Problem 63

2.2.60 Problem 63

2.2.60.1 Solved using ﬁrst_order_ode_LIE
2.2.60.2 Solved using ﬁrst_order_ode_riccati
2.2.60.3 Solved using ﬁrst_order_ode_riccati_by_guessing_particular_solution
2.2.60.4 ✓Maple
2.2.60.5 ✓Mathematica
2.2.60.6 ✓Sympy

Internal problem ID [13266]
Book : Handbook of exact solutions for ordinary diﬀerential equations. By Polyanin and Zaitsev. Second edition
Section : Chapter 1, section 1.2. Riccati Equation. 1.2.2. Equations Containing Power Functions
Problem number : 63
Date solved : Wednesday, December 31, 2025 at 12:43:05 PM
CAS classiﬁcation : [_rational, [_1st_order, `_with_symmetry_[F(x),G(x)]`], _Riccati]

2.2.60.1 Solved using ﬁrst_order_ode_LIE

9.057 (sec)

Entering ﬁrst order ode LIE solver

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

Writing the ode as

\begin{align*} y^{\prime }&=-\frac {a b k -a k x -a k y -b k x -b k y +k \,x^{2}+2 k x y +k \,y^{2}+y^{2}}{\left (b -x \right ) \left (a -x \right )}\\ 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 2 to use as anstaz gives

\begin{align*} \tag{1E} \xi &= x^{2} a_{4}+x y a_{5}+y^{2} a_{6}+x a_{2}+y a_{3}+a_{1} \\ \tag{2E} \eta &= x^{2} b_{4}+x y b_{5}+y^{2} b_{6}+x b_{2}+y b_{3}+b_{1} \\ \end{align*}

Where the unknown coeﬃcients are

\[ \{a_{1}, a_{2}, a_{3}, a_{4}, a_{5}, a_{6}, b_{1}, b_{2}, b_{3}, b_{4}, b_{5}, b_{6}\} \]

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

\begin{equation} \tag{5E} 2 x b_{4}+y b_{5}+b_{2}-\frac {\left (a b k -a k x -a k y -b k x -b k y +k \,x^{2}+2 k x y +k \,y^{2}+y^{2}\right ) \left (-2 x a_{4}+x b_{5}-y a_{5}+2 y b_{6}-a_{2}+b_{3}\right )}{\left (b -x \right ) \left (a -x \right )}-\frac {\left (a b k -a k x -a k y -b k x -b k y +k \,x^{2}+2 k x y +k \,y^{2}+y^{2}\right )^{2} \left (x a_{5}+2 y a_{6}+a_{3}\right )}{\left (b -x \right )^{2} \left (a -x \right )^{2}}-\left (-\frac {-a k -b k +2 k x +2 k y}{\left (b -x \right ) \left (a -x \right )}-\frac {a b k -a k x -a k y -b k x -b k y +k \,x^{2}+2 k x y +k \,y^{2}+y^{2}}{\left (b -x \right )^{2} \left (a -x \right )}-\frac {a b k -a k x -a k y -b k x -b k y +k \,x^{2}+2 k x y +k \,y^{2}+y^{2}}{\left (b -x \right ) \left (a -x \right )^{2}}\right ) \left (x^{2} a_{4}+x y a_{5}+y^{2} a_{6}+x a_{2}+y a_{3}+a_{1}\right )+\frac {\left (-a k -b k +2 k x +2 k y +2 y \right ) \left (x^{2} b_{4}+x y b_{5}+y^{2} b_{6}+x b_{2}+y b_{3}+b_{1}\right )}{\left (b -x \right ) \left (a -x \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} \text {Expression too large to display} \end{equation}

