Internal problem ID [11051]
Internal file name [OUTPUT/10308_Wednesday_January_24_2024_10_07_01_PM_84430845/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 2.1.2-7 Equation of form
\((a_4 x^4+a_3 x^3+a_2 x^2 x+a_1 x+a_0) y''+f(x)y'+g(x)y=0\)
Problem number: 227.
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]]
Unable to solve or complete the solution.
\[ \boxed {a \left (x^{2}-1\right )^{2} y^{\prime \prime }+b x \left (x^{2}-1\right ) y^{\prime }+\left (c \,x^{2}+d x +e \right ) y=0} \]
\[ \begin {array}{lll} & {} & \textrm {Let's solve}\hspace {3pt} \\ {} & {} & \left (\frac {d}{d x}y^{\prime }\right ) a \left (x -1\right )^{2} \left (x +1\right )^{2}+b \left (x^{3}-x \right ) y^{\prime }+\left (c \,x^{2}+d x +e \right ) y=0 \\ \bullet & {} & \textrm {Highest derivative means the order of the ODE is}\hspace {3pt} 2 \\ {} & {} & \frac {d}{d x}y^{\prime } \\ \bullet & {} & \textrm {Isolate 2nd derivative}\hspace {3pt} \\ {} & {} & \frac {d}{d x}y^{\prime }=-\frac {\left (c \,x^{2}+d x +e \right ) y}{a \left (x -1\right )^{2} \left (x +1\right )^{2}}-\frac {b x y^{\prime }}{\left (x +1\right ) \left (x -1\right ) a} \\ \bullet & {} & \textrm {Group terms with}\hspace {3pt} y\hspace {3pt}\textrm {on the lhs of the ODE and the rest on the rhs of the ODE; ODE is linear}\hspace {3pt} \\ {} & {} & \frac {d}{d x}y^{\prime }+\frac {b x y^{\prime }}{\left (x +1\right ) \left (x -1\right ) a}+\frac {\left (c \,x^{2}+d x +e \right ) y}{a \left (x -1\right )^{2} \left (x +1\right )^{2}}=0 \\ \square & {} & \textrm {Check to see if}\hspace {3pt} x_{0}\hspace {3pt}\textrm {is a regular singular point}\hspace {3pt} \\ {} & \circ & \textrm {Define functions}\hspace {3pt} \\ {} & {} & \left [P_{2}\left (x \right )=\frac {x b}{\left (x +1\right ) \left (x -1\right ) a}, P_{3}\left (x \right )=\frac {c \,x^{2}+d x +e}{a \left (x -1\right )^{2} \left (x +1\right )^{2}}\right ] \\ {} & \circ & \left (x +1\right )\cdot P_{2}\left (x \right )\textrm {is analytic at}\hspace {3pt} x =-1 \\ {} & {} & \left (\left (x +1\right )\cdot P_{2}\left (x \right )\right )\bigg | {\mstack {}{_{x \hiderel {=}-1}}}=\frac {b}{2 a} \\ {} & \circ & \left (x +1\right )^{2}\cdot P_{3}\left (x \right )\textrm {is analytic at}\hspace {3pt} x =-1 \\ {} & {} & \left (\left (x +1\right )^{2}\cdot P_{3}\left (x \right )\right )\bigg | {\mstack {}{_{x \hiderel {=}-1}}}=\frac {e -d +c}{4 a} \\ {} & \circ & x =-1\textrm {is a regular singular point}\hspace {3pt} \\ & {} & \textrm {Check to see if}\hspace {3pt} x_{0}\hspace {3pt}\textrm {is a regular singular point}\hspace {3pt} \\ {} & {} & x_{0}=-1 \\ \bullet & {} & \textrm {Multiply by denominators}\hspace {3pt} \\ {} & {} & \left (\frac {d}{d x}y^{\prime }\right ) a \left (x -1\right )^{2} \left (x +1\right )^{2}+b x y^{\prime } \left (x -1\right ) \left (x +1\right )+\left (c \,x^{2}+d x +e \right ) y=0 \\ \bullet & {} & \textrm {Change variables using}\hspace {3pt} x =u -1\hspace {3pt}\textrm {so that the regular singular point is at}\hspace {3pt} u =0 \\ {} & {} & \left (a \,u^{4}-4 a \,u^{3}+4 a \,u^{2}\right ) \left (\frac {d}{d u}\frac {d}{d u}y \left (u \right )\right )+\left (b \,u^{3}-3 b \,u^{2}+2 b u \right ) \left (\frac {d}{d u}y \left (u \right )\right )+\left (c \,u^{2}-2 c u +d u +c -d +e \right ) y \left (u \right )=0 \\ \bullet & {} & \textrm {Assume series solution for}\hspace {3pt} y \left (u \right ) \\ {} & {} & y \left (u \right )=\moverset {\infty }{\munderset {k =0}{\sum }}a_{k} u^{k +r} \\ \square & {} & \textrm {Rewrite ODE with series expansions}\hspace {3pt} \\ {} & \circ & \textrm {Convert}\hspace {3pt} u^{m}\cdot y \left (u \right )\hspace {3pt}\textrm {to series expansion for}\hspace {3pt} m =0..2 \\ {} & {} & u^{m}\cdot y \left (u \right )=\moverset {\infty }{\munderset {k =0}{\sum }}a_{k} u^{k +r +m} \\ {} & \circ & \textrm {Shift index using}\hspace {3pt} k \mathrm {->}k -m \\ {} & {} & u^{m}\cdot y \left (u \right )=\moverset {\infty }{\munderset {k =m}{\sum }}a_{k -m} u^{k +r} \\ {} & \circ & \textrm {Convert}\hspace {3pt} u^{m}\cdot \left (\frac {d}{d u}y \left (u \right )\right )\hspace {3pt}\textrm {to series expansion for}\hspace {3pt} m =1..3 \\ {} & {} & u^{m}\cdot \left (\frac {d}{d u}y \left (u \right )\right )=\moverset {\infty }{\munderset {k =0}{\sum }}a_{k} \left (k +r \right ) u^{k +r -1+m} \\ {} & \circ & \textrm {Shift index using}\hspace {3pt} k \mathrm {->}k +1-m \\ {} & {} & u^{m}\cdot \left (\frac {d}{d u}y \left (u \right )\right )=\moverset {\infty }{\munderset {k =-1+m}{\sum }}a_{k +1-m} \left (k +1-m +r \right ) u^{k +r} \\ {} & \circ & \textrm {Convert}\hspace {3pt} u^{m}\cdot \left (\frac {d}{d u}\frac {d}{d u}y \left (u \right )\right )\hspace {3pt}\textrm {to series expansion for}\hspace {3pt} m =2..4 \\ {} & {} & u^{m}\cdot \left (\frac {d}{d u}\frac {d}{d u}y \left (u \right )\right )=\moverset {\infty }{\munderset {k =0}{\sum }}a_{k} \left (k +r \right ) \left (k +r -1\right ) u^{k +r -2+m} \\ {} & \circ & \textrm {Shift index using}\hspace {3pt} k \mathrm {->}k +2-m \\ {} & {} & u^{m}\cdot \left (\frac {d}{d u}\frac {d}{d u}y \left (u \right )\right )=\moverset {\infty }{\munderset {k =-2+m}{\sum }}a_{k +2-m} \left (k +2-m +r \right ) \left (k +1-m +r \right ) u^{k +r} \\ & {} & \textrm {Rewrite ODE with series expansions}\hspace {3pt} \\ {} & {} & a_{0} \left (4 a \,r^{2}-4 a r +2 b r +c -d +e \right ) u^{r}+\left (\left (4 a \,r^{2}+4 a r +2 b r +2 b +c -d +e \right ) a_{1}-a_{0} \left (4 a \,r^{2}-4 a r +3 b r +2 c -d \right )\right ) u^{1+r}+\left (\moverset {\infty }{\munderset {k =2}{\sum }}\left (a_{k} \left (4 a \,k^{2}+8 a k r +4 a \,r^{2}-4 a k -4 a r +2 b k +2 b r +c -d +e \right )-a_{k -1} \left (4 a \left (k -1\right )^{2}+8 a \left (k -1\right ) r +4 a \,r^{2}-4 a \left (k -1\right )-4 a r +3 b \left (k -1\right )+3 b r +2 c -d \right )+a_{k -2} \left (a \left (k -2\right )^{2}+2 a \left (k -2\right ) r +a \,r^{2}-a \left (k -2\right )-a r +b \left (k -2\right )+b r +c \right )\right ) u^{k +r}\right )=0 \\ \bullet & {} & a_{0}\textrm {cannot be 0 by assumption, giving the indicial equation}\hspace {3pt} \\ {} & {} & 4 a \,r^{2}-4 a r +2 b r +c -d +e =0 \\ \bullet & {} & \textrm {Values of r that satisfy the indicial equation}\hspace {3pt} \\ {} & {} & r \in \left \{-\frac {-2 a +b +\sqrt {4 a^{2}-4 b a -4 a c +4 d a -4 a e +b^{2}}}{4 a}, \frac {2 a -b +\sqrt {4 a^{2}-4 b a -4 a c +4 d a -4 a e +b^{2}}}{4 a}\right \} \\ \bullet & {} & \textrm {Each term must be 0}\hspace {3pt} \\ {} & {} & \left (4 a \,r^{2}+4 a r +2 b r +2 b +c -d +e \right ) a_{1}-a_{0} \left (4 a \,r^{2}-4 a r +3 b r +2 c -d \right )=0 \\ \bullet & {} & \textrm {Solve for the dependent coefficient(s)}\hspace {3pt} \\ {} & {} & a_{1}=\frac {a_{0} \left (4 a \,r^{2}-4 a r +3 b r +2 c -d \right )}{4 a \,r^{2}+4 a r +2 b r +2 b +c -d +e} \\ \bullet & {} & \textrm {Each term in the series must be 0, giving the recursion relation}\hspace {3pt} \\ {} & {} & \left (\left (-4 a_{k -1}+4 a_{k}+a_{k -2}\right ) k^{2}+\left (\left (-8 a_{k -1}+8 a_{k}+2 a_{k -2}\right ) r +12 a_{k -1}-4 a_{k}-5 a_{k -2}\right ) k +\left (-4 a_{k -1}+4 a_{k}+a_{k -2}\right ) r^{2}+\left (12 a_{k -1}-4 a_{k}-5 a_{k -2}\right ) r -8 a_{k -1}+6 a_{k -2}\right ) a -3 \left (a_{k -1}-\frac {2 a_{k}}{3}-\frac {a_{k -2}}{3}\right ) b k -3 \left (a_{k -1}-\frac {2 a_{k}}{3}-\frac {a_{k -2}}{3}\right ) b r +\left (3 b -2 c +d \right ) a_{k -1}+\left (-2 b +c \right ) a_{k -2}+a_{k} \left (e -d +c \right )=0 \\ \bullet & {} & \textrm {Shift index using}\hspace {3pt} k \mathrm {->}k +2 \\ {} & {} & \left (\left (-4 a_{k +1}+4 a_{k +2}+a_{k}\right ) \left (k +2\right )^{2}+\left (\left (-8 a_{k +1}+8 a_{k +2}+2 a_{k}\right ) r +12 a_{k +1}-4 a_{k +2}-5 a_{k}\right ) \left (k +2\right )+\left (-4 a_{k +1}+4 a_{k +2}+a_{k}\right ) r^{2}+\left (12 a_{k +1}-4 a_{k +2}-5 a_{k}\right ) r -8 a_{k +1}+6 a_{k}\right ) a -3 \left (a_{k +1}-\frac {2 a_{k +2}}{3}-\frac {a_{k}}{3}\right ) b \left (k +2\right )-3 \left (a_{k +1}-\frac {2 a_{k +2}}{3}-\frac {a_{k}}{3}\right ) b r +\left (3 b -2 c +d \right ) a_{k +1}+\left (-2 b +c \right ) a_{k}+a_{k +2} \left (e -d +c \right )=0 \\ \bullet & {} & \textrm {Recursion relation that defines series solution to ODE}\hspace {3pt} \\ {} & {} & a_{k +2}=-\frac {a \,k^{2} a_{k}-4 a \,k^{2} a_{k +1}+2 a k r a_{k}-8 a k r a_{k +1}+a \,r^{2} a_{k}-4 a \,r^{2} a_{k +1}-a k a_{k}-4 a k a_{k +1}-a r a_{k}-4 a r a_{k +1}+a_{k} b k -3 b k a_{k +1}+b r a_{k}-3 b r a_{k +1}-3 b a_{k +1}+a_{k} c -2 c a_{k +1}+d a_{k +1}}{4 a \,k^{2}+8 a k r +4 a \,r^{2}+12 a k +12 a r +2 b k +2 b r +8 a +4 b +c -d +e} \\ \bullet & {} & \textrm {Recursion relation for}\hspace {3pt} r =-\frac {-2 a +b +\sqrt {4 a^{2}-4 b a -4 a c +4 d a -4 a e +b^{2}}}{4 a} \\ {} & {} & a_{k +2}=-\frac {a \,k^{2} a_{k}-4 a \,k^{2} a_{k +1}-\frac {k \left (-2 a +b +\sqrt {4 a^{2}-4 b a -4 a c +4 d a -4 a e +b^{2}}\right ) a_{k}}{2}+2 k \left (-2 a +b +\sqrt {4 a^{2}-4 b a -4 a c +4 d a -4 a e +b^{2}}\right ) a_{k +1}+\frac {\left (-2 a +b +\sqrt {4 a^{2}-4 b a -4 a c +4 d a -4 a e +b^{2}}\right )^{2} a_{k}}{16 a}-\frac {\left (-2 a +b +\sqrt {4 a^{2}-4 b a -4 a c +4 d a -4 a e +b^{2}}\right )^{2} a_{k +1}}{4 a}-a k a_{k}-4 a k a_{k +1}+\frac {\left (-2 a +b +\sqrt {4 a^{2}-4 b a -4 a c +4 d a -4 a e +b^{2}}\right ) a_{k}}{4}+\left (-2 a +b +\sqrt {4 a^{2}-4 b a -4 a c +4 d a -4 a e +b^{2}}\right ) a_{k +1}+a_{k} b k -3 b k a_{k +1}-\frac {a_{k} b \left (-2 a +b +\sqrt {4 a^{2}-4 b a -4 a c +4 d a -4 a e +b^{2}}\right )}{4 a}+\frac {3 b \left (-2 a +b +\sqrt {4 a^{2}-4 b a -4 a c +4 d a -4 a e +b^{2}}\right ) a_{k +1}}{4 a}-3 b a_{k +1}+a_{k} c -2 c a_{k +1}+d a_{k +1}}{4 a \,k^{2}-2 k \left (-2 a +b +\sqrt {4 a^{2}-4 b a -4 a c +4 d a -4 a e +b^{2}}\right )+\frac {\left (-2 a +b +\sqrt {4 a^{2}-4 b a -4 a c +4 d a -4 a e +b^{2}}\right )^{2}}{4 a}+12 a k +14 a +b -3 \sqrt {4 a^{2}-4 b a -4 a c +4 d a -4 a e +b^{2}}+2 b