Internal problem ID [11079]
Internal file name [OUTPUT/10336_Wednesday_January_24_2024_10_17_58_PM_43209986/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: 256.
ODE order: 2.
ODE degree: 1.
The type(s) of ODE detected by this program : "unknown"
Maple gives the following as the ode type
[[_2nd_order, _with_linear_symmetries]]
\[ \boxed {\left (x^{n} a +b \right )^{2} y^{\prime \prime }+\left (x^{n} a +b \right ) \left (c \,x^{n}+d \right ) y^{\prime }+n \left (-a d +b c \right ) x^{n -1} y=0} \]
In normal form the ode \begin {align*} y^{\prime \prime } \left (x^{2 n} a^{2}+2 x^{n} a b +b^{2}\right )+\left (x^{2 n} a c +d \,x^{n} a +x^{n} b c +b d \right ) y^{\prime }+n \left (-a d +b c \right ) x^{n -1} y = 0 \tag {1} \end {align*}
Becomes \begin {align*} y^{\prime \prime }+p \left (x \right ) y^{\prime }+q \left (x \right ) y&=r \left (x \right ) \tag {2} \end {align*}
Where \begin {align*} p \left (x \right )&=\frac {x^{2 n} a c +d \,x^{n} a +x^{n} b c +b d}{x^{2 n} a^{2}+2 x^{n} a b +b^{2}}\\ q \left (x \right )&=\frac {n \,x^{n -1} \left (-a d +b c \right )}{x^{2 n} a^{2}+2 x^{n} a b +b^{2}}\\ r \left (x \right )&=0 \end {align*}
The Lagrange adjoint ode is given by \begin {align*} \xi ^{''}-(\xi \, p)'+\xi q &= 0\\ \xi ^{''}-\left (\frac {\left (x^{2 n} a c +d \,x^{n} a +x^{n} b c +b d \right ) \xi \left (x \right )}{x^{2 n} a^{2}+2 x^{n} a b +b^{2}}\right )' + \left (\frac {n \,x^{n -1} \left (-a d +b c \right ) \xi \left (x \right )}{x^{2 n} a^{2}+2 x^{n} a b +b^{2}}\right ) &= 0\\ \xi ^{\prime \prime }\left (x \right )-\frac {\left (x^{2 n} a c +d \,x^{n} a +x^{n} b c +b d \right ) \xi ^{\prime }\left (x \right )}{x^{2 n} a^{2}+2 x^{n} a b +b^{2}}+\left (-\frac {\frac {2 x^{2 n} n a c}{x}+\frac {x^{n} a d n}{x}+\frac {x^{n} b c n}{x}}{x^{2 n} a^{2}+2 x^{n} a b +b^{2}}+\frac {\left (x^{2 n} a c +d \,x^{n} a +x^{n} b c +b d \right ) \left (\frac {2 x^{2 n} n \,a^{2}}{x}+\frac {2 n \,x^{n} a b}{x}\right )}{\left (x^{2 n} a^{2}+2 x^{n} a b +b^{2}\right )^{2}}+\frac {n \,x^{n -1} \left (-a d +b c \right )}{x^{2 n} a^{2}+2 x^{n} a b +b^{2}}\right ) \xi \left (x \right )&= 0 \end {align*}
Which is solved for \(\xi (x)\). This is second order ode with missing dependent variable \(\xi \left (x \right )\). Let \begin {align*} p(x) &= \xi ^{\prime }\left (x \right ) \end {align*}
Then \begin {align*} p'(x) &= \xi ^{\prime \prime }\left (x \right ) \end {align*}
Hence the ode becomes \begin {align*} x \left (-x^{4 n} a^{4}-4 b \,x^{3 n} a^{3}-6 x^{2 n} a^{2} b^{2}-4 x^{n} a \,b^{3}-b^{4}\right ) p^{\prime }\left (x \right )+x \left (\left (3 d b \,a^{2}+3 c \,b^{2} a \right ) x^{2 n}+\left (d \,a^{3}+3 c b \,a^{2}\right ) x^{3 n}+x^{4 n} a^{3} c +\left (3 a \,b^{2} d +c \,b^{3}\right ) x^{n}+b^{3} d \right ) p \left (x \right ) = 0 \end {align*}
