problem 261

Internal problem ID [11084]
Internal ﬁle name [OUTPUT/10341_Wednesday_January_24_2024_10_18_04_PM_71204093/index.tex]

Book: Handbook of exact solutions for ordinary diﬀerential equations. By Polyanin and Zaitsev. Second edition
Section: Chapter 2, Second-Order Diﬀerential Equations. section 2.1.2-8. Other equations.
Problem number: 261.
ODE order: 2.
ODE degree: 1.

The type(s) of ODE detected by this program : "second_order_ode_non_constant_coeﬀ_transformation_on_B"

\[ \boxed {\left (x^{n} a +b \,x^{m}+c \right ) y^{\prime \prime }+\left (\lambda -x \right ) y^{\prime }+y=0} \]

33.23.1 Solving as second order ode non constant coeﬀ transformation on B ode

Given an ode of the form \begin {align*} A y^{\prime \prime } + B y^{\prime } + C y &= F(x) \end {align*}

This method reduces the order ode the ODE by one by applying the transformation \begin {align*} y&= B v \end {align*}

This results in \begin {align*} y' &=B' v+ v' B \\ y'' &=B'' v+ B' v' +v'' B + v' B' \\ &=v'' B+2 v'+ B'+B'' v \end {align*}

And now the original ode becomes \begin {align*} A\left ( v'' B+2v'B'+ B'' v\right )+B\left ( B'v+ v' B\right ) +CBv & =0\\ ABv'' +\left ( 2AB'+B^{2}\right ) v'+\left (AB''+BB'+CB\right ) v & =0 \tag {1} \end {align*}

If the term \(AB''+BB'+CB\) is zero, then this method works and can be used to solve \[ ABv''+\left ( 2AB' +B^{2}\right ) v'=0 \] By Using \(u=v'\) which reduces the order of the above ode to one. The new ode is \[ ABu'+\left ( 2AB'+B^{2}\right ) u=0 \] The above ode is ﬁrst order ode which is solved for \(u\). Now a new ode \(v'=u\) is solved for \(v\) as ﬁrst order ode. Then the ﬁnal solution is obtain from \(y=Bv\).

This method works only if the term \(AB''+BB'+CB\) is zero. The given ODE shows that \begin {align*} A &= x^{n} a +b \,x^{m}+c\\ B &= \lambda -x\\ C &= 1\\ F &= 0 \end {align*}

The above shows that for this ode \begin {align*} AB''+BB'+CB &= \left (x^{n} a +b \,x^{m}+c\right ) \left (0\right ) + \left (\lambda -x\right ) \left (-1\right ) + \left (1\right ) \left (\lambda -x\right ) \\ &=0 \end {align*}

Hence the ode in \(v\) given in (1) now simpliﬁes to \begin {align*} \left (x^{n} a +b \,x^{m}+c \right ) \left (\lambda -x \right ) v'' +\left ( -2 x^{n} a -2 b \,x^{m}-2 c +\left (\lambda -x \right )^{2}\right ) v' & =0 \end {align*}

Now by applying \(v'=u\) the above becomes \begin {align*} \left (x^{n} a +b \,x^{m}+c \right ) \left (\lambda -x \right ) u^{\prime }\left (x \right )-2 u \left (x \right ) \left (x^{n} a +b \,x^{m}-\frac {x^{2}}{2}+\lambda x -\frac {\lambda ^{2}}{2}+c \right ) = 0 \end {align*}

Which is now solved for \(u\). In canonical form the ODE is \begin {align*} u' &= F(x,u)\\ &= f( x) g(u)\\ &= \frac {u \left (2 x^{n} a +2 b \,x^{m}-\lambda ^{2}+2 \lambda x -x^{2}+2 c \right )}{x^{n} a \lambda -x^{n} a x +x^{m} b \lambda -x^{m} b x +c \lambda -c x} \end {align*}

