear_symmetries], _rational, [_Abel, `2nd type`, `class B`]]

2.25.14.1 Solved using first_order_ode_LIE

14.881 (sec)

Entering first order ode LIE solver

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

Writing the ode as

\begin{align*} y^{\prime }&=\frac {-B a k y +B b x k -A b y +y^{2} A -B b x +B y x}{A a y k -A b k x -A a y +A x y -B a x +B \,x^{2}}\\ y^{\prime }&= \omega \left ( x,y\right ) \end{align*}

The condition of Lie symmetry is the linearized PDE given by

\begin{align*} \eta _{x}+\omega \left ( \eta _{y}-\xi _{x}\right ) -\omega ^{2}\xi _{y}-\omega _{x}\xi -\omega _{y}\eta =0\tag {A} \end{align*}

To determine \(\xi ,\eta \) then (A) is solved using ansatz. Making bivariate polynomials of degree 1 to use as anstaz gives

\begin{align*} \tag{1E} \xi &= x a_{2}+y a_{3}+a_{1} \\ \tag{2E} \eta &= x b_{2}+y b_{3}+b_{1} \\ \end{align*}

Where the unknown coefficients are

\[ \{a_{1}, a_{2}, a_{3}, b_{1}, b_{2}, b_{3}\} \]

Substituting equations (1E,2E) and \(\omega \) into (A) gives

\begin{equation} \tag{5E} b_{2}+\frac {\left (-B a k y +B b x k -A b y +y^{2} A -B b x +B y x \right ) \left (b_{3}-a_{2}\right )}{A a y k -A b k x -A a y +A x y -B a x +B \,x^{2}}-\frac {\left (-B a k y +B b x k -A b y +y^{2} A -B b x +B y x \right )^{2} a_{3}}{\left (A a y k -A b k x -A a y +A x y -B a x +B \,x^{2}\right )^{2}}-\left (\frac {B b k -B b +B y}{A a y k -A b k x -A a y +A x y -B a x +B \,x^{2}}-\frac {\left (-B a k y +B b x k -A b y +y^{2} A -B b x +B y x \right ) \left (-A k b +y A -B a +2 B x \right )}{\left (A a y k -A b k x -A a y +A x y -B a x +B \,x^{2}\right )^{2}}\right ) \left (x a_{2}+y a_{3}+a_{1}\right )-\left (\frac {-B a k -A b +2 y A +B x}{A a y k -A b k x -A a y +A x y -B a x +B \,x^{2}}-\frac {\left (-B a k y +B b x k -A b y +y^{2} A -B b x +B y x \right ) \left (A a k -A a +A x \right )}{\left (A a y k -A b k x -A a y +A x y -B a x +B \,x^{2}\right )^{2}}\right ) \left (x b_{2}+y b_{3}+b_{1}\right ) = 0 \end{equation}

Putting the above in normal form gives

\[ \text {Expression too large to display} \]

Setting the numerator to zero gives

\begin{equation} \tag{6E} \text {Expression too large to display} \end{equation}

Looking at the above PDE shows the following are all the terms with \(\{x, y\}\) in them.

\[ \{x, y\} \]

The following substitution is now made to be able to collect on all terms with \(\{x, y\}\) in them

\[ \{x = v_{1}, y = v_{2}\} \]

The above PDE (6E) now becomes

\begin{equation} \tag{7E} \text {Expression too large to display} \end{equation}

Collecting the above on the terms \(v_i\) introduced, and these are

\[ \{v_{1}, v_{2}\} \]