Which is now solve for \(p(x)\) as first order ode. In canonical form the ODE is \begin {align*} p' &= F(x,p)\\ &= f( x) g(p)\\ &= \frac {p \left (x^{4 n} a^{3} c +d \,x^{3 n} a^{3}+3 b c \,x^{3 n} a^{2}+3 x^{2 n} a^{2} b d +3 x^{2 n} a \,b^{2} c +3 x^{n} a \,b^{2} d +x^{n} b^{3} c +b^{3} d \right )}{x^{4 n} a^{4}+4 b \,x^{3 n} a^{3}+6 x^{2 n} a^{2} b^{2}+4 x^{n} a \,b^{3}+b^{4}} \end {align*}
Where \(f(x)=\frac {x^{4 n} a^{3} c +d \,x^{3 n} a^{3}+3 b c \,x^{3 n} a^{2}+3 x^{2 n} a^{2} b d +3 x^{2 n} a \,b^{2} c +3 x^{n} a \,b^{2} d +x^{n} b^{3} c +b^{3} d}{x^{4 n} a^{4}+4 b \,x^{3 n} a^{3}+6 x^{2 n} a^{2} b^{2}+4 x^{n} a \,b^{3}+b^{4}}\) and \(g(p)=p\). Integrating both sides gives \begin {align*} \frac {1}{p} \,dp &= \frac {x^{4 n} a^{3} c +d \,x^{3 n} a^{3}+3 b c \,x^{3 n} a^{2}+3 x^{2 n} a^{2} b d +3 x^{2 n} a \,b^{2} c +3 x^{n} a \,b^{2} d +x^{n} b^{3} c +b^{3} d}{x^{4 n} a^{4}+4 b \,x^{3 n} a^{3}+6 x^{2 n} a^{2} b^{2}+4 x^{n} a \,b^{3}+b^{4}} \,d x\\ \int { \frac {1}{p} \,dp} &= \int {\frac {x^{4 n} a^{3} c +d \,x^{3 n} a^{3}+3 b c \,x^{3 n} a^{2}+3 x^{2 n} a^{2} b d +3 x^{2 n} a \,b^{2} c +3 x^{n} a \,b^{2} d +x^{n} b^{3} c +b^{3} d}{x^{4 n} a^{4}+4 b \,x^{3 n} a^{3}+6 x^{2 n} a^{2} b^{2}+4 x^{n} a \,b^{3}+b^{4}} \,d x}\\ \ln \left (p \right )&=\int \frac {x^{4 n} a^{3} c +d \,x^{3 n} a^{3}+3 b c \,x^{3 n} a^{2}+3 x^{2 n} a^{2} b d +3 x^{2 n} a \,b^{2} c +3 x^{n} a \,b^{2} d +x^{n} b^{3} c +b^{3} d}{x^{4 n} a^{4}+4 b \,x^{3 n} a^{3}+6 x^{2 n} a^{2} b^{2}+4 x^{n} a \,b^{3}+b^{4}}d x +c_{1}\\ p&={\mathrm e}^{\int \frac {x^{4 n} a^{3} c +d \,x^{3 n} a^{3}+3 b c \,x^{3 n} a^{2}+3 x^{2 n} a^{2} b d +3 x^{2 n} a \,b^{2} c +3 x^{n} a \,b^{2} d +x^{n} b^{3} c +b^{3} d}{x^{4 n} a^{4}+4 b \,x^{3 n} a^{3}+6 x^{2 n} a^{2} b^{2}+4 x^{n} a \,b^{3}+b^{4}}d x +c_{1}}\\ &=c_{1} {\mathrm e}^{\int \frac {x^{4 n} a^{3} c +d \,x^{3 n} a^{3}+3 b c \,x^{3 n} a^{2}+3 x^{2 n} a^{2} b d +3 x^{2 n} a \,b^{2} c +3 x^{n} a \,b^{2} d +x^{n} b^{3} c +b^{3} d}{x^{4 n} a^{4}+4 b \,x^{3 n} a^{3}+6 x^{2 n} a^{2} b^{2}+4 x^{n} a \,b^{3}+b^{4}}d x} \end {align*}
Since \(p=\xi ^{\prime }\left (x \right )\) then the new first order ode to solve is \begin {align*} \xi ^{\prime }\left (x \right ) = c_{1} {\mathrm e}^{\int \frac {x^{4 n} a^{3} c +d \,x^{3 n} a^{3}+3 b c \,x^{3 n} a^{2}+3 x^{2 n} a^{2} b d +3 x^{2 n} a \,b^{2} c +3 x^{n} a \,b^{2} d +x^{n} b^{3} c +b^{3} d}{x^{4 n} a^{4}+4 b \,x^{3 n} a^{3}+6 x^{2 n} a^{2} b^{2}+4 x^{n} a \,b^{3}+b^{4}}d x} \end {align*}
Writing the ode as \begin {align*} \xi ^{\prime }\left (x \right )&=c_{1} {\mathrm e}^{\int \frac {x^{4 n} a^{3} c +d \,x^{3 n} a^{3}+3 b c \,x^{3 n} a^{2}+3 x^{2 n} a^{2} b d +3 x^{2 n} a \,b^{2} c +3 x^{n} a \,b^{2} d +x^{n} b^{3} c +b^{3} d}{x^{4 n} a^{4}+4 b \,x^{3 n} a^{3}+6 x^{2 n} a^{2} b^{2}+4 x^{n} a \,b^{3}+b^{4}}d x}\\ \xi ^{\prime }\left (x \right )&= \omega \left ( x,\xi \right ) \end {align*}