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

\begin{align*} -2 k^{2} a_{6}-4 k a_{6}-2 a_{6}&=0\\ -4 k^{2} a_{5}-12 k^{2} a_{6}-5 k a_{5}-6 k a_{6}-a_{5}&=0\\ -k^{2} a_{5}+2 k a_{4}+2 k b_{4}-k b_{5}+2 b_{4}&=0\\ -k^{2} a_{5}-8 k^{2} a_{6}-2 k a_{5}-10 k a_{6}-a_{5}-2 a_{6}&=0\\ -6 k^{2} a_{5}-8 k^{2} a_{6}-2 k a_{5}+k b_{5}-2 k b_{6}+b_{5}&=0\\ -a^{2} b^{2} k^{2} a_{3}+a^{2} b^{2} k a_{2}-a^{2} b^{2} k b_{3}+a^{2} b^{2} b_{2}-a^{2} b k b_{1}-a \,b^{2} k b_{1}&=0\\ -4 k^{2} a_{5}-2 k^{2} a_{6}+2 k a_{4}+k a_{5}+2 k b_{4}-2 k b_{6}+2 b_{4}+b_{5}&=0\\ 4 a \,k^{2} a_{6}+4 b \,k^{2} a_{6}+5 a k a_{6}+5 b k a_{6}-k^{2} a_{3}+a a_{6}+b a_{6}-2 k a_{3}-a_{3}&=0\\ 2 a \,k^{2} a_{5}+12 a \,k^{2} a_{6}+2 b \,k^{2} a_{5}+12 b \,k^{2} a_{6}+2 a k a_{5}+6 a k a_{6}+2 b k a_{5}+6 b k a_{6}-4 k^{2} a_{3}-6 k a_{3}-2 a_{3}&=0\\ -2 a^{2} b^{2} k^{2} a_{6}+a^{2} b^{2} k a_{5}-2 a^{2} b^{2} k b_{6}+2 a^{2} b \,k^{2} a_{3}+2 a \,b^{2} k^{2} a_{3}+a^{2} b^{2} b_{5}-a^{2} b k a_{2}-a \,b^{2} k a_{2}-a^{2} k a_{1}+2 a b k b_{1}-b^{2} k a_{1}+2 a b b_{1}&=0\\ -2 a^{2} k^{2} a_{6}-8 a b \,k^{2} a_{6}-2 b^{2} k^{2} a_{6}-a^{2} k a_{6}+a b k a_{5}-4 a b k a_{6}+2 a \,k^{2} a_{3}-b^{2} k a_{6}+2 b \,k^{2} a_{3}+a b a_{5}+3 a k a_{3}+3 b k a_{3}+a a_{3}+b a_{3}&=0\\ 2 a \,k^{2} a_{5}+2 b \,k^{2} a_{5}-4 a k a_{4}-3 a k b_{4}+2 a k b_{5}-4 b k a_{4}-3 b k b_{4}+2 b k b_{5}-k^{2} a_{3}-4 a b_{4}-4 b b_{4}+k a_{2}+2 k b_{2}-k b_{3}+b_{2}&=0\\ -a^{2} b^{2} k^{2} a_{5}+2 a^{2} b^{2} k a_{4}-a^{2} b^{2} k b_{5}+2 a^{2} b \,k^{2} a_{3}+2 a \,b^{2} k^{2} a_{3}+2 a^{2} b^{2} b_{4}-2 a^{2} b k a_{2}-a^{2} b k b_{2}+2 a^{2} b k b_{3}-2 a \,b^{2} k a_{2}-a \,b^{2} k b_{2}+2 a \,b^{2} k b_{3}-2 a^{2} b b_{2}+a^{2} k b_{1}-2 a \,b^{2} b_{2}+4 a b k b_{1}+b^{2} k b_{1}&=0\\ 6 a \,k^{2} a_{5}+4 a \,k^{2} a_{6}+6 b \,k^{2} a_{5}+4 b \,k^{2} a_{6}-4 a k a_{4}-2 a k a_{5}-2 a k b_{4}+4 a k b_{6}-4 b k a_{4}-2 b k a_{5}-2 b k b_{4}+4 b k b_{6}-4 k^{2} a_{3}-2 a b_{4}-2 a b_{5}-2 b b_{4}-2 b b_{5}+2 k b_{2}+2 b_{2}&=0\\ 4 a^{2} b \,k^{2} a_{6}+4 a \,b^{2} k^{2} a_{6}-a^{2} b k a_{5}+a^{2} b k b_{6}-a^{2} k^{2} a_{3}-a \,b^{2} k a_{5}+a \,b^{2} k b_{6}-4 a b \,k^{2} a_{3}-b^{2} k^{2} a_{3}-a^{2} k a_{3}+a b k a_{2}-2 a b k a_{3}+a b k b_{3}-b^{2} k a_{3}+a b a_{2}+a b b_{3}+a k a_{1}+b k a_{1}+a a_{1}+b a_{1}&=0\\ 6 a \,k^{2} a_{5}+12 a \,k^{2} a_{6}+6 b \,k^{2} a_{5}+12 b \,k^{2} a_{6}-a k a_{4}+a k a_{5}-a k b_{5}+3 a k b_{6}-b k a_{4}+b k a_{5}-b k b_{5}+3 b k b_{6}-6 k^{2} a_{3}-a a_{4}-a b_{5}-b a_{4}-b b_{5}-k a_{2}-4 k a_{3}+k b_{3}-a_{2}+b_{3}&=0\\ -a^{2} k^{2} a_{5}-4 a^{2} k^{2} a_{6}-4 a b \,k^{2} a_{5}-16 a b \,k^{2} a_{6}-b^{2} k^{2} a_{5}-4 b^{2} k^{2} a_{6}-a^{2} k b_{6}+2 a b k a_{4}+2 a b k a_{5}+a b k b_{5}-4 a b k b_{6}+6 a \,k^{2} a_{3}-b^{2} k b_{6}+6 b \,k^{2} a_{3}+2 a b a_{4}+a b b_{5}+4 a k a_{3}-a k b_{3}+4 b k a_{3}-b k b_{3}-a b_{3}-b b_{3}-2 k a_{1}-2 a_{1}&=0\\ 2 a^{2} b \,k^{2} a_{5}+4 a^{2} b \,k^{2} a_{6}+2 a \,b^{2} k^{2} a_{5}+4 a \,b^{2} k^{2} a_{6}-2 a^{2} b k a_{4}-2 a^{2} b k a_{5}+4 a^{2} b k b_{6}-2 a^{2} k^{2} a_{3}-2 a \,b^{2} k a_{4}-2 a \,b^{2} k a_{5}+4 a \,b^{2} k b_{6}-8 a b \,k^{2} a_{3}-2 b^{2} k^{2} a_{3}-2 a^{2} b b_{5}-2 a \,b^{2} b_{5}+4 a b k a_{2}+2 a b k b_{2}+2 a b b_{2}+2 a k a_{1}-2 a k b_{1}+2 b k a_{1}-2 b k b_{1}-2 a b_{1}-2 b b_{1}&=0\\ -a^{2} k^{2} a_{5}-4 a b \,k^{2} a_{5}-b^{2} k^{2} a_{5}+2 a^{2} k a_{4}+a^{2} k b_{4}-a^{2} k b_{5}+8 a b k a_{4}+4 a b k b_{4}-4 a b k b_{5}+2 a \,k^{2} a_{3}+2 b^{2} k a_{4}+b^{2} k b_{4}-b^{2} k b_{5}+2 b \,k^{2} a_{3}+2 a^{2} b_{4}+8 a b b_{4}-2 a k a_{2}-3 a k b_{2}+2 a k b_{3}+2 b^{2} b_{4}-2 b k a_{2}-3 b k b_{2}+2 b k b_{3}-2 a b_{2}-2 b b_{2}+2 k b_{1}&=0\\ 2 a^{2} b \,k^{2} a_{5}+2 a \,b^{2} k^{2} a_{5}-4 a^{2} b k a_{4}-a^{2} b k b_{4}+2 a^{2} b k b_{5}-a^{2} k^{2} a_{3}-4 a \,b^{2} k a_{4}-a \,b^{2} k b_{4}+2 a \,b^{2} k b_{5}-4 a b \,k^{2} a_{3}-b^{2} k^{2} a_{3}-4 a^{2} b b_{4}+a^{2} k a_{2}+a^{2} k b_{2}-a^{2} k b_{3}-4 a \,b^{2} b_{4}+4 a b k a_{2}+4 a b k b_{2}-4 a b k b_{3}+b^{2} k a_{2}+b^{2} k b_{2}-b^{2} k b_{3}+a^{2} b_{2}+4 a b b_{2}-3 a k b_{1}+b^{2} b_{2}-3 b k b_{1}&=0\\ -2 a^{2} k^{2} a_{5}-2 a^{2} k^{2} a_{6}-8 a b \,k^{2} a_{5}-8 a b \,k^{2} a_{6}-2 b^{2} k^{2} a_{5}-2 b^{2} k^{2} a_{6}+a^{2} k a_{4}+a^{2} k a_{5}-2 a^{2} k b_{6}+8 a b k a_{4}+4 a b k a_{5}+2 a b k b_{4}-8 a b k b_{6}+6 a \,k^{2} a_{3}+b^{2} k a_{4}+b^{2} k a_{5}-2 b^{2} k b_{6}+6 b \,k^{2} a_{3}+a^{2} b_{5}+2 a b b_{4}+4 a b b_{5}-a k a_{2}-2 a k b_{2}+b^{2} b_{5}-b k a_{2}-2 b k b_{2}-2 a b_{2}-2 b b_{2}-2 k a_{1}+2 k b_{1}+2 b_{1}&=0 \end{align*}