Equation (7E) now becomes

\begin{equation} \tag{8E} \left (B^{2} a k b_{2}+B^{2} b k b_{3}-B^{2} a b_{2}-B^{2} b b_{3}-B^{2} b_{1}\right ) v_{1}^{3}+\left (2 A B a k b_{2}+2 A B b k b_{3}-B^{2} a k a_{2}-B^{2} b k a_{3}-2 A B a b_{2}-2 A B b b_{3}+B^{2} a a_{2}+B^{2} b a_{3}-2 A B b_{1}+B^{2} a_{1}\right ) v_{1}^{2} v_{2}+\left (A^{2} b^{2} k^{2} b_{2}+A B \,b^{2} k^{2} a_{2}-A B \,b^{2} k^{2} b_{3}-B^{2} b^{2} k^{2} a_{3}-A^{2} b^{2} k b_{2}-A B \,b^{2} k a_{2}+A B \,b^{2} k b_{3}-B^{2} a^{2} k b_{2}+B^{2} a b k a_{2}-B^{2} a b k b_{3}+2 B^{2} b^{2} k a_{3}+2 A B b k b_{1}+B^{2} a^{2} b_{2}-B^{2} a b a_{2}+B^{2} a b b_{3}+B^{2} a k b_{1}-B^{2} b^{2} a_{3}+B^{2} b k a_{1}+B^{2} a b_{1}-B^{2} b a_{1}\right ) v_{1}^{2}+\left (A^{2} a k b_{2}+A^{2} b k b_{3}-2 A B a k a_{2}-2 A B b k a_{3}-A^{2} a b_{2}-A^{2} b b_{3}+2 A B a a_{2}+2 A B b a_{3}-A^{2} b_{1}+2 A B a_{1}\right ) v_{1} v_{2}^{2}+\left (-2 A^{2} a b \,k^{2} b_{2}-2 A B a b \,k^{2} a_{2}+2 A B a b \,k^{2} b_{3}+2 B^{2} a b \,k^{2} a_{3}+2 A^{2} a b k b_{2}-2 A B \,a^{2} k b_{2}+4 A B a b k a_{2}-4 A B a b k b_{3}+2 A B \,b^{2} k a_{3}-2 B^{2} a b k a_{3}+2 A^{2} b k b_{1}+2 A B \,a^{2} b_{2}-2 A B a b a_{2}+2 A B a b b_{3}-2 A B \,b^{2} a_{3}-2 B^{2} a k a_{1}+2 A B a b_{1}-2 A B b a_{1}\right ) v_{1} v_{2}+\left (-A^{2} b^{2} k b_{1}-2 A B a b k b_{1}-B^{2} a^{2} k b_{1}\right ) v_{1}+\left (-A^{2} a k a_{2}-A^{2} b k a_{3}+A^{2} a a_{2}+A^{2} b a_{3}+A^{2} a_{1}\right ) v_{2}^{3}+\left (A^{2} a^{2} k^{2} b_{2}+A B \,a^{2} k^{2} a_{2}-A B \,a^{2} k^{2} b_{3}-B^{2} a^{2} k^{2} a_{3}-2 A^{2} a^{2} k b_{2}+A^{2} a b k a_{2}-A^{2} a b k b_{3}+A^{2} b^{2} k a_{3}-A B \,a^{2} k a_{2}+A B \,a^{2} k b_{3}+B^{2} a^{2} k a_{3}+A^{2} a^{2} b_{2}-A^{2} a b a_{2}+A^{2} a b b_{3}-A^{2} a k b_{1}-A^{2} b^{2} a_{3}-A^{2} b k a_{1}-2 A B a k a_{1}+A^{2} a b_{1}-A^{2} b a_{1}\right ) v_{2}^{2}+\left (A^{2} b^{2} k a_{1}+2 A B a b k a_{1}+B^{2} a^{2} k a_{1}\right ) v_{2} = 0 \end{equation}