The condition of Lie symmetry is the linearized PDE given by \begin {align*} \eta _{x}+\omega \left ( \eta _{\xi }-\xi _{x}\right ) -\omega ^{2}\xi _{\xi }-\omega _{x}\xi -\omega _{\xi }\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 &= x a_{2}+\xi a_{3}+a_{1} \\ \tag{2E} \eta &= x b_{2}+\xi 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}+c_{1} {\mathrm e}^{\int \frac {x^{4 n} a^{3} c +d \,x^{3 n} a^{3}+3 b c \,x^{3 n} a^{2}+3 x^{2 n} a^{2} b d +3 x^{2 n} a \,b^{2} c +3 x^{n} a \,b^{2} d +x^{n} b^{3} c +b^{3} d}{x^{4 n} a^{4}+4 b \,x^{3 n} a^{3}+6 x^{2 n} a^{2} b^{2}+4 x^{n} a \,b^{3}+b^{4}}d x} \left (b_{3}-a_{2}\right )-c_{1}^{2} {\mathrm e}^{\int \frac {2 x^{4 n} a^{3} c +2 d \,x^{3 n} a^{3}+6 b c \,x^{3 n} a^{2}+6 x^{2 n} a^{2} b d +6 x^{2 n} a \,b^{2} c +6 x^{n} a \,b^{2} d +2 x^{n} b^{3} c +2 b^{3} d}{x^{4 n} a^{4}+4 b \,x^{3 n} a^{3}+6 x^{2 n} a^{2} b^{2}+4 x^{n} a \,b^{3}+b^{4}}d x} a_{3}-\frac {c_{1} \left (x^{4 n} a^{3} c +d \,x^{3 n} a^{3}+3 b c \,x^{3 n} a^{2}+3 x^{2 n} a^{2} b d +3 x^{2 n} a \,b^{2} c +3 x^{n} a \,b^{2} d +x^{n} b^{3} c +b^{3} d \right ) {\mathrm e}^{\int \frac {x^{4 n} a^{3} c +d \,x^{3 n} a^{3}+3 b c \,x^{3 n} a^{2}+3 x^{2 n} a^{2} b d +3 x^{2 n} a \,b^{2} c +3 x^{n} a \,b^{2} d +x^{n} b^{3} c +b^{3} d}{x^{4 n} a^{4}+4 b \,x^{3 n} a^{3}+6 x^{2 n} a^{2} b^{2}+4 x^{n} a \,b^{3}+b^{4}}d x} \left (x a_{2}+\xi a_{3}+a_{1}\right )}{x^{4 n} a^{4}+4 b \,x^{3 n} a^{3}+6 x^{2 n} a^{2} b^{2}+4 x^{n} a \,b^{3}+b^{4}} = 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 \(\{x, \xi \}\) in them. \[ \left \{x, \xi , x^{n}, x^{2 n}, x^{3 n}, x^{4 n}, \int \frac {x^{4 n} a^{3} c +d \,x^{3 n} a^{3}+3 b c \,x^{3 n} a^{2}+3 x^{2 n} a^{2} b d +3 x^{2 n} a \,b^{2} c +3 x^{n} a \,b^{2} d +x^{n} b^{3} c +b^{3} d}{x^{4 n} a^{4}+4 b \,x^{3 n} a^{3}+6 x^{2 n} a^{2} b^{2}+4 x^{n} a \,b^{3}+b^{4}}d x, {\mathrm e}^{2 \left (\int \frac {x^{4 n} a^{3} c +d \,x^{3 n} a^{3}+3 b c \,x^{3 n} a^{2}+3 x^{2 n} a^{2} b d +3 x^{2 n} a \,b^{2} c +3 x^{n} a \,b^{2} d +x^{n} b^{3} c +b^{3} d}{x^{4 n} a^{4}+4 b \,x^{3 n} a^{3}+6 x^{2 n} a^{2} b^{2}+4 x^{n} a \,b^{3}+b^{4}}d x \right )}, {\mathrm e}^{\int \frac {x^{4 n} a^{3} c +d \,x^{3 n} a^{3}+3 b c \,x^{3 n} a^{2}+3 x^{2 n} a^{2} b d +3 x^{2 n} a \,b^{2} c +3 x^{n} a \,b^{2} d +x^{n} b^{3} c +b^{3} d}{x^{4 n} a^{4}+4 b \,x^{3 n} a^{3}+6 x^{2 n} a^{2} b^{2}+4 x^{n} a \,b^{3}+b^{4}}d x}\right \} \] The following substitution is now made to be able to collect on all terms with \(\{x, \xi \}\) in them \[ \left \{x = v_{1}, \xi = v_{2}, x^{n} = v_{3}, x^{2 n} = v_{4}, x^{3 n} = v_{5}, x^{4 n} = v_{6}, \int \frac {x^{4 n} a^{3} c +d \,x^{3 n} a^{3}+3 b c \,x^{3 n} a^{2}+3 x^{2 n} a^{2} b d +3 x^{2 n} a \,b^{2} c +3 x^{n} a \,b^{2} d +x^{n} b^{3} c +b^{3} d}{x^{4 n} a^{4}+4 b \,x^{3 n} a^{3}+6 x^{2 n} a^{2} b^{2}+4 x^{n} a \,b^{3}+b^{4}}d x = v_{7}, {\mathrm e}^{2 \left (\int \frac {x^{4 n} a^{3} c +d \,x^{3 n} a^{3}+3 b c \,x^{3 n} a^{2}+3 x^{2 n} a^{2} b d +3 x^{2 n} a \,b^{2} c +3 x^{n} a \,b^{2} d +x^{n} b^{3} c +b^{3} d}{x^{4 n} a^{4}+4 b \,x^{3 n} a^{3}+6 x^{2 n} a^{2} b^{2}+4 x^{n} a \,b^{3}+b^{4}}d x \right )} = v_{8}, {\mathrm e}^{\int \frac {x^{4 n} a^{3} c +d \,x^{3 n} a^{3}+3 b c \,x^{3 n} a^{2}+3 x^{2 n} a^{2} b d +3 x^{2 n} a \,b^{2} c +3 x^{n} a \,b^{2} d +x^{n} b^{3} c +b^{3} d}{x^{4 n} a^{4}+4 b \,x^{3 n} a^{3}+6 x^{2 n} a^{2} b^{2}+4 x^{n} a \,b^{3}+b^{4}}d x} = v_{9}\right \} \] The above PDE (6E) now becomes \begin{equation} \tag{7E} -a^{4} c_{1}^{2} a_{3} v_{3}^{4} v_{8}-a^{3} c c_{1} a_{2} v_{1} v_{3}^{4} v_{9}-a^{3} c c_{1} a_{3} v_{2} v_{3}^{4} v_{9}-a^{4} c_{1} a_{2} v_{3}^{4} v_{9}+a^{4} c_{1} b_{3} v_{3}^{4} v_{9}-4 a^{3} b \,c_{1}^{2} a_{3} v_{3}^{3} v_{8}-a^{3} c c_{1} a_{1} v_{3}^{4} v_{9}-a^{3} c_{1} d a_{2} v_{1} v_{3}^{3} v_{9}-a^{3} c_{1} d a_{3} v_{2} v_{3}^{3} v_{9}-3 a^{2} b c c_{1} a_{2} v_{1} v_{3}^{3} v_{9}-3 a^{2} b c c_{1} a_{3} v_{2} v_{3}^{3} v_{9}-4 a^{3} b c_{1} a_{2} v_{3}^{3} v_{9}+4 a^{3} b c_{1} b_{3} v_{3}^{3} v_{9}-a^{3} c_{1} d a_{1} v_{3}^{3} v_{9}-6 a^{2} b^{2} c_{1}^{2} a_{3} v_{3}^{2} v_{8}-3 a^{2} b c c_{1} a_{1} v_{3}^{3} v_{9}-3 a^{2} b c_{1} d a_{2} v_{1} v_{3}^{2} v_{9}-3 a^{2} b c_{1} d a_{3} v_{2} v_{3}^{2} v_{9}-3 a \,b^{2} c c_{1} a_{2} v_{1} v_{3}^{2} v_{9}-3 a \,b^{2} c c_{1} a_{3} v_{2} v_{3}^{2} v_{9}+v_{3}^{4} a^{4} b_{2}-6 a^{2} b^{2} c_{1} a_{2} v_{3}^{2} v_{9}+6 a^{2} b^{2} c_{1} b_{3} v_{3}^{2} v_{9}-3 a^{2} b c_{1} d a_{1} v_{3}^{2} v_{9}-4 a \,b^{3} c_{1}^{2} a_{3} v_{3} v_{8}-3 a \,b^{2} c c_{1} a_{1} v_{3}^{2} v_{9}-3 a \,b^{2} c_{1} d a_{2} v_{1} v_{3} v_{9}-3 a \,b^{2} c_{1} d a_{3} v_{2} v_{3} v_{9}-b^{3} c c_{1} a_{2} v_{1} v_{3} v_{9}-b^{3} c c_{1} a_{3} v_{2} v_{3} v_{9}+4 v_{3}^{3} a^{3} b b_{2}-4 a \,b^{3} c_{1} a_{2} v_{3} v_{9}+4 a \,b^{3} c_{1} b_{3} v_{3} v_{9}-3 a \,b^{2} c_{1} d a_{1} v_{3} v_{9}-b^{4} c_{1}^{2} a_{3} v_{8}-b^{3} c c_{1} a_{1} v_{3} v_{9}-b^{3} c_{1} d a_{2} v_{1} v_{9}-b^{3} c_{1} d a_{3} v_{2} v_{9}+6 v_{3}^{2} a^{2} b^{2} b_{2}-b^{4} c_{1} a_{2} v_{9}+b^{4} c_{1} b_{3} v_{9}-b^{3} c_{1} d a_{1} v_{9}+4 v_{3} a \,b^{3} b_{2}+b^{4} b_{2} = 0 \end{equation} Collecting the above on the terms \(v_i\) introduced, and these are \[ \{v_{1}, v_{2}, v_{3}, v_{4}, v_{5}, v_{6}, v_{7}, v_{8}, v_{9}\} \] Equation (7E) now becomes \begin{equation} \tag{8E} -6 a^{2} b^{2} c_{1}^{2} a_{3} v_{3}^{2} v_{8}-4 a \,b^{3} c_{1}^{2} a_{3} v_{3} v_{8}-b^{3} c_{1} d a_{2} v_{1} v_{9}-b^{3} c_{1} d a_{3} v_{2} v_{9}+\left (-b^{4} c_{1} a_{2}+b^{4} c_{1} b_{3}-b^{3} c_{1} d a_{1}\right ) v_{9}-4 a^{3} b \,c_{1}^{2} a_{3} v_{3}^{3} v_{8}-a^{4} c_{1}^{2} a_{3} v_{3}^{4} v_{8}-a^{3} c c_{1} a_{2} v_{1} v_{3}^{4} v_{9}-a^{3} c c_{1} a_{3} v_{2} v_{3}^{4} v_{9}+v_{3}^{4} a^{4} b_{2}+b^{4} b_{2}-b^{4} c_{1}^{2} a_{3} v_{8}+\left (-3 a^{2} b c_{1} d a_{2}-3 a \,b^{2} c c_{1} a_{2}\right ) v_{1} v_{3}^{2} v_{9}+\left (-3 a \,b^{2} c_{1} d a_{2}-b^{3} c