Where \(f(x)=\frac {2 x^{n} a +2 b \,x^{m}-\lambda ^{2}+2 \lambda x -x^{2}+2 c}{x^{n} a \lambda -x^{n} a x +x^{m} b \lambda -x^{m} b x +c \lambda -c x}\) and \(g(u)=u\). Integrating both sides gives \begin {align*} \frac {1}{u} \,du &= \frac {2 x^{n} a +2 b \,x^{m}-\lambda ^{2}+2 \lambda x -x^{2}+2 c}{x^{n} a \lambda -x^{n} a x +x^{m} b \lambda -x^{m} b x +c \lambda -c x} \,d x\\ \int { \frac {1}{u} \,du} &= \int {\frac {2 x^{n} a +2 b \,x^{m}-\lambda ^{2}+2 \lambda x -x^{2}+2 c}{x^{n} a \lambda -x^{n} a x +x^{m} b \lambda -x^{m} b x +c \lambda -c x} \,d x}\\ \ln \left (u \right )&=\int \frac {2 x^{n} a +2 b \,x^{m}-\lambda ^{2}+2 \lambda x -x^{2}+2 c}{x^{n} a \lambda -x^{n} a x +x^{m} b \lambda -x^{m} b x +c \lambda -c x}d x +c_{1}\\ u&={\mathrm e}^{\int \frac {2 x^{n} a +2 b \,x^{m}-\lambda ^{2}+2 \lambda x -x^{2}+2 c}{x^{n} a \lambda -x^{n} a x +x^{m} b \lambda -x^{m} b x +c \lambda -c x}d x +c_{1}}\\ &=c_{1} {\mathrm e}^{\int \frac {2 x^{n} a +2 b \,x^{m}-\lambda ^{2}+2 \lambda x -x^{2}+2 c}{x^{n} a \lambda -x^{n} a x +x^{m} b \lambda -x^{m} b x +c \lambda -c x}d x} \end {align*}

The ode for \(v\) now becomes \begin {align*} v' &= u\\ &=c_{1} {\mathrm e}^{\int \frac {2 x^{n} a +2 b \,x^{m}-\lambda ^{2}+2 \lambda x -x^{2}+2 c}{x^{n} a \lambda -x^{n} a x +x^{m} b \lambda -x^{m} b x +c \lambda -c x}d x} \end {align*}

Which is now solved for \(v\). Writing the ode as \begin {align*} v^{\prime }\left (x \right )&=c_{1} {\mathrm e}^{\int \frac {2 x^{n} a +2 b \,x^{m}-\lambda ^{2}+2 \lambda x -x^{2}+2 c}{x^{n} a \lambda -x^{n} a x +x^{m} b \lambda -x^{m} b x +c \lambda -c x}d x}\\ v^{\prime }\left (x \right )&= \omega \left ( x,v\right ) \end {align*}

The condition of Lie symmetry is the linearized PDE given by \begin {align*} \eta _{x}+\omega \left ( \eta _{v}-\xi _{x}\right ) -\omega ^{2}\xi _{v}-\omega _{x}\xi -\omega _{v}\eta =0\tag {A} \end {align*}