k -\frac {b \left (-2 a +b +\sqrt {4 a^{2}-4 b a -4 a c +4 d a -4 a e +b^{2}}\right )}{2 a}+c -d +e} \\ \bullet & {} & \textrm {Solution for}\hspace {3pt} r =-\frac {-2 a +b +\sqrt {4 a^{2}-4 b a -4 a c +4 d a -4 a e +b^{2}}}{4 a} \\ {} & {} & \left [y \left (u \right )=\moverset {\infty }{\munderset {k =0}{\sum }}a_{k} u^{k -\frac {-2 a +b +\sqrt {4 a^{2}-4 b a -4 a c +4 d a -4 a e +b^{2}}}{4 a}}, a_{k +2}=-\frac {a \,k^{2} a_{k}-4 a \,k^{2} a_{k +1}-\frac {k \left (-2 a +b +\sqrt {4 a^{2}-4 b a -4 a c +4 d a -4 a e +b^{2}}\right ) a_{k}}{2}+2 k \left (-2 a +b +\sqrt {4 a^{2}-4 b a -4 a c +4 d a -4 a e +b^{2}}\right ) a_{k +1}+\frac {\left (-2 a +b +\sqrt {4 a^{2}-4 b a -4 a c +4 d a -4 a e +b^{2}}\right )^{2} a_{k}}{16 a}-\frac {\left (-2 a +b +\sqrt {4 a^{2}-4 b a -4 a c +4 d a -4 a e +b^{2}}\right )^{2} a_{k +1}}{4 a}-a k a_{k}-4 a k a_{k +1}+\frac {\left (-2 a +b +\sqrt {4 a^{2}-4 b a -4 a c +4 d a -4 a e +b^{2}}\right ) a_{k}}{4}+\left (-2 a +b +\sqrt {4 a^{2}-4 b a -4 a c +4 d a -4 a e +b^{2}}\right ) a_{k +1}+a_{k} b k -3 b k a_{k +1}-\frac {a_{k} b \left (-2 a +b +\sqrt {4 a^{2}-4 b a -4 a c +4 d a -4 a e +b^{2}}\right )}{4 a}+\frac {3 b \left (-2 a +b +\sqrt {4 a^{2}-4 b a -4 a c +4 d a -4 a e +b^{2}}\right ) a_{k +1}}{4 a}-3 b a_{k +1}+a_{k} c -2 c a_{k +1}+d a_{k +1}}{4 a \,k^{2}-2 k \left (-2 a +b +\sqrt {4 a^{2}-4 b a -4 a c +4 d a -4 a e +b^{2}}\right )+\frac {\left (-2 a +b +\sqrt {4 a^{2}-4 b a -4 a c +4 d a -4 a e +b^{2}}\right )^{2}}{4 a}+12 a k +14 a +b -3 \sqrt {4 a^{2}-4 b a -4 a c +4 d a -4 a e +b^{2}}+2 b k -\frac {b \left (-2 a +b +\sqrt {4 a^{2}-4 b a -4 a c +4 d a -4 a e +b^{2}}\right )}{2 a}+c -d +e}, a_{1}=\frac {a_{0} \left (\frac {\left (-2 a +b +\sqrt {4 a^{2}-4 b a -4 a c +4 d a -4 a e +b^{2}}\right )^{2}}{4 a}-2 a +b +\sqrt {4 a^{2}-4 b a -4 a c +4 d a -4 a e +b^{2}}-\frac {3 b \left (-2 a +b +\sqrt {4 a^{2}-4 b a -4 a c +4 d a -4 a e +b^{2}}\right )}{4 a}+2 c -d \right )}{\frac {\left (-2 a +b +\sqrt {4 a^{2}-4 b a -4 a c +4 d a -4 a e +b^{2}}\right )^{2}}{4 a}+2 a +b -\sqrt {4 a^{2}-4 b a -4 a c +4 d a -4 a e +b^{2}}-\frac {b \left (-2 a +b +\sqrt {4 a^{2}-4 b a -4 a c +4 d a -4 a e +b^{2}}\right )}{2 a}+c -d +e}\right ] \\ \bullet & {} & \textrm {Revert the change of variables}\hspace {3pt} u =x +1 \\ {} & {} & \left [y=\moverset {\infty }{\munderset {k =0}{\sum }}a_{k} \left (x +1\right )^{k -\frac {-2 a +b +\sqrt {4 a^{2}-4 b a -4 a c +4 d a -4 a e +b^{2}}}{4 a}}, a_{k +2}=-\frac {a \,k^{2} a_{k}-4 