c_{1} a_{2}\right ) v_{1} v_{3} v_{9}+\left (-a^{3} c_{1} d a_{3}-3 a^{2} b c c_{1} a_{3}\right ) v_{2} v_{3}^{3} v_{9}+\left (-3 a^{2} b c_{1} d a_{3}-3 a \,b^{2} c c_{1} a_{3}\right ) v_{2} v_{3}^{2} v_{9}+\left (-3 a \,b^{2} c_{1} d a_{3}-b^{3} c c_{1} a_{3}\right ) v_{2} v_{3} v_{9}+\left (-a^{3} c_{1} d a_{2}-3 a^{2} b c c_{1} a_{2}\right ) v_{1} v_{3}^{3} v_{9}+\left (-a^{4} c_{1} a_{2}+a^{4} c_{1} b_{3}-a^{3} c c_{1} a_{1}\right ) v_{3}^{4} v_{9}+\left (-4 a^{3} b c_{1} a_{2}+4 a^{3} b c_{1} b_{3}-a^{3} c_{1} d a_{1}-3 a^{2} b c c_{1} a_{1}\right ) v_{3}^{3} v_{9}+\left (-6 a^{2} b^{2} c_{1} a_{2}+6 a^{2} b^{2} c_{1} b_{3}-3 a^{2} b c_{1} d a_{1}-3 a \,b^{2} c c_{1} a_{1}\right ) v_{3}^{2} v_{9}+\left (-4 a \,b^{3} c_{1} a_{2}+4 a \,b^{3} c_{1} b_{3}-3 a \,b^{2} c_{1} d a_{1}-b^{3} c c_{1} a_{1}\right ) v_{3} v_{9}+4 v_{3} a \,b^{3} b_{2}+4 v_{3}^{3} a^{3} b b_{2}+6 v_{3}^{2} a^{2} b^{2} b_{2} = 0 \end{equation} Setting each coefficients in (8E) to zero gives the following equations to solve \begin {align*} a^{4} b_{2}&=0\\ b^{4} b_{2}&=0\\ -c_{1}^{2} a^{4} a_{3}&=0\\ -c_{1}^{2} b^{4} a_{3}&=0\\ 4 a \,b^{3} b_{2}&=0\\ 6 a^{2} b^{2} b_{2}&=0\\ 4 a^{3} b b_{2}&=0\\ -c_{1} a^{3} c a_{2}&=0\\ -c_{1} a^{3} c a_{3}&=0\\ -c_{1} b^{3} d a_{2}&=0\\ -c_{1} b^{3} d a_{3}&=0\\ -4 c_{1}^{2} a \,b^{3} a_{3}&=0\\ -6 c_{1}^{2} a^{2} b^{2} a_{3}&=0\\ -4 c_{1}^{2} a^{3} b a_{3}&=0\\ -3 a \,b^{2} c_{1} d a_{2}-b^{3} c c_{1} a_{2}&=0\\ -3 a^{2} b c_{1} d a_{2}-3 a \,b^{2} c c_{1} a_{2}&=0\\ -a^{3} c_{1} d a_{2}-3 a^{2} b c c_{1} a_{2}&=0\\ -3 a \,b^{2} c_{1} d a_{3}-b^{3} c c_{1} a_{3}&=0\\ -3 a^{2} b c_{1} d a_{3}-3 a \,b^{2} c c_{1} a_{3}&=0\\ -a^{3} c_{1} d a_{3}-3 a^{2} b c c_{1} a_{3}&=0\\ -a^{4} c_{1} a_{2}+a^{4} c_{1} b_{3}-a^{3} c c_{1} a_{1}&=0\\ -b^{4} c_{1} a_{2}+b^{4} c_{1} b_{3}-b^{3} c_{1} d a_{1}&=0\\ -4 a \,b^{3} c_{1} a_{2}+4 a \,b^{3} c_{1} b_{3}-3 a \,b^{2} c_{1} d a_{1}-b^{3} c c_{1} a_{1}&=0\\ -6 a^{2} b^{2} c_{1} a_{2}+6 a^{2} b^{2} c_{1} b_{3}-3 a^{2} b c_{1} d a_{1}-3 a \,b^{2} c c_{1} a_{1}&=0\\ -4 a^{3} b c_{1} a_{2}+4 a^{3} b c_{1} b_{3}-a^{3} c_{1} d a_{1}-3 a^{2} b 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 \(\left ( x,\xi \right ) \to \left ( R,S \right )\) 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 \xi }{\eta } = dS \tag {1} \end {align*}
The above comes from the requirements that \(\left ( \xi \frac {\partial }{\partial x} + \eta \frac {\partial }{\partial \xi }\right ) S(x,\xi ) = 1\). 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*}
Which results in \begin {align*} S&= \xi \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,\xi ) S_{\xi } }{ R_{x} + \omega (x,\xi ) R_{\xi } }\tag {2} \end {align*}