Setting each coefficients in (8E) to zero gives the following equations to solve

\begin{align*} A^{2} b^{2} k a_{1}+2 A B a b k a_{1}+B^{2} a^{2} k a_{1}&=0\\ -A^{2} b^{2} k b_{1}-2 A B a b k b_{1}-B^{2} a^{2} k b_{1}&=0\\ -A^{2} a k a_{2}-A^{2} b k a_{3}+A^{2} a a_{2}+A^{2} b a_{3}+A^{2} a_{1}&=0\\ B^{2} a k b_{2}+B^{2} b k b_{3}-B^{2} a b_{2}-B^{2} b b_{3}-B^{2} b_{1}&=0\\ 2 A B a k b_{2}+2 A B b k b_{3}-B^{2} a k a_{2}-B^{2} b k a_{3}-2 A B a b_{2}-2 A B b b_{3}+B^{2} a a_{2}+B^{2} b a_{3}-2 A B b_{1}+B^{2} a_{1}&=0\\ A^{2} a k b_{2}+A^{2} b k b_{3}-2 A B a k a_{2}-2 A B b k a_{3}-A^{2} a b_{2}-A^{2} b b_{3}+2 A B a a_{2}+2 A B b a_{3}-A^{2} b_{1}+2 A B a_{1}&=0\\ -2 A^{2} a b \,k^{2} b_{2}-2 A B a b \,k^{2} a_{2}+2 A B a b \,k^{2} b_{3}+2 B^{2} a b \,k^{2} a_{3}+2 A^{2} a b k b_{2}-2 A B \,a^{2} k b_{2}+4 A B a b k a_{2}-4 A B a b k b_{3}+2 A B \,b^{2} k a_{3}-2 B^{2} a b k a_{3}+2 A^{2} b k b_{1}+2 A B \,a^{2} b_{2}-2 A B a b a_{2}+2 A B a b b_{3}-2 A B \,b^{2} a_{3}-2 B^{2} a k a_{1}+2 A B a b_{1}-2 A B b a_{1}&=0\\ A^{2} b^{2} k^{2} b_{2}+A B \,b^{2} k^{2} a_{2}-A B \,b^{2} k^{2} b_{3}-B^{2} b^{2} k^{2} a_{3}-A^{2} b^{2} k b_{2}-A B \,b^{2} k a_{2}+A B \,b^{2} k b_{3}-B^{2} a^{2} k b_{2}+B^{2} a b k a_{2}-B^{2} a b k b_{3}+2 B^{2} b^{2} k a_{3}+2 A B b k b_{1}+B^{2} a^{2} b_{2}-B^{2} a b a_{2}+B^{2} a b b_{3}+B^{2} a k b_{1}-B^{2} b^{2} a_{3}+B^{2} b k a_{1}+B^{2} a b_{1}-B^{2} b a_{1}&=0\\ A^{2} a^{2} k^{2} b_{2}+A B \,a^{2} k^{2} a_{2}-A B \,a^{2} k^{2} b_{3}-B^{2} a^{2} k^{2} a_{3}-2 A^{2} a^{2} k b_{2}+A^{2} a b k a_{2}-A^{2} a b k b_{3}+A^{2} b^{2} k a_{3}-A B \,a^{2} k a_{2}+A B \,a^{2} k b_{3}+B^{2} a^{2} k a_{3}+A^{2} a^{2} b_{2}-A^{2} a b a_{2}+A^{2} a b b_{3}-A^{2} a k b_{1}-A^{2} b^{2} a_{3}-A^{2} b k a_{1}-2 A B a k a_{1}+A^{2} a b_{1}-A^{2} b a_{1}&=0 \end{align*}

Solving the above equations for the unknowns gives

\begin{align*} a_{1}&=0\\ a_{2}&=\frac {A b b_{3}}{B a}\\ a_{3}&=-\frac {A b_{3}}{B}\\ b_{1}&=0\\ b_{2}&=-\frac {b b_{3}}{a}\\ b_{3}&=b_{3} \end{align*}

Substituting the above solution in the anstaz (1E,2E) (using \(1\) as arbitrary value for any unknown in the RHS) gives

\begin{align*} \xi &= -\frac {A \left (a y -b x \right )}{B a} \\ \eta &= \frac {a y -b x}{a} \\ \end{align*}