The type of this ode is not in the lookup table. 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 &= v a_{3}+x a_{2}+a_{1} \\ \tag{2E} \eta &= v b_{3}+x b_{2}+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}+c_{1} {\mathrm e}^{\int \frac {2 x^{n} a +2 b \,x^{m}-\lambda ^{2}+2 \lambda x -x^{2}+2 c}{x^{n} a \lambda -x^{n} a x +x^{m} b \lambda -x^{m} b x +c \lambda -c x}d x} \left (b_{3}-a_{2}\right )-c_{1}^{2} {\mathrm e}^{\int \frac {4 x^{n} a +4 b \,x^{m}-2 \lambda ^{2}+4 \lambda x -2 x^{2}+4 c}{x^{n} a \lambda -a \,x^{1+n}+x^{m} b \lambda -b \,x^{m +1}+c \lambda -c x}d x} a_{3}-\frac {c_{1} \left (2 x^{n} a +2 b \,x^{m}-\lambda ^{2}+2 \lambda x -x^{2}+2 c \right ) {\mathrm e}^{\int \frac {2 x^{n} a +2 b \,x^{m}-\lambda ^{2}+2 \lambda x -x^{2}+2 c}{x^{n} a \lambda -x^{n} a x +x^{m} b \lambda -x^{m} b x +c \lambda -c x}d x} \left (v a_{3}+x a_{2}+a_{1}\right )}{x^{n} a \lambda -x^{n} a x +x^{m} b \lambda -x^{m} b x +c \lambda -c x} = 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} Simplifying the above 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 \(\{v, x\}\) in them. \[ \left \{v, x, x^{m}, x^{n}, \int \frac {2 x^{n} a +2 b \,x^{m}-\lambda ^{2}+2 \lambda x -x^{2}+2 c}{x^{n} a \lambda -x^{n} a x +x^{m} b \lambda -x^{m} b x +c \lambda -c x}d x, {\mathrm e}^{2 \left (\int \frac {2 x^{n} a +2 b \,x^{m}-\lambda ^{2}+2 \lambda x -x^{2}+2 c}{x^{n} a \lambda -x^{n} a x +x^{m} b \lambda -x^{m} b x +c \lambda -c x}d x \right )}, {\mathrm e}^{\int \frac {2 x^{n} a +2 b \,x^{m}-\lambda ^{2}+2 \lambda x -x^{2}+2 c}{x^{n} a \lambda -x^{n} a x +x^{m} b \lambda -x^{m} b x +c \lambda -c x}d x}\right \} \] The following substitution is now made to be able to collect on all terms with \(\{v, x\}\) in them \[ \left \{v = v_{1}, x = v_{2}, x^{m} = v_{3}, x^{n} = v_{4}, \int \frac {2 x^{n} a +2 b \,x^{m}-\lambda ^{2}+2 \lambda x -x^{2}+2 c}{x^{n} a \lambda -x^{n} a x +x^{m} b \lambda -x^{m} b x +c \lambda -c x}d x = v_{5}, {\mathrm e}^{2 \left (\int \frac {2 x^{n} a +2 b \,x^{m}-\lambda ^{2}+2 \lambda x -x^{2}+2 c}{x^{n} a \lambda -x^{n} a x +x^{m} b \lambda -x^{m} b x +c \lambda -c x}d x \right )} = v_{6}, {\mathrm e}^{\int \frac {2 x^{n} a +2 b \,x^{m}-\lambda ^{2}+2 \lambda x -x^{2}+2 c}{x^{n} a \lambda -x^{n} a x +x^{m} b \lambda -x^{m} b x +c \lambda -c x}d x} = v_{7}\right \} \] The above PDE (6E) now becomes \begin{equation} \tag{7E} -a \,c_{1}^{2} \lambda a_{3} v_{4} v_{6}+a \,c_{1}^{2} a_{3} v_{2} v_{4} v_{6}-b \,c_{1}^{2} \lambda a_{3} v_{3} v_{6}+b \,c_{1}^{2} a_{3} v_{2} v_{3} v_{6}-a c_{1} \lambda a_{2} v_{4} v_{7}+a c_{1} \lambda b_{3} v_{4} v_{7}-a c_{1} a_{2} v_{2} v_{4} v_{7}-2 a c_{1} a_{3} v_{1} v_{4} v_{7}-a c_{1} b_{3} v_{2} v_{4} v_{7}-b c_{1} \lambda a_{2} v_{3} v_{7}+b c_{1} \lambda b_{3} v_{3} v_{7}-b c_{1} a_{2} v_{2} v_{3} v_{7}-2 b c_{1} a_{3} v_{1} v_{3} v_{7}-b c_{1} b_{3} v_{2} v_{3} v_{7}-c \,c_{1}^{2} \lambda a_{3} v_{6}+c \,c_{1}^{2} a_{3} v_{2} v_{6}+c_{1} \lambda ^{2} a_{2} v_{2} v_{7}+c_{1} \lambda ^{2} a_{3} v_{1} v_{7}-2 c_{1} \lambda a_{2} v_{2}^{2} v_{7}-2 c_{1} \lambda a_{3} v_{1} v_{2} v_{7}+c_{1} a_{2} v_{2}^{3} v_{7}+c_{1} a_{3} v_{1} v_{2}^{2} v_{7}-2 a c_{1} a_{1} v_{4} v_{7}-2 b c_{1} a_{1} v_{3} v_{7}-c c_{1} \lambda a_{2} v_{7}+c c_{1} \lambda b_{3} v_{7}-c c_{1} a_{2} v_{2} v_{7}-2 c c_{1} a_{3} v_{1} v_{7}-c c_{1} b_{3} v_{2} v_{7}+c_{1} \lambda ^{2} a_{1} v_{7}-2 c_{1} \lambda a_{1} v_{2} v_{7}+c_{1} a_{1} v_{2}^{2} v_{7}+v_{4} a \lambda b_{2}-v_{4} a v_{2} b_{2}+v_{3} b \lambda b_{2}-v_{3} b v_{2} b_{2}-2 c c_{1} a_{1} v_{7}+c \lambda b_{2}-c v_{2} b_{2} = 0 \end{equation} Collecting the above on the terms \(v_i\) introduced, and these are \[ \{v_{1}, v_{2}, v_{3}, v_{4}, v_{5}, v_{6}, v_{7}\} \] Equation (7E) now becomes \begin{equation} \tag{8E} \left (c_{1} \lambda ^{2} a_{3}-2 c c_{1} a_{3}\right ) v_{1} v_{7}+\left (-2 c_{1} \lambda a_{2}+c_{1} a_{1}\right ) v_{2}^{2} v_{7}+\left (c_{1} \lambda ^{2} a_{2}-c c_{1} a_{2}-c c_{1} b_{3}-2 c_{1} \lambda a_{1}\right ) v_{2} v_{7}+\left (-b c_{1} \lambda a_{2}+b c_{1} \lambda b_{3}-2 b c_{1} a_{1}\right ) v_{3} v_{7}+\left (-a c_{1} \lambda a_{2}+a c_{1} \lambda b_{3}-2 a c_{1} a_{1}\right ) v_{4} v_{7}-a \,c_{1}^{2} \lambda a_{3} v_{4} v_{6}+a \,c_{1}^{2} a_{3} v_{2} v_{4} v_{6}-b \,c_{1}^{2} \lambda a_{3} v_{3} v_{6}+b \,c_{1}^{2} a_{3} v_{2} v_{3} v_{6}-2 a c_{1} a_{3} v_{1} v_{4} v_{7}-2 b c_{1} a_{3} v_{1} v_{3} v_{7}-2 c_{1} \lambda a_{3} v_{1} v_{2} v_{7}-c v_{2} b_{2}+\left (-b c_{1} a_{2}-b c_{1} b_{3}\right ) v_{2} v_{3} v_{7}+\left (-a c_{1} a_{2}-a c_{1} b_{3}\right ) v_{2} v_{4} v_{7}+v_{4} a \lambda b_{2}-v_{4} a v_{2} b_{2}+v_{3} b \lambda b_{2}-v_{3} b v_{2} b_{2}+c_{1} a_{2} v_{2}^{3} v_{7}+\left (-c c_{1} \lambda a_{2}+c c_{1} \lambda b_{3}+c_{1} \lambda ^{2} a_{1}-2 c c_{1} a_{1}\right ) v_{7}-c \,c_{1}^{2} \lambda a_{3} v_{6}+c \,c_{1}^{2} a_{3} v_{2} v_{6}+c_{1} a_{3} v_{1} v_{2}^{2} v_{7}+c \lambda b_{2} = 0 \end{equation} Setting each coeﬃcients in (8E) to zero gives the following equations to solve \begin {align*} c_{1} a_{2}&=0\\ c_{1} a_{3}&=0\\ c_{1}^{2} a a_{3}&=0\\ c_{1}^{2} b a_{3}&=0\\ c_{1}^{2} c a_{3}&=0\\ a \lambda b_{2}&=0\\ b \lambda b_{2}&=0\\ c \lambda b_{2}&=0\\ -a b_{2}&=0\\ -b b_{2}&=0\\ -c b_{2}&=0\\ -2 c_{1} a a_{3}&=0\\ -2 c_{1} b a_{3}&=0\\ -2 c_{1} \lambda a_{3}&=0\\ -c_{1}^{2} a \lambda a_{3}&=0\\ -c_{1}^{2} b \lambda a_{3}&=0\\ -c_{1}^{2} c \lambda a_{3}&=0\\ -a c_{1} a_{2}-a c_{1} b_{3}&=0\\ -b c_{1} a_{2}-b c_{1} b_{3}&=0\\ c_{1} \lambda ^{2} a_{3}-2 c c_{1} a_{3}&=0\\ -2 c_{1} \lambda a_{2}+c_{1} a_{1}&=0\\ -a c_{1} \lambda a_{2}+a c_{1} \lambda b_{3}-2 a c_{1} a_{1}&=0\\ -b c_{1} \lambda a_{2}+b c_{1} \lambda b_{3}-2 b c_{1} a_{1}&=0\\ c_{1} \lambda ^{2} a_{2}-c c_{1} a_{2}-c c_{1} b_{3}-2 c_{1} \lambda a_{1}&=0\\ -c c_{1} \lambda a_{2}+c c_{1} \lambda b_{3}+c_{1} \lambda ^{2} a_{1}-2 c c_{1} a_{1}&=0 \end {align*}