Solving the above equations for the unknowns gives

\begin{align*} a_{1}&=a b a_{4}\\ a_{2}&=-\left (a +b \right ) a_{4}\\ a_{3}&=0\\ a_{4}&=a_{4}\\ a_{5}&=0\\ a_{6}&=0\\ b_{1}&=a b b_{4}\\ b_{2}&=-\left (a +b \right ) b_{4}\\ b_{3}&=-\frac {a k a_{4}+a k b_{4}+b k a_{4}+b k b_{4}+a b_{4}+b b_{4}}{k}\\ b_{4}&=b_{4}\\ b_{5}&=\frac {2 k a_{4}+2 k b_{4}+2 b_{4}}{k}\\ b_{6}&=\frac {\left (k +1\right ) \left (k a_{4}+k b_{4}+b_{4}\right )}{k^{2}} \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 &= 0 \\ \eta &= \frac {a b \,k^{2}-a \,k^{2} x -a \,k^{2} y -b \,k^{2} x -b \,k^{2} y +x^{2} k^{2}+2 y x \,k^{2}+y^{2} k^{2}-a k y -b k y +2 k x y +2 k \,y^{2}+y^{2}}{k^{2}} \\ \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 {a b \,k^{2}-a \,k^{2} x -a \,k^{2} y -b \,k^{2} x -b \,k^{2} y +x^{2} k^{2}+2 y x \,k^{2}+y^{2} k^{2}-a k y -b k y +2 k x y +2 k \,y^{2}+y^{2}}{k^{2}}}} dy \end{align*}

Which results in

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

Evaluating all the partial derivatives gives

\begin{align*} R_{x} &= 1\\ R_{y} &= 0\\ S_{x} &= \frac {k^{3}}{\left (k \left (b -x -y \right )-y \right ) \left (k +1\right ) \left (k \left (a -x -y \right )-y \right )}\\ S_{y} &= \frac {k^{2}}{\left (k \left (b -x -y \right )-y \right ) \left (k \left (a -x -y \right )-y \right )} \end{align*}

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

\begin{align*} \frac {dS}{dR} &= -\frac {k^{2}}{\left (k +1\right ) \left (b -x \right ) \left (a -x \right )}\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} &= -\frac {k^{2}}{\left (k +1\right ) \left (-R +b \right ) \left (-R +a \right )} \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 {-\frac {k^{2}}{\left (k +1\right ) \left (R -b \right ) \left (R -a \right )}\, dR}\\ S \left (R \right ) &= -\frac {k^{2} \left (\frac {\ln \left (R -a \right )}{a -b}-\frac {\ln \left (R -b \right )}{a -b}\right )}{k +1} + c_2 \end{align*}

\begin{align*} S \left (R \right )&= \frac {k^{2} \left (-\ln \left (R -a \right )+\ln \left (R -b \right )\right )}{\left (k +1\right ) \left (a -b \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 {k \left (-\ln \left (k \left (y+x -b \right )+y\right )+\ln \left (k \left (y+x -a \right )+y\right )\right )}{\left (a -b \right ) \left (k +1\right )} = \frac {k^{2} \left (-\ln \left (-a +x \right )+\ln \left (x -b \right )\right )}{\left (k +1\right ) \left (a -b \right )}+c_2 \end{align*}

Summary of solutions found

\begin{align*} \frac {k \left (-\ln \left (k \left (y+x -b \right )+y\right )+\ln \left (k \left (y+x -a \right )+y\right )\right )}{\left (a -b \right ) \left (k +1\right )} &= \frac {k^{2} \left (-\ln \left (-a +x \right )+\ln \left (x -b \right )\right )}{\left (k +1\right ) \left (a -b \right )}+c_2 \\ \end{align*}

2.2.60.2 Solved using ﬁrst_order_ode_riccati

5.825 (sec)

Entering ﬁrst order ode riccati solver

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

In canonical form the ODE is

\begin{align*} y' &= F(x,y)\\ &= -\frac {a b k -a k x -a k y-b k x -b k y+k \,x^{2}+2 k x y+k y^{2}+y^{2}}{\left (b -x \right ) \left (a -x \right )} \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 \(f_0(x)=-\frac {a b k}{\left (b -x \right ) \left (a -x \right )}+\frac {a k x}{\left (b -x \right ) \left (a -x \right )}+\frac {b k x}{\left (b -x \right ) \left (a -x \right )}-\frac {k \,x^{2}}{\left (b -x \right ) \left (a -x \right )}\), \(f_1(x)=\frac {a k}{\left (a -x \right ) \left (b -x \right )}+\frac {b k}{\left (a -x \right ) \left (b -x \right )}-\frac {2 x k}{\left (a -x \right ) \left (b -x \right )}\) and \(f_2(x)=-\frac {k}{\left (a -x \right ) \left (b -x \right )}-\frac {1}{\left (b -x \right ) \left (a -x \right )}\). Let

\begin{align*} y &= \frac {-u'}{f_2 u} \\ &= \frac {-u'}{u \left (-\frac {k}{\left (a -x \right ) \left (b -x \right )}-\frac {1}{\left (b -x \right ) \left (a -x \right )}\right )} \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 {k}{\left (a -x \right )^{2} \left (b -x \right )}-\frac {k}{\left (a -x \right ) \left (b -x \right )^{2}}-\frac {1}{\left (b -x \right )^{2} \left (a -x \right )}-\frac {1}{\left (b -x \right ) \left (a -x \right )^{2}}\\ f_1 f_2 &=\left (\frac {a k}{\left (a -x \right ) \left (b -x \right )}+\frac {b k}{\left (a -x \right ) \left (b -x \right )}-\frac {2 x k}{\left (a -x \right ) \left (b -x \right )}\right ) \left (-\frac {k}{\left (a -x \right ) \left (b -x \right )}-\frac {1}{\left (b -x \right ) \left (a -x \right )}\right )\\ f_2^2 f_0 &=\left (-\frac {k}{\left (a -x \right ) \left (b -x \right )}-\frac {1}{\left (b -x \right ) \left (a -x \right )}\right )^{2} \left (-\frac {a b k}{\left (b -x \right ) \left (a -x \right )}+\frac {a k x}{\left (b -x \right ) \left (a -x \right )}+\frac {b k x}{\left (b -x \right ) \left (a -x \right )}-\frac {k \,x^{2}}{\left (b -x \right ) \left (a -x \right )}\right ) \end{align*}