Where in the above \(R_{x},R_{\xi },S_{x},S_{\xi }\) are all partial derivatives and \(\omega (x,\xi )\) is the right hand side of the original ode given by \begin {align*} \omega (x,\xi ) &= c_{1} {\mathrm e}^{\int \frac {x^{4 n} a^{3} c +d \,x^{3 n} a^{3}+3 b c \,x^{3 n} a^{2}+3 x^{2 n} a^{2} b d +3 x^{2 n} a \,b^{2} c +3 x^{n} a \,b^{2} d +x^{n} b^{3} c +b^{3} d}{x^{4 n} a^{4}+4 b \,x^{3 n} a^{3}+6 x^{2 n} a^{2} b^{2}+4 x^{n} a \,b^{3}+b^{4}}d x} \end {align*}
Evaluating all the partial derivatives gives \begin {align*} R_{x} &= 1\\ R_{\xi } &= 0\\ S_{x} &= 0\\ S_{\xi } &= 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 {\left (3 d b \,a^{2}+3 c \,b^{2} a \right ) x^{2 n}+\left (d \,a^{3}+3 c b \,a^{2}\right ) x^{3 n}+x^{4 n} a^{3} c +\left (3 a \,b^{2} d +c \,b^{3}\right ) x^{n}+b^{3} d}{x^{4 n} a^{4}+4 b \,x^{3 n} a^{3}+6 x^{2 n} a^{2} b^{2}+4 x^{n} a \,b^{3}+b^{4}}d x}\tag {2A} \end {align*}
We now need to express the RHS as function of \(R\) only. This is done by solving for \(x,\xi \) 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 {\left (3 d b \,a^{2}+3 c \,b^{2} a \right ) R^{2 n}+\left (d \,a^{3}+3 c b \,a^{2}\right ) R^{3 n}+R^{4 n} a^{3} c +\left (3 a \,b^{2} d +c \,b^{3}\right ) R^{n}+b^{3} d}{R^{4 n} a^{4}+4 b \,R^{3 n} a^{3}+6 R^{2 n} a^{2} b^{2}+4 R^{n} a \,b^{3}+b^{4}}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}^{\int \frac {3 R^{2 n} a^{2} b d +3 R^{2 n} a \,b^{2} c +R^{3 n} a^{3} d +3 R^{3 n} a^{2} b c +R^{4 n} a^{3} c +3 R^{n} a \,b^{2} d +R^{n} b^{3} c +b^{3} d}{R^{4 n} a^{4}+4 b \,R^{3 n} a^{3}+6 R^{2 n} a^{2} b^{2}+4 R^{n} a \,b^{3}+b^{4}}d R}d R +c_{2}\tag {4} \end {align*}
To complete the solution, we just need to transform (4) back to \(x,\xi \) coordinates. This results in \begin {align*} \xi \left (x \right ) = \int c_{1} {\mathrm e}^{\int \frac {x^{4 n} a^{3} c +d \,x^{3 n} a^{3}+3 b c \,x^{3 n} a^{2}+3 x^{2 n} a^{2} b d +3 x^{2 n} a \,b^{2} c +3 x^{n} a \,b^{2} d +x^{n} b^{3} c +b^{3} d}{x^{4 n} a^{4}+4 b \,x^{3 n} a^{3}+6 x^{2 n} a^{2} b^{2}+4 x^{n} a \,b^{3}+b^{4}}d x}d x +c_{2} \end {align*}
Which simplifies to \begin {align*} \xi \left (x \right )-c_{1} \left (\int {\mathrm e}^{\int \frac {\left (3 d b \,a^{2}+3 c \,b^{2} a \right ) x^{2 n}+\left (d \,a^{3}+3 c b \,a^{2}\right ) x^{3 n}+x^{4 n} a^{3} c +\left (3 a \,b^{2} d +c \,b^{3}\right ) x^{n}+b^{3} d}{x^{4 n} a^{4}+4 b \,x^{3 n} a^{3}+6 x^{2 n} a^{2} b^{2}+4 x^{n} a \,b^{3}+b^{4}}d x}d x \right )-c_{2} = 0 \end {align*}
Which gives \begin {align*} \xi \left (x \right ) = c_{1} \left (\int {\mathrm e}^{\int \frac {\left (3 d b \,a^{2}+3 c \,b^{2} a \right ) x^{2 n}+\left (d \,a^{3}+3 c b \,a^{2}\right ) x^{3 n}+x^{4 n} a^{3} c +\left (3 a \,b^{2} d +c \,b^{3}\right ) x^{n}+b^{3} d}{x^{4 n} a^{4}+4 b \,x^{3 n} a^{3}+6 x^{2 n} a^{2} b^{2}+4 x^{n} a \,b^{3}+b^{4}}d x}d x \right )+c_{2} \end {align*}