Shifting is now applied to make \(\xi =0\) in order to simplify the rest of the computation

\begin{align*} \eta &= \eta - \omega \left (x,y\right ) \xi \\ &= \frac {a y -b x}{a} - \left (\frac {-B a k y +B b x k -A b y +y^{2} A -B b x +B y x}{A a y k -A b k x -A a y +A x y -B a x +B \,x^{2}}\right ) \left (-\frac {A \left (a y -b x \right )}{B a}\right ) \\ &= -\frac {\left (\left (b -y \right ) A +B \left (a -x \right )\right ) \left (y A +B x \right ) \left (a y -b x \right )}{B a \left (\left (\left (-b k +y \right ) x +a y \left (k -1\right )\right ) A -B x \left (a -x \right )\right )}\\ \xi &= 0 \end{align*}

The next step is to determine the canonical coordinates \(R,S\). The canonical coordinates map \(\left ( x,y\right ) \to \left ( R,S \right )\) where \(\left ( R,S \right )\) are the canonical coordinates which make the original ode become a quadrature and hence solved by integration.

The characteristic pde which is used to find the canonical coordinates is

\begin{align*} \frac {d x}{\xi } &= \frac {d y}{\eta } = dS \tag {1} \end{align*}

The above comes from the requirements that \(\left ( \xi \frac {\partial }{\partial x} + \eta \frac {\partial }{\partial y}\right ) S(x,y) = 1\). Starting with the first pair of ode’s in (1) gives an ode to solve for the independent variable \(R\) in the canonical coordinates, where \(S(R)\). Since \(\xi =0\) then in this special case

\begin{align*} R = x \end{align*}

\(S\) is found from

\begin{align*} S &= \int { \frac {1}{\eta }} dy\\ &= \int { \frac {1}{-\frac {\left (\left (b -y \right ) A +B \left (a -x \right )\right ) \left (y A +B x \right ) \left (a y -b x \right )}{B a \left (\left (\left (-b k +y \right ) x +a y \left (k -1\right )\right ) A -B x \left (a -x \right )\right )}}} dy \end{align*}

Which results in

\begin{align*} S&= -B a \left (\frac {k \ln \left (y A +B x \right )}{A b +B a}-\frac {\left (k -1\right ) \ln \left (-A b +y A -B a +B x \right )}{A b +B a}-\frac {\ln \left (a y -b x \right )}{A b +B a}\right ) \end{align*}

Now that \(R,S\) are found, we need to setup the ode in these coordinates. This is done by evaluating

\begin{align*} \frac {dS}{dR} &= \frac { S_{x} + \omega (x,y) S_{y} }{ R_{x} + \omega (x,y) R_{y} }\tag {2} \end{align*}

Where in the above \(R_{x},R_{y},S_{x},S_{y}\) are all partial derivatives and \(\omega (x,y)\) is the right hand side of the original ode given by

\begin{align*} \omega (x,y) &= \frac {-B a k y +B b x k -A b y +y^{2} A -B b x +B y x}{A a y k -A b k x -A a y +A x y -B a x +B \,x^{2}} \end{align*}

Evaluating all the partial derivatives gives

\begin{align*} R_{x} &= 1\\ R_{y} &= 0\\ S_{x} &= \frac {B a \left (\frac {\left (k -1\right ) B}{\left (-b +y \right ) A -B \left (a -x \right )}-\frac {k B}{y A +B x}-\frac {b}{a y -b x}\right )}{A b +B a}\\ S_{y} &= \frac {B a \left (\frac {\left (k -1\right ) A}{\left (-b +y \right ) A -B \left (a -x \right )}-\frac {k A}{y A +B x}+\frac {a}{a y -b x}\right )}{A b +B a} \end{align*}

Substituting all the above in (2) and simplifying gives the ode in canonical coordinates.

\begin{align*} \frac {dS}{dR} &= 0\tag {2A} \end{align*}

We now need to express the RHS as function of \(R\) only. This is done by solving for \(x,y\) in terms of \(R,S\) from the result obtained earlier and simplifying. This gives

\begin{align*} \frac {dS}{dR} &= 0 \end{align*}

The above is a quadrature ode. This is the whole point of Lie symmetry method. It converts an ode, no matter how complicated it is, to one that can be solved by integration when the ode is in the canonical coordiates \(R,S\).