Substituting the above terms back in equation (2) gives

\[ \left (-\frac {k}{\left (a -x \right ) \left (b -x \right )}-\frac {1}{\left (b -x \right ) \left (a -x \right )}\right ) u^{\prime \prime }\left (x \right )-\left (-\frac {k}{\left (a -x \right )^{2} \left (b -x \right )}-\frac {k}{\left (a -x \right ) \left (b -x \right )^{2}}-\frac {1}{\left (b -x \right )^{2} \left (a -x \right )}-\frac {1}{\left (b -x \right ) \left (a -x \right )^{2}}+\left (\frac {a k}{\left (a -x \right ) \left (b -x \right )}+\frac {b k}{\left (a -x \right ) \left (b -x \right )}-\frac {2 x k}{\left (a -x \right ) \left (b -x \right )}\right ) \left (-\frac {k}{\left (a -x \right ) \left (b -x \right )}-\frac {1}{\left (b -x \right ) \left (a -x \right )}\right )\right ) u^{\prime }\left (x \right )+\left (-\frac {k}{\left (a -x \right ) \left (b -x \right )}-\frac {1}{\left (b -x \right ) \left (a -x \right )}\right )^{2} \left (-\frac {a b k}{\left (b -x \right ) \left (a -x \right )}+\frac {a k x}{\left (b -x \right ) \left (a -x \right )}+\frac {b k x}{\left (b -x \right ) \left (a -x \right )}-\frac {k \,x^{2}}{\left (b -x \right ) \left (a -x \right )}\right ) u \left (x \right ) = 0 \]

Entering kovacic solverWriting the ode as

\begin{align*} \frac {\left (-k -1\right ) \left (\frac {d^{2}u}{d x^{2}}\right )}{\left (b -x \right ) \left (a -x \right )}+\frac {\left (k +1\right )^{2} \left (a +b -2 x \right ) \left (\frac {d u}{d x}\right )}{\left (b -x \right )^{2} \left (a -x \right )^{2}}-\frac {k \left (k +1\right )^{2} u}{\left (b -x \right )^{2} \left (a -x \right )^{2}} &= 0 \tag {1} \\ A \frac {d^{2}u}{d x^{2}} + B \frac {d u}{d x} + C u &= 0 \tag {2} \end{align*}

Comparing (1) and (2) shows that

\begin{align*} A &= \frac {-k -1}{\left (b -x \right ) \left (a -x \right )} \\ B &= \frac {\left (k +1\right )^{2} \left (a +b -2 x \right )}{\left (b -x \right )^{2} \left (a -x \right )^{2}}\tag {3} \\ C &= -\frac {k \left (k +1\right )^{2}}{\left (b -x \right )^{2} \left (a -x \right )^{2}} \end{align*}

Applying the Liouville transformation on the dependent variable gives

\begin{align*} z(x) &= u 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 {\left (a^{2} k -2 a b k +b^{2} k -a^{2}+2 a b -b^{2}\right ) \left (k +1\right )}{4 \left (a b -a x -b x +x^{2}\right )^{2}}\tag {6} \end{align*}

Comparing the above to (5) shows that

\begin{align*} s &= \left (a^{2} k -2 a b k +b^{2} k -a^{2}+2 a b -b^{2}\right ) \left (k +1\right )\\ t &= 4 \left (a b -a x -b x +x^{2}\right )^{2} \end{align*}

Therefore eq. (4) becomes

\begin{align*} z''(x) &= \left ( \frac {\left (a^{2} k -2 a b k +b^{2} k -a^{2}+2 a b -b^{2}\right ) \left (k +1\right )}{4 \left (a b -a x -b x +x^{2}\right )^{2}}\right ) z(x)\tag {7} \end{align*}

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

\begin{align*} u &= 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 2.9: 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 - 0 \\ &= 4 \end{align*}

The poles of \(r\) in eq. (7) and the order of each pole are determined by solving for the roots of \(t=4 \left (a b -a x -b x +x^{2}\right )^{2}\). There is a pole at \(x=b\) of order \(2\). There is a pole at \(x=a\) of order \(2\). Since there is no odd order pole larger than \(2\) and the order at \(\infty \) is \(4\) 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 \(4\) 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 \(n=1\).

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

\[ r = \frac {\left (k^{2}-1\right ) \left (a -b \right )}{\left (4 a -4 b \right ) \left (x -b \right )^{2}}+\frac {-k^{2}+1}{\left (-2 a +2 b \right ) \left (x -b \right )}+\frac {\left (k^{2}-1\right ) \left (a -b \right )}{\left (4 a -4 b \right ) \left (-a +x \right )^{2}}+\frac {k^{2}-1}{\left (-2 a +2 b \right ) \left (-a +x \right )} \]

For the pole at \(x=b\) let \(b\) be the coeﬃcient of \(\frac {1}{ \left (x -b \right )^{2}}\) in the partial fractions decomposition of \(r\) given above. Therefore \(b=\frac {k^{2}}{4}-\frac {1}{4}\). Hence

\begin{alignat*}{2} [\sqrt r]_c &= 0 \\ \alpha _c^{+} &= \frac {1}{2} + \sqrt {1+4 b} &&= \frac {1}{2}+\frac {k}{2}\\ \alpha _c^{-} &= \frac {1}{2} - \sqrt {1+4 b} &&= \frac {1}{2}-\frac {k}{2} \end{alignat*}