The original ode (2) now reduces to first order ode \begin {align*} \xi \left (x \right ) y^{\prime }-y \xi ^{\prime }\left (x \right )+\xi \left (x \right ) p \left (x \right ) y&=\int \xi \left (x \right ) r \left (x \right )d x\\ y^{\prime }+y \left (p \left (x \right )-\frac {\xi ^{\prime }\left (x \right )}{\xi \left (x \right )}\right )&=\frac {\int \xi \left (x \right ) r \left (x \right )d x}{\xi \left (x \right )}\\ y^{\prime }+y \left (\frac {x^{2 n} a c +d \,x^{n} a +x^{n} b c +b d}{x^{2 n} a^{2}+2 x^{n} a b +b^{2}}-\frac {c_{1} {\mathrm e}^{\int \frac {\left (3 d b \,a^{2}+3 c \,b^{2} a \right ) x^{2 n}+\left (d \,a^{3}+3 c b \,a^{2}\right ) x^{3 n}+x^{4 n} a^{3} c +\left (3 a \,b^{2} d +c \,b^{3}\right ) x^{n}+b^{3} d}{x^{4 n} a^{4}+4 b \,x^{3 n} a^{3}+6 x^{2 n} a^{2} b^{2}+4 x^{n} a \,b^{3}+b^{4}}d x}}{c_{1} \left (\int {\mathrm e}^{\int \frac {\left (3 d b \,a^{2}+3 c \,b^{2} a \right ) x^{2 n}+\left (d \,a^{3}+3 c b \,a^{2}\right ) x^{3 n}+x^{4 n} a^{3} c +\left (3 a \,b^{2} d +c \,b^{3}\right ) x^{n}+b^{3} d}{x^{4 n} a^{4}+4 b \,x^{3 n} a^{3}+6 x^{2 n} a^{2} b^{2}+4 x^{n} a \,b^{3}+b^{4}}d x}d x \right )+c_{2}}\right )&=0 \end {align*}
Which is now a first order ode. This is now solved for \(y\).
Entering Linear first order ODE solver. In canonical form a linear first order is \begin {align*} y^{\prime } + p(x)y &= q(x) \end {align*}
Where here \begin {align*} p(x) &=\text {Expression too large to display}\\ q(x) &=0 \end {align*}
Hence the ode is \begin {align*} \text {Expression too large to display} \end {align*}
The integrating factor \(\mu \) is \[ \mu = \text {Expression too large to display} \] The ode becomes \begin {align*} \frac {\mathop {\mathrm {d}}}{ \mathop {\mathrm {d}x}} \mu y &= 0 \\ \frac {\mathop {\mathrm {d}}}{ \mathop {\mathrm {d}x}} \left (\text {Expression too large to display}\right ) &= 0 \end {align*}
Integrating gives \begin {align*} \text {Expression too large to display} &= c_{3} \end {align*}
Dividing both sides by the integrating factor \(\mu =\text {Expression too large to display}\) results in \begin {align*} y &= c_{3} {\mathrm e}^{-\left (\int \frac {\left (\left (a d +b c \right ) x^{n}+x^{2 n} a c +b d \right ) c_{1} \left (\int {\mathrm e}^{\int \frac {c \,x^{n}+d}{x^{n} a +b}d x}d x \right )-c_{1} \left (x^{2 n} a^{2}+2 x^{n} a b +b^{2}\right ) {\mathrm e}^{\int \frac {c \,x^{n}+d}{x^{n} a +b}d x}+\left (\left (a d +b c \right ) x^{n}+x^{2 n} a c +b d \right ) c_{2}}{\left (c_{1} \left (\int {\mathrm e}^{\int \frac {c \,x^{n}+d}{x^{n} a +b}d x}d x \right )+c_{2} \right ) \left (x^{2 n} a^{2}+2 x^{n} a b +b^{2}\right )}d x \right )} \end {align*}
Hence, the solution found using Lagrange adjoint equation method is \[ y = c_{3} {\mathrm e}^{-\left (\int \frac {\left (\left (a d +b c \right ) x^{n}+x^{2 n} a c +b d \right ) c_{1} \left (\int {\mathrm e}^{\int \frac {c \,x^{n}+d}{x^{n} a +b}d x}d x \right )-c_{1} \left (x^{2 n} a^{2}+2 x^{n} a b +b^{2}\right ) {\mathrm e}^{\int \frac {c \,x^{n}+d}{x^{n} a +b}d x}+\left (\left (a d +b c \right ) x^{n}+x^{2 n} a c +b d \right ) c_{2}}{\left (c_{1} \left (\int {\mathrm e}^{\int \frac {c \,x^{n}+d}{x^{n} a +b}d x}d x \right )+c_{2} \right ) \left (x^{2 n} a^{2}+2 x^{n} a b +b^{2}\right )}d x \right )} \]