Solving the above equations for the unknowns gives \begin {align*} a_{1}&=0\\ a_{2}&=0\\ a_{3}&=0\\ b_{1}&=b_{1}\\ b_{2}&=0\\ b_{3}&=0 \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 &= 1 \\ \end{align*} The next step is to determine the canonical coordinates \(R,S\). The canonical coordinates map \(\left ( x,v\right ) \to \left ( R,S \right )\) 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 v}{\eta } = dS \tag {1} \end {align*}

The above comes from the requirements that \(\left ( \xi \frac {\partial }{\partial x} + \eta \frac {\partial }{\partial v}\right ) S(x,v) = 1\). 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}{1}} dy \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,v) S_{v} }{ R_{x} + \omega (x,v) R_{v} }\tag {2} \end {align*}

Where in the above \(R_{x},R_{v},S_{x},S_{v}\) are all partial derivatives and \(\omega (x,v)\) is the right hand side of the original ode given by \begin {align*} \omega (x,v) &= c_{1} {\mathrm e}^{\int \frac {2 x^{n} a +2 b \,x^{m}-\lambda ^{2}+2 \lambda x -x^{2}+2 c}{x^{n} a \lambda -x^{n} a x +x^{m} b \lambda -x^{m} b x +c \lambda -c x}d x} \end {align*}

Evaluating all the partial derivatives gives \begin {align*} R_{x} &= 1\\ R_{v} &= 0\\ S_{x} &= 0\\ S_{v} &= 1 \end {align*}

Substituting all the above in (2) and simplifying gives the ode in canonical coordinates. \begin {align*} \frac {dS}{dR} &= c_{1} {\mathrm e}^{\int \frac {-2 x^{n} a -2 b \,x^{m}+x^{2}-2 \lambda x +\lambda ^{2}-2 c}{\left (x^{n} a +b \,x^{m}+c \right ) \left (-\lambda +x \right )}d x}\tag {2A} \end {align*}

We now need to express the RHS as function of \(R\) only. This is done by solving for \(x,v\) in terms of \(R,S\) from the result obtained earlier and simplifying. This gives \begin {align*} \frac {dS}{dR} &= c_{1} {\mathrm e}^{\int \frac {-2 R^{n} a -2 b \,R^{m}+R^{2}-2 \lambda R +\lambda ^{2}-2 c}{\left (R^{n} a +b \,R^{m}+c \right ) \left (-\lambda +R \right )}d R} \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 \(R,S\). Integrating the above gives \begin {align*} S \left (R \right ) = \int c_{1} {\mathrm e}^{-\left (\int \frac {2 R^{n} a +2 b \,R^{m}-R^{2}+2 \lambda R -\lambda ^{2}+2 c}{\left (R^{n} a +b \,R^{m}+c \right ) \left (-\lambda +R \right )}d R \right )}d R +c_{2}\tag {4} \end {align*}

To complete the solution, we just need to transform (4) back to \(x,v\) coordinates. This results in \begin {align*} v \left (x \right ) = \int c_{1} {\mathrm e}^{-\left (\int \frac {2 x^{n} a +2 b \,x^{m}-\lambda ^{2}+2 \lambda x -x^{2}+2 c}{\left (x^{n} a +b \,x^{m}+c \right ) \left (-\lambda +x \right )}d x \right )}d x +c_{2} \end {align*}