For the pole at \(x=a\) let \(b\) be the coeﬃcient of \(\frac {1}{ \left (-a +x \right )^{2}}\) in the partial fractions decomposition of \(r\) given above. Therefore \(b=\frac {k^{2}}{4}-\frac {1}{4}\). Hence

Since the order of \(r\) at \(\infty \) is \(4 > 2\) then

\begin{alignat*}{2} [\sqrt r]_\infty &= 0 \\ \alpha _{\infty }^{+} &= 0 \\ \alpha _{\infty }^{-} &= 1 \end{alignat*}

The following table summarizes the ﬁndings so far for poles and for the order of \(r\) at \(\infty \) where \(r\) is

\[ r=\frac {\left (a^{2} k -2 a b k +b^{2} k -a^{2}+2 a b -b^{2}\right ) \left (k +1\right )}{4 \left (a b -a x -b x +x^{2}\right )^{2}} \]


pole \(c\) location	pole order	\([\sqrt r]_c\)	\(\alpha _c^{+}\)	\(\alpha _c^{-}\)

\(b\)	\(2\)	\(0\)	\(\frac {1}{2}+\frac {k}{2}\)	\(\frac {1}{2}-\frac {k}{2}\)

\(a\)	\(2\)	\(0\)	\(\frac {1}{2}+\frac {k}{2}\)	\(\frac {1}{2}-\frac {k}{2}\)


Order of \(r\) at \(\infty \)	\([\sqrt r]_\infty \)	\(\alpha _\infty ^{+}\)	\(\alpha _\infty ^{-}\)

\(4\)	\(0\)	\(0\)	\(1\)

Now that the all \([\sqrt r]_c\) and its associated \(\alpha _c^{\pm }\) have been determined for all the poles in the set \(\Gamma \) and \([\sqrt r]_\infty \) and its associated \(\alpha _\infty ^{\pm }\) have also been found, the next step is to determine possible non negative integer \(d\) from these using

\begin{align*} d &= \alpha _\infty ^{s(\infty )} - \sum _{c \in \Gamma } \alpha _c^{s(c)} \end{align*}

Where \(s(c)\) is either \(+\) or \(-\) and \(s(\infty )\) is the sign of \(\alpha _\infty ^{\pm }\). This is done by trial over all set of families \(s=(s(c))_{c \in \Gamma \cup {\infty }}\) until such \(d\) is found to work in ﬁnding candidate \(\omega \). Trying \(\alpha _\infty ^{-} = 1\) then

\begin{align*} d &= \alpha _\infty ^{-} - \left ( \alpha _{c_1}^{-}+\alpha _{c_2}^{+} \right ) \\ &= 1 - \left ( 1 \right ) \\ &= 0 \end{align*}

Since \(d\) an integer and \(d \geq 0\) then it can be used to ﬁnd \(\omega \) using

\begin{align*} \omega &= \sum _{c \in \Gamma } \left ( s(c) [\sqrt r]_c + \frac {\alpha _c^{s(c)}}{x-c} \right ) + s(\infty ) [\sqrt r]_\infty \end{align*}

The above gives

\begin{align*} \omega &= \left ( (-)[\sqrt r]_{c_1} + \frac { \alpha _{c_1}^{-} }{x- c_1}\right )+\left ( (+) [\sqrt r]_{c_2} + \frac { \alpha _{c_2}^{+} }{x- c_2}\right ) + (-) [\sqrt r]_\infty \\ &= \frac {\frac {1}{2}-\frac {k}{2}}{x -b}+\frac {\frac {1}{2}+\frac {k}{2}}{-a +x} + (-) \left ( 0 \right ) \\ &= \frac {\frac {1}{2}-\frac {k}{2}}{x -b}+\frac {\frac {1}{2}+\frac {k}{2}}{-a +x}\\ &= \frac {\left (k -1\right ) a -b \left (k +1\right )+2 x}{2 \left (b -x \right ) \left (a -x \right )} \end{align*}

Now that \(\omega \) is determined, the next step is ﬁnd a corresponding minimal polynomial \(p(x)\) of degree \(d=0\) to solve the ode. The polynomial \(p(x)\) needs to satisfy the equation

\begin{align*} p'' + 2 \omega p' + \left ( \omega ' +\omega ^2 -r\right ) p = 0 \tag {1A} \end{align*}

Let

\begin{align*} p(x) &= 1\tag {2A} \end{align*}

Substituting the above in eq. (1A) gives

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

The equation is satisﬁed since both sides are zero. Therefore the ﬁrst solution to the ode \(z'' = r z\) is

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

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

\begin{align*} u_1 &= z_1 e^{ \int -\frac {1}{2} \frac {B}{A} \,dx} \\ &= z_1 e^{ -\int \frac {1}{2} \frac {\frac {\left (k +1\right )^{2} \left (a +b -2 x \right )}{\left (b -x \right )^{2} \left (a -x \right )^{2}}}{\frac {-k -1}{\left (b -x \right ) \left (a -x \right )}} \,dx} \\ &= z_1 e^{\frac {\left (k +1\right )^{2} \ln \left (\left (-a +x \right ) \left (x -b \right )\right )}{-2 k -2}} \\ &= z_1 \left (\left (\left (b -x \right ) \left (a -x \right )\right )^{-\frac {1}{2}-\frac {k}{2}}\right ) \\ \end{align*}

Which simpliﬁes to

\[ u_1 = \left (\left (b -x \right ) \left (a -x \right )\right )^{-\frac {1}{2}-\frac {k}{2}} \left (-a +x \right )^{\frac {1}{2}+\frac {k}{2}} \left (x -b \right )^{\frac {1}{2}-\frac {k}{2}} \]

The second solution \(u_2\) to the original ode is found using reduction of order

\[ u_2 = u_1 \int \frac { e^{\int -\frac {B}{A} \,dx}}{u_1^2} \,dx \]

Substituting gives

\begin{align*} u_2 &= u_1 \int \frac { e^{\int -\frac {\frac {\left (k +1\right )^{2} \left (a +b -2 x \right )}{\left (b -x \right )^{2} \left (a -x \right )^{2}}}{\frac {-k -1}{\left (b -x \right ) \left (a -x \right )}} \,dx}}{\left (u_1\right )^2} \,dx \\ &= u_1 \int \frac { e^{-\left (k +1\right ) \ln \left (\left (b -x \right ) \left (a -x \right )\right )}}{\left (u_1\right )^2} \,dx \\ &= u_1 \left (-\frac {\left (b -x \right ) \left (a -x \right ) {\mathrm e}^{-\left (k +1\right ) \ln \left (\left (b -x \right ) \left (a -x \right )\right )} \left (\left (b -x \right ) \left (a -x \right )\right )^{k +1} \left (-a +x \right )^{-k -1} \left (x -b \right )^{k -1}}{\left (a -b \right ) k}\right ) \\ \end{align*}