The solution(s) found are the following \begin{align*} \tag{1} y &= c_{3} {\mathrm e}^{-\left (\int \frac {\left (\left (a d +b c \right ) x^{n}+x^{2 n} a c +b d \right ) c_{1} \left (\int {\mathrm e}^{\int \frac {c \,x^{n}+d}{x^{n} a +b}d x}d x \right )-c_{1} \left (x^{2 n} a^{2}+2 x^{n} a b +b^{2}\right ) {\mathrm e}^{\int \frac {c \,x^{n}+d}{x^{n} a +b}d x}+\left (\left (a d +b c \right ) x^{n}+x^{2 n} a c +b d \right ) c_{2}}{\left (c_{1} \left (\int {\mathrm e}^{\int \frac {c \,x^{n}+d}{x^{n} a +b}d x}d x \right )+c_{2} \right ) \left (x^{2 n} a^{2}+2 x^{n} a b +b^{2}\right )}d x \right )} \\ \end{align*}
Verification of solutions
\[ y = c_{3} {\mathrm e}^{-\left (\int \frac {\left (\left (a d +b c \right ) x^{n}+x^{2 n} a c +b d \right ) c_{1} \left (\int {\mathrm e}^{\int \frac {c \,x^{n}+d}{x^{n} a +b}d x}d x \right )-c_{1} \left (x^{2 n} a^{2}+2 x^{n} a b +b^{2}\right ) {\mathrm e}^{\int \frac {c \,x^{n}+d}{x^{n} a +b}d x}+\left (\left (a d +b c \right ) x^{n}+x^{2 n} a c +b d \right ) c_{2}}{\left (c_{1} \left (\int {\mathrm e}^{\int \frac {c \,x^{n}+d}{x^{n} a +b}d x}d x \right )+c_{2} \right ) \left (x^{2 n} a^{2}+2 x^{n} a b +b^{2}\right )}d x \right )} \] 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) trying to convert to a linear ODE with constant coefficients trying 2nd order, integrating factor of the form mu(x,y) 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 to convert to an ODE of Bessel type -> trying reduction of order to Riccati trying Riccati sub-methods: trying Riccati_symmetries -> trying a symmetry pattern of the form [F(x)*G(y), 0] -> trying a symmetry pattern of the form [0, F(x)*G(y)] -> trying a symmetry pattern of the form [F(x),G(x)*y+H(x)] --- Trying Lie symmetry methods, 2nd order --- `, `-> Computing symmetries using: way = 3`[0, y]
✓ Solution by Maple
Time used: 0.406 (sec). Leaf size: 53
dsolve((a*x^n+b)^2*diff(y(x),x$2)+(a*x^n+b)*(c*x^n+d)*diff(y(x),x)+n*(b*c-a*d)*x^(n-1)*y(x)=0,y(x), singsol=all)
\[ y \left (x \right ) = {\mathrm e}^{-\left (\int \frac {c \,x^{n}+d}{a \,x^{n}+b}d x \right )} \left (c_{1} +\left (\int {\mathrm e}^{\int \frac {c \,x^{n}+d}{a \,x^{n}+b}d x}d x \right ) c_{2} \right ) \]
✓ Solution by Mathematica
Time used: 0.923 (sec). Leaf size: 106
DSolve[(a*x^n+b)^2*y''[x]+(a*x^n+b)*(c*x^n+d)*y'[x]+n*(b*c-a*d)*x^(n-1)*y[x]==0,y[x],x,IncludeSingularSolutions -> True]
\[ y(x)\to \exp \left (-\frac {x \left ((a d-b c) \operatorname {Hypergeometric2F1}\left (1,\frac {1}{n},1+\frac {1}{n},-\frac {a x^n}{b}\right )+b c\right )}{a b}\right ) \left (\int _1^x\exp \left (\frac {\left (b c+(a d-b c) \operatorname {Hypergeometric2F1}\left (1,\frac {1}{n},1+\frac {1}{n},-\frac {a K[1]^n}{b}\right )\right ) K[1]}{a b}\right ) c_1dK[1]+c_2\right ) \]