Which simpliﬁes to \begin {align*} v \left (x \right )-c_{1} \left (\int {\mathrm e}^{\int \frac {-2 x^{n} a -2 b \,x^{m}+x^{2}-2 \lambda x +\lambda ^{2}-2 c}{\left (x^{n} a +b \,x^{m}+c \right ) \left (-\lambda +x \right )}d x}d x \right )-c_{2} = 0 \end {align*}

Which gives \begin {align*} v \left (x \right ) = c_{1} \left (\int {\mathrm e}^{\int \frac {-2 x^{n} a -2 b \,x^{m}+x^{2}-2 \lambda x +\lambda ^{2}-2 c}{\left (x^{n} a +b \,x^{m}+c \right ) \left (-\lambda +x \right )}d x}d x \right )+c_{2} \end {align*}

Therefore the solution is \begin {align*} y(x) &= B v\\ &= \left (\lambda -x\right ) \left (c_{1} \left (\int {\mathrm e}^{\int \frac {-2 x^{n} a -2 b \,x^{m}+x^{2}-2 \lambda x +\lambda ^{2}-2 c}{\left (x^{n} a +b \,x^{m}+c \right ) \left (-\lambda +x \right )}d x}d x \right )+c_{2}\right ) \\ &= \left (\lambda -x \right ) \left (c_{1} \left (\int {\mathrm e}^{\int \frac {-2 x^{n} a -2 b \,x^{m}+x^{2}-2 \lambda x +\lambda ^{2}-2 c}{\left (x^{n} a +b \,x^{m}+c \right ) \left (-\lambda +x \right )}d x}d x \right )+c_{2} \right ) \end {align*}

Summary

The solution(s) found are the following \begin{align*} \tag{1} y &= \left (\lambda -x \right ) \left (c_{1} \left (\int {\mathrm e}^{\int \frac {-2 x^{n} a -2 b \,x^{m}+x^{2}-2 \lambda x +\lambda ^{2}-2 c}{\left (x^{n} a +b \,x^{m}+c \right ) \left (-\lambda +x \right )}d x}d x \right )+c_{2} \right ) \\ \end{align*}

Veriﬁcation of solutions

\[ y = \left (\lambda -x \right ) \left (c_{1} \left (\int {\mathrm e}^{\int \frac {-2 x^{n} a -2 b \,x^{m}+x^{2}-2 \lambda x +\lambda ^{2}-2 c}{\left (x^{n} a +b \,x^{m}+c \right ) \left (-\lambda +x \right )}d x}d x \right )+c_{2} \right ) \] Veriﬁed OK.

Maple trace

`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 
-> Trying an equivalence, under non-integer power transformations, 
   to LODEs admitting Liouvillian solutions. 
-> Trying a solution in terms of special functions: 
   -> Bessel 
   -> elliptic 
   -> Legendre 
   -> Whittaker 
      -> hyper3: Equivalence to 1F1 under a power @ Moebius 
   -> hypergeometric 
      -> heuristic approach 
      -> hyper3: Equivalence to 2F1, 1F1 or 0F1 under a power @ Moebius 
   -> Mathieu 
      -> Equivalence to the rational form of Mathieu ODE under a power @ Moebius 
-> Heun: Equivalence to the GHE or one of its 4 confluent cases under a power @ Moebius 
-> 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 
<- unable to find a useful change of variables 
   trying a symmetry of the form [xi=0, eta=F(x)] 
   trying differential order: 2; exact nonlinear 
   trying symmetries linear in x and y(x) 
   <- linear symmetries successful`

\[ y \left (x \right ) = -\left (\left (\int {\mathrm e}^{\int \frac {-2 a \,x^{n}-2 b \,x^{m}-2 c +x^{2}-2 x \lambda +\lambda ^{2}}{\left (a \,x^{n}+b \,x^{m}+c \right ) \left (-\lambda +x \right )}d x}d x \right ) c_{1} +c_{2} \right ) \left (\lambda -x \right ) \]

33.23 problem 261

33.23.1 Solving as second order ode non constant coeﬀ transformation on B ode