Therefore the solution is

\begin{align*} u &= c_1 u_1 + c_2 u_2 \\ &= c_1 \left (\left (\left (b -x \right ) \left (a -x \right )\right )^{-\frac {1}{2}-\frac {k}{2}} \left (-a +x \right )^{\frac {1}{2}+\frac {k}{2}} \left (x -b \right )^{\frac {1}{2}-\frac {k}{2}}\right ) + c_2 \left (\left (\left (b -x \right ) \left (a -x \right )\right )^{-\frac {1}{2}-\frac {k}{2}} \left (-a +x \right )^{\frac {1}{2}+\frac {k}{2}} \left (x -b \right )^{\frac {1}{2}-\frac {k}{2}}\left (-\frac {\left (b -x \right ) \left (a -x \right ) {\mathrm e}^{-\left (k +1\right ) \ln \left (\left (b -x \right ) \left (a -x \right )\right )} \left (\left (b -x \right ) \left (a -x \right )\right )^{k +1} \left (-a +x \right )^{-k -1} \left (x -b \right )^{k -1}}{\left (a -b \right ) k}\right )\right ) \\ \end{align*}

Taking derivative gives

\begin{equation} \tag{4} u^{\prime }\left (x \right ) = \frac {c_1 \left (\left (b -x \right ) \left (a -x \right )\right )^{-\frac {1}{2}-\frac {k}{2}} \left (-\frac {1}{2}-\frac {k}{2}\right ) \left (-a -b +2 x \right ) \left (-a +x \right )^{\frac {1}{2}+\frac {k}{2}} \left (x -b \right )^{\frac {1}{2}-\frac {k}{2}}}{\left (b -x \right ) \left (a -x \right )}+\frac {c_1 \left (\left (b -x \right ) \left (a -x \right )\right )^{-\frac {1}{2}-\frac {k}{2}} \left (-a +x \right )^{\frac {1}{2}+\frac {k}{2}} \left (\frac {1}{2}+\frac {k}{2}\right ) \left (x -b \right )^{\frac {1}{2}-\frac {k}{2}}}{-a +x}+\frac {c_1 \left (\left (b -x \right ) \left (a -x \right )\right )^{-\frac {1}{2}-\frac {k}{2}} \left (-a +x \right )^{\frac {1}{2}+\frac {k}{2}} \left (x -b \right )^{\frac {1}{2}-\frac {k}{2}} \left (\frac {1}{2}-\frac {k}{2}\right )}{x -b}-\frac {c_2 \left (-a +x \right )^{\frac {1}{2}-\frac {k}{2}} \left (\frac {1}{2}-\frac {k}{2}\right ) \left (x -b \right )^{\frac {1}{2}+\frac {k}{2}} \left (\left (b -x \right ) \left (a -x \right )\right )^{-\frac {1}{2}-\frac {k}{2}}}{k \left (a -b \right ) \left (-a +x \right )}-\frac {c_2 \left (-a +x \right )^{\frac {1}{2}-\frac {k}{2}} \left (x -b \right )^{\frac {1}{2}+\frac {k}{2}} \left (\frac {1}{2}+\frac {k}{2}\right ) \left (\left (b -x \right ) \left (a -x \right )\right )^{-\frac {1}{2}-\frac {k}{2}}}{k \left (a -b \right ) \left (x -b \right )}-\frac {c_2 \left (-a +x \right )^{\frac {1}{2}-\frac {k}{2}} \left (x -b \right )^{\frac {1}{2}+\frac {k}{2}} \left (\left (b -x \right ) \left (a -x \right )\right )^{-\frac {1}{2}-\frac {k}{2}} \left (-\frac {1}{2}-\frac {k}{2}\right ) \left (-a -b +2 x \right )}{k \left (a -b \right ) \left (b -x \right ) \left (a -x \right )} \end{equation}

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

\begin{align*} y &= \frac {-u'}{f_2 u} \\ y &= \frac {-u'}{u \left (-\frac {k}{\left (a -x \right ) \left (b -x \right )}-\frac {1}{\left (b -x \right ) \left (a -x \right )}\right )} \\ y &= -\frac {\frac {c_1 \left (\left (b -x \right ) \left (a -x \right )\right )^{-\frac {1}{2}-\frac {k}{2}} \left (-\frac {1}{2}-\frac {k}{2}\right ) \left (-a -b +2 x \right ) \left (-a +x \right )^{\frac {1}{2}+\frac {k}{2}} \left (x -b \right )^{\frac {1}{2}-\frac {k}{2}}}{\left (b -x \right ) \left (a -x \right )}+\frac {c_1 \left (\left (b -x \right ) \left (a -x \right )\right )^{-\frac {1}{2}-\frac {k}{2}} \left (-a +x \right )^{\frac {1}{2}+\frac {k}{2}} \left (\frac {1}{2}+\frac {k}{2}\right ) \left (x -b \right )^{\frac {1}{2}-\frac {k}{2}}}{-a +x}+\frac {c_1 \left (\left (b -x \right ) \left (a -x \right )\right )^{-\frac {1}{2}-\frac {k}{2}} \left (-a +x \right )^{\frac {1}{2}+\frac {k}{2}} \left (x -b \right )^{\frac {1}{2}-\frac {k}{2}} \left (\frac {1}{2}-\frac {k}{2}\right )}{x -b}-\frac {c_2 \left (-a +x \right )^{\frac {1}{2}-\frac {k}{2}} \left (\frac {1}{2}-\frac {k}{2}\right ) \left (x -b \right )^{\frac {1}{2}+\frac {k}{2}} \left (\left (b -x \right ) \left (a -x \right )\right )^{-\frac {1}{2}-\frac {k}{2}}}{k \left (a -b \right ) \left (-a +x \right )}-\frac {c_2 \left (-a +x \right )^{\frac {1}{2}-\frac {k}{2}} \left (x -b \right )^{\frac {1}{2}+\frac {k}{2}} \left (\frac {1}{2}+\frac {k}{2}\right ) \left (\left (b -x \right ) \left (a -x \right )\right )^{-\frac {1}{2}-\frac {k}{2}}}{k \left (a -b \right ) \left (x -b \right )}-\frac {c_2 \left (-a +x \right )^{\frac {1}{2}-\frac {k}{2}} \left (x -b \right )^{\frac {1}{2}+\frac {k}{2}} \left (\left (b -x \right ) \left (a -x \right )\right )^{-\frac {1}{2}-\frac {k}{2}} \left (-\frac {1}{2}-\frac {k}{2}\right ) \left (-a -b +2 x \right )}{k \left (a -b \right ) \left (b -x \right ) \left (a -x \right )}}{\left (-\frac {k}{\left (a -x \right ) \left (b -x \right )}-\frac {1}{\left (b -x \right ) \left (a -x \right )}\right ) \left (c_1 \left (\left (b -x \right ) \left (a -x \right )\right )^{-\frac {1}{2}-\frac {k}{2}} \left (-a +x \right )^{\frac {1}{2}+\frac {k}{2}} \left (x -b \right )^{\frac {1}{2}-\frac {k}{2}}-\frac {c_2 \left (-a +x \right )^{\frac {1}{2}-\frac {k}{2}} \left (x -b \right )^{\frac {1}{2}+\frac {k}{2}} \left (\left (b -x \right ) \left (a -x \right )\right )^{-\frac {1}{2}-\frac {k}{2}}}{k \left (a -b \right )}\right )} \\ \end{align*}