Since the ode has the form \(\frac {d}{d R}S \left (R \right )=f(R)\), then we only need to integrate \(f(R)\).

\begin{align*} \int {dS} &= \int {0\, dR} + c_2 \\ S \left (R \right ) &= c_2 \end{align*}

To complete the solution, we just need to transform the above back to \(x,y\) coordinates. This results in

\begin{align*} \frac {\left (\left (k -1\right ) \ln \left (\left (y-b \right ) A -B \left (a -x \right )\right )-k \ln \left (y A +B x \right )+\ln \left (a y-b x \right )\right ) a B}{A b +B a} = c_2 \end{align*}

Summary of solutions found

\begin{align*} \frac {\left (\left (k -1\right ) \ln \left (\left (y-b \right ) A -B \left (a -x \right )\right )-k \ln \left (y A +B x \right )+\ln \left (a y-b x \right )\right ) a B}{A b +B a} &= c_2 \\ \end{align*}

2.25.14.2 ✓ Maple. Time used: 0.058 (sec). Leaf size: 50

ode:=(A*x*y(x)+B*x^2+(-1+k)*A*a*y(x)-(A*b*k+B*a)*x)*diff(y(x),x) = A*y(x)^2+B*x*y(x)-(B*a*k+A*b)*y(x)+(-1+k)*B*b*x; 
dsolve(ode,y(x), singsol=all);
 

\[ \left (y A +B x \right )^{-k} \left (a y-b x \right ) \left (y A -A b -B \left (a -x \right )\right )^{k -1}-c_1 = 0 \]

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 
differential order: 1; found: 1 linear symmetries. Trying reduction of order 
1st order, trying the canonical coordinates of the invariance group 
   -> Calling odsolve with the ODE, diff(y(x),x) = (-a*y(x)+b*x)/A/(a*y(x)-b*x) 
*B, y(x) 
      *** Sublevel 2 *** 
      Methods for first order ODEs: 
      --- Trying classification methods --- 
      trying a quadrature 
      trying 1st order linear 
      <- 1st order linear successful 
<- 1st order, canonical coordinates successful
 

Maple step by step

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

2.25.14.3 ✓ Mathematica. Time used: 0.939 (sec). Leaf size: 62

ode=(A*x*y[x]+B*x^2+(k-1)*A*a*y[x]-(A*b*k+B*a)*x)*D[y[x],x]==A*y[x]^2+B*x*y[x]-(A*b+B*a*k)*y[x]+(k-1)*B*b*x; 
ic={}; 
DSolve[{ode,ic},y[x],x,IncludeSingularSolutions->True]
 

\[ \text {Solve}\left [\frac {A b ((k-1) \log (-a B-A b+A y(x)+B x)+\log (b x-a y(x))-k \log (A y(x)+B x))}{a B+A b}=c_1,y(x)\right ] \]

2.25.14.4 ✗ Sympy

from sympy import * 
x = symbols("x") 
A = symbols("A") 
B = symbols("B") 
k = symbols("k") 
a = symbols("a") 
b = symbols("b") 
y = Function("y") 
ode = Eq(-A*y(x)**2 - B*b*x*(k - 1) - B*x*y(x) + (A*b + B*a*k)*y(x) + (A*a*(k - 1)*y(x) + A*x*y(x) + B*x**2 - x*(A*b*k + B*a))*Derivative(y(x), x),0) 
ics = {} 
dsolve(ode,func=y(x),ics=ics)
 

Timed Out