Doing change of constants, the above solution becomes

\[ y = -\frac {\frac {\left (\left (b -x \right ) \left (a -x \right )\right )^{-\frac {1}{2}-\frac {k}{2}} \left (-\frac {1}{2}-\frac {k}{2}\right ) \left (-a -b +2 x \right ) \left (-a +x \right )^{\frac {1}{2}+\frac {k}{2}} \left (x -b \right )^{\frac {1}{2}-\frac {k}{2}}}{\left (b -x \right ) \left (a -x \right )}+\frac {\left (\left (b -x \right ) \left (a -x \right )\right )^{-\frac {1}{2}-\frac {k}{2}} \left (-a +x \right )^{\frac {1}{2}+\frac {k}{2}} \left (\frac {1}{2}+\frac {k}{2}\right ) \left (x -b \right )^{\frac {1}{2}-\frac {k}{2}}}{-a +x}+\frac {\left (\left (b -x \right ) \left (a -x \right )\right )^{-\frac {1}{2}-\frac {k}{2}} \left (-a +x \right )^{\frac {1}{2}+\frac {k}{2}} \left (x -b \right )^{\frac {1}{2}-\frac {k}{2}} \left (\frac {1}{2}-\frac {k}{2}\right )}{x -b}-\frac {c_3 \left (-a +x \right )^{\frac {1}{2}-\frac {k}{2}} \left (\frac {1}{2}-\frac {k}{2}\right ) \left (x -b \right )^{\frac {1}{2}+\frac {k}{2}} \left (\left (b -x \right ) \left (a -x \right )\right )^{-\frac {1}{2}-\frac {k}{2}}}{k \left (a -b \right ) \left (-a +x \right )}-\frac {c_3 \left (-a +x \right )^{\frac {1}{2}-\frac {k}{2}} \left (x -b \right )^{\frac {1}{2}+\frac {k}{2}} \left (\frac {1}{2}+\frac {k}{2}\right ) \left (\left (b -x \right ) \left (a -x \right )\right )^{-\frac {1}{2}-\frac {k}{2}}}{k \left (a -b \right ) \left (x -b \right )}-\frac {c_3 \left (-a +x \right )^{\frac {1}{2}-\frac {k}{2}} \left (x -b \right )^{\frac {1}{2}+\frac {k}{2}} \left (\left (b -x \right ) \left (a -x \right )\right )^{-\frac {1}{2}-\frac {k}{2}} \left (-\frac {1}{2}-\frac {k}{2}\right ) \left (-a -b +2 x \right )}{k \left (a -b \right ) \left (b -x \right ) \left (a -x \right )}}{\left (-\frac {k}{\left (a -x \right ) \left (b -x \right )}-\frac {1}{\left (b -x \right ) \left (a -x \right )}\right ) \left (\left (\left (b -x \right ) \left (a -x \right )\right )^{-\frac {1}{2}-\frac {k}{2}} \left (-a +x \right )^{\frac {1}{2}+\frac {k}{2}} \left (x -b \right )^{\frac {1}{2}-\frac {k}{2}}-\frac {c_3 \left (-a +x \right )^{\frac {1}{2}-\frac {k}{2}} \left (x -b \right )^{\frac {1}{2}+\frac {k}{2}} \left (\left (b -x \right ) \left (a -x \right )\right )^{-\frac {1}{2}-\frac {k}{2}}}{k \left (a -b \right )}\right )} \]

Simplifying the above gives

\begin{align*} y &= -\frac {k \left (\left (x -b \right )^{1+\frac {k}{2}} c_3 \left (-a +x \right )^{-\frac {k}{2}}+k \left (x -b \right )^{-\frac {k}{2}} \left (a -x \right ) \left (a -b \right ) \left (-a +x \right )^{\frac {k}{2}}\right )}{\left (\left (-a +x \right )^{-\frac {k}{2}} \left (x -b \right )^{\frac {k}{2}} c_3 -k \left (x -b \right )^{-\frac {k}{2}} \left (a -b \right ) \left (-a +x \right )^{\frac {k}{2}}\right ) \left (k +1\right )} \\ \end{align*}

Summary of solutions found

2.2.60.3 Solved using ﬁrst_order_ode_riccati_by_guessing_particular_solution

0.166 (sec)

Entering ﬁrst order ode riccati guess solver

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

This is a Riccati ODE. Comparing the above ODE to solve with the Riccati standard form

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

Shows that

\begin{align*} f_0(x) & =-\frac {a b k}{\left (b -x \right ) \left (a -x \right )}+\frac {a k x}{\left (b -x \right ) \left (a -x \right )}+\frac {b k x}{\left (b -x \right ) \left (a -x \right )}-\frac {k \,x^{2}}{\left (b -x \right ) \left (a -x \right )}\\ f_1(x) & =\frac {a k}{\left (a -x \right ) \left (b -x \right )}+\frac {b k}{\left (a -x \right ) \left (b -x \right )}-\frac {2 x k}{\left (a -x \right ) \left (b -x \right )}\\ f_2(x) &=-\frac {k}{\left (a -x \right ) \left (b -x \right )}-\frac {1}{\left (b -x \right ) \left (a -x \right )} \end{align*}

Using trial and error, the following particular solution was found

\[ y_p = -\frac {k x}{k +1}+\frac {b k}{k +1} \]

Since a particular solution is known, then the general solution is given by

\begin{align*} y &= y_p + \frac { \phi (x) }{ c_1 - \int { \phi (x) f_2 \,dx} } \end{align*}

Where

\begin{align*} \phi (x) &= e^{ \int 2 f_2 y_p + f_1 \,dx } \end{align*}

Evaluating the above gives the general solution as

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

Simplifying the above gives

\begin{align*} y &= \frac {\left (-\left (-a +x \right )^{k +1} \left (k +1\right )+k c_1 \left (b -x \right ) \left (a -b \right ) \left (x -b \right )^{k}\right ) k}{\left (k c_1 \left (a -b \right ) \left (x -b \right )^{k}+\left (-a +x \right )^{k} \left (k +1\right )\right ) \left (k +1\right )} \\ \end{align*}

Summary of solutions found

2.2.60.4 ✓ Maple. Time used: 0.003 (sec). Leaf size: 54

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

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

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 to 2nd Order 
   -> Calling odsolve with the ODE, diff(diff(y(x),x),x) = (a*k+b*k-2*k*x+a+b-2 
*x)/(a*b-a*x-b*x+x^2)*diff(y(x),x)-(k+1)*k/(a*b-a*x-b*x+x^2)*y(x), y(x) 
      *** Sublevel 2 *** 
      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)] 
      checking if the LODE is missing y 
      -> Trying a Liouvillian solution using Kovacics algorithm 
         A Liouvillian solution exists 
         Reducible group (found an exponential solution) 
         Reducible group (found another exponential solution) 
      <- Kovacics algorithm successful 
   <- Riccati to 2nd Order successful

Maple step by step

\[ \begin {array}{lll} & {} & \textrm {Let's solve}\hspace {3pt} \\ {} & {} & \left (x -a \right ) \left (x -b \right ) \left (\frac {d}{d x}y \left (x \right )\right )+y \left (x \right )^{2}+k \left (y \left (x \right )+x -a \right ) \left (y \left (x \right )+x -b \right )=0 \\ \bullet & {} & \textrm {Highest derivative means the order of the ODE is}\hspace {3pt} 1 \\ {} & {} & \frac {d}{d x}y \left (x \right ) \\ \bullet & {} & \textrm {Solve for the highest derivative}\hspace {3pt} \\ {} & {} & \frac {d}{d x}y \left (x \right )=-\frac {y \left (x \right )^{2}+k \left (y \left (x \right )+x -a \right ) \left (y \left (x \right )+x -b \right )}{\left (x -a \right ) \left (x -b \right )} \end {array} \]

2.2.60.5 ✓ Mathematica. Time used: 2.427 (sec). Leaf size: 101

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

\begin{align*} y(x)&\to \frac {1}{2} \left (\frac {k (a+b-2 x)}{k+1}+\sqrt {-\frac {k^2 (a-b)^2}{(k+1)^2}} \tan \left (\frac {1}{2} \sqrt {-\frac {k^2 (a-b)^2}{(k+1)^2}} \int _1^x\frac {k+1}{(a-K[5]) (K[5]-b)}dK[5]+c_1\right )\right ) \end{align*}

2.2.60.6 ✓ Sympy. Time used: 9.437 (sec). Leaf size: 525

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

\[ y{\left (x \right )} = - \frac {k \left (a e^{\frac {k \left (C_{1} a^{2} k + C_{1} a^{2} + C_{1} b^{2} k + C_{1} b^{2} + a k \log {\left (- a + x \right )} + a \log {\left (- a + x \right )} + b k \log {\left (- b + x \right )} + b \log {\left (- b + x \right )}\right )}{a k + a - b k - b}} - b e^{\frac {k \left (2 C_{1} a b k + 2 C_{1} a b + a k \log {\left (- b + x \right )} + a \log {\left (- b + x \right )} + b k \log {\left (- a + x \right )} + b \log {\left (- a + x \right )}\right )}{a k + a - b k - b}} + x e^{\frac {k \left (2 C_{1} a b k + 2 C_{1} a b + a k \log {\left (- b + x \right )} + a \log {\left (- b + x \right )} + b k \log {\left (- a + x \right )} + b \log {\left (- a + x \right )}\right )}{a k + a - b k - b}} - x e^{\frac {k \left (C_{1} a^{2} k + C_{1} a^{2} + C_{1} b^{2} k + C_{1} b^{2} + a k \log {\left (- a + x \right )} + a \log {\left (- a + x \right )} + b k \log {\left (- b + x \right )} + b \log {\left (- b + x \right )}\right )}{a k + a - b k - b}}\right )}{k e^{\frac {k \left (2 C_{1} a b k + 2 C_{1} a b + a k \log {\left (- b + x \right )} + a \log {\left (- b + x \right )} + b k \log {\left (- a + x \right )} + b \log {\left (- a + x \right )}\right )}{a k + a - b k - b}} - k e^{\frac {k \left (C_{1} a^{2} k + C_{1} a^{2} + C_{1} b^{2} k + C_{1} b^{2} + a k \log {\left (- a + x \right )} + a \log {\left (- a + x \right )} + b k \log {\left (- b + x \right )} + b \log {\left (- b + x \right )}\right )}{a k + a - b k - b}} + e^{\frac {k \left (2 C_{1} a b k + 2 C_{1} a b + a k \log {\left (- b + x \right )} + a \log {\left (- b + x \right )} + b k \log {\left (- a + x \right )} + b \log {\left (- a + x \right )}\right )}{a k + a - b k - b}} - e^{\frac {k \left (C_{1} a^{2} k + C_{1} a^{2} + C_{1} b^{2} k + C_{1} b^{2} + a k \log {\left (- a + x \right )} + a \log {\left (- a + x \right )} + b k \log {\left (- b + x \right )} + b \log {\left (- b + x \right )}\right )}{a k + a - b k - b}}} \]

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