a \,k^{2} a_{k +1}-\frac {k \left (-2 a +b +\sqrt {4 a^{2}-4 b a -4 a c +4 d a -4 a e +b^{2}}\right ) a_{k}}{2}+2 k \left (-2 a +b +\sqrt {4 a^{2}-4 b a -4 a c +4 d a -4 a e +b^{2}}\right ) a_{k +1}+\frac {\left (-2 a +b +\sqrt {4 a^{2}-4 b a -4 a c +4 d a -4 a e +b^{2}}\right )^{2} a_{k}}{16 a}-\frac {\left (-2 a +b +\sqrt {4 a^{2}-4 b a -4 a c +4 d a -4 a e +b^{2}}\right )^{2} a_{k +1}}{4 a}-a k a_{k}-4 a k a_{k +1}+\frac {\left (-2 a +b +\sqrt {4 a^{2}-4 b a -4 a c +4 d a -4 a e +b^{2}}\right ) a_{k}}{4}+\left (-2 a +b +\sqrt {4 a^{2}-4 b a -4 a c +4 d a -4 a e +b^{2}}\right ) a_{k +1}+a_{k} b k -3 b k a_{k +1}-\frac {a_{k} b \left (-2 a +b +\sqrt {4 a^{2}-4 b a -4 a c +4 d a -4 a e +b^{2}}\right )}{4 a}+\frac {3 b \left (-2 a +b +\sqrt {4 a^{2}-4 b a -4 a c +4 d a -4 a e +b^{2}}\right ) a_{k +1}}{4 a}-3 b a_{k +1}+a_{k} c -2 c a_{k +1}+d a_{k +1}}{4 a \,k^{2}-2 k \left (-2 a +b +\sqrt {4 a^{2}-4 b a -4 a c +4 d a -4 a e +b^{2}}\right )+\frac {\left (-2 a +b +\sqrt {4 a^{2}-4 b a -4 a c +4 d a -4 a e +b^{2}}\right )^{2}}{4 a}+12 a k +14 a +b -3 \sqrt {4 a^{2}-4 b a -4 a c +4 d a -4 a e +b^{2}}+2 b k -\frac {b \left (-2 a +b +\sqrt {4 a^{2}-4 b a -4 a c +4 d a -4 a e +b^{2}}\right )}{2 a}+c -d +e}, a_{1}=\frac {a_{0} \left (\frac {\left (-2 a +b +\sqrt {4 a^{2}-4 b a -4 a c +4 d a -4 a e +b^{2}}\right )^{2}}{4 a}-2 a +b +\sqrt {4 a^{2}-4 b a -4 a c +4 d a -4 a e +b^{2}}-\frac {3 b \left (-2 a +b +\sqrt {4 a^{2}-4 b a -4 a c +4 d a -4 a e +b^{2}}\right )}{4 a}+2 c -d \right )}{\frac {\left (-2 a +b +\sqrt {4 a^{2}-4 b a -4 a c +4 d a -4 a e +b^{2}}\right )^{2}}{4 a}+2 a +b -\sqrt {4 a^{2}-4 b a -4 a c +4 d a -4 a e +b^{2}}-\frac {b \left (-2 a +b +\sqrt {4 a^{2}-4 b a -4 a c +4 d a -4 a e +b^{2}}\right )}{2 a}+c -d +e}\right ] \\ \bullet & {} & \textrm {Recursion relation for}\hspace {3pt} r =\frac {2 a -b +\sqrt {4 a^{2}-4 b a -4 a c +4 d a -4 a e +b^{2}}}{4 a} \\ {} & {} & a_{k +2}=-\frac {a \,k^{2} a_{k}-4 a \,k^{2} a_{k +1}+\frac {k \left (2 a -b +\sqrt {4 a^{2}-4 b a -4 a c +4 d a -4 a e +b^{2}}\right ) a_{k}}{2}-2 k \left (2 a -b +\sqrt {4 a^{2}-4 b a -4 a c +4 d a -4 a e +b^{2}}\right ) a_{k +1}+\frac {\left (2 a -b +\sqrt {4 a^{2}-4 b a -4 a c +4 d a -4 a e +b^{2}}\right )^{2} a_{k}}{16 a}-\frac {\left (2 a -b +\sqrt {4 a^{2}-4 b a -4 a c +4 d a -4 a e +b^{2}}\right )^{2} a_{k +1}}{4 a}-a k a_{k}-4 a k a_{k +1}-\frac {\left (2 a -b +\sqrt {4 a^{2}-4 b a -4 a c +4 d a -4 a e +b^{2}}\right ) a_{k}}{4}-\left (2 a -b +\sqrt {4 a^{2}-4 b a -4 a c +4 d a -4 a e +b^{2}}\right ) a_{k +1}+a_{k} b k -3 b k a_{k +1}+\frac {a_{k} b \left (2 a -b +\sqrt {4 a^{2}-4 b a -4 a c +4 d a -4 a e +b^{2}}\right )}{4 a}-\frac {3 b \left (2 a -b +\sqrt {4 a^{2}-4 b a -4 a c +4 d a -4 a e +b^{2}}\right ) a_{k +1}}{4 a}-3 b a_{k +1}+a_{k} c -2 c a_{k +1}+d a_{k +1}}{4 a \,k^{2}+2 k \left (2 a -b +\sqrt {4 a^{2}-4 b a -4 a c +4 d a -4 a e +b^{2}}\right )+\frac {\left (2 a -b +\sqrt {4 a^{2}-4 b a -4 a c +4 d a -4 a e +b^{2}}\right )^{2}}{4 a}+12 a k +14 a +b +3 \sqrt {4 a^{2}-4 b a -4 a c +4 d a -4 a e +b^{2}}+2 b k +\frac {b \left (2 a -b +\sqrt {4 a^{2}-4 b a -4 a c +4 d a -4 a e +b^{2}}\right )}{2 a}+c -d +e} \\ \bullet & {} & \textrm {Solution for}\hspace {3pt} r =\frac {2 a -b +\sqrt {4 a^{2}-4 b a -4 a c +4 d a -4 a e +b^{2}}}{4 a} \\ {} & {} & \left [y \left (u \right )=\moverset {\infty }{\munderset {k =0}{\sum }}a_{k} u^{k +\frac {2 a -b +\sqrt {4 a^{2}-4 b a -4 a c +4 d a -4 a e +b^{2}}}{4 a}}, a_{k +2}=-\frac {a \,k^{2} a_{k}-4 a \,k^{2} a_{k +1}+\frac {k \left (2 a -b +\sqrt {4 a^{2}-4 b a -4 a c +4 d a -4 a e +b^{2}}\right ) a_{k}}{2}-2 k \left (2 a -b +\sqrt {4 a^{2}-4 b a -4 a c +4 d a -4 a e +b^{2}}\right ) a_{k +1}+\frac {\left (2 a -b +\sqrt {4 a^{2}-4 b a -4 a c +4 d a -4 a e +b^{2}}\right )^{2} a_{k}}{16 a}-\frac {\left (2 a -b +\sqrt {4 a^{2}-4 b a -4 a c +4 d a -4 a e +b^{2}}\right )^{2} a_{k +1}}{4 a}-a k a_{k}-4 a k a_{k +1}-\frac {\left (2 a -b +\sqrt {4 a^{2}-4 b a -4 a c +4 d a -4 a e +b^{2}}\right ) a_{k}}{4}-\left (2 a -b +\sqrt {4 a^{2}-4 b a -4 a c +4 d a -4 a e +b^{2}}\right ) a_{k +1}+a_{k} b k -3 b k a_{k +1}+\frac {a_{k} b \left (2 a -b +\sqrt {4 a^{2}-4 b a -4 a c +4 d a -4 a e +b^{2}}\right )}{4 a}-\frac {3 b \left (2 a -b +\sqrt {4 a^{2}-4 b a -4 a c +4 d a -4 a e +b^{2}}\right ) a_{k +1}}{4 a}-3 b a_{k +1}+a_{k} c -2 c a_{k +1}+d a_{k +1}}{4 a \,k^{2}+2 k \left (2 a -b +\sqrt {4 a^{2}-4 b a -4 a c +4 d a -4 a e +b^{2}}\right )+\frac {\left (2 a -b +\sqrt {4 a^{2}-4 b a -4 a c +4 d a -4 a e +b^{2}}\right )^{2}}{4 a}+12 a k +14 a +b +3 \sqrt {4 a^{2}-4 b a -4 a c +4 d a -4 a e +b^{2}}+2 b k +\frac {b \left (2 a -b +\sqrt {4 a^{2}-4 b a -4 a c +4 d a -4 a e +b^{2}}\right )}{2 a}+c -d +e}, a_{1}=\frac {a_{0} \left (\frac {\left (2 a -b +\sqrt {4 a^{2}-4 b a -4 a c +4 d a -4 a e +b^{2}}\right )^{2}}{4 a}-2 a +b -\sqrt {4 a^{2}-4 b a -4 a c +4 d a -4 a e +b^{2}}+\frac {3 b \left (2 a -b +\sqrt {4 a^{2}-4 b a -4 a c +4 d a -4 a e +b^{2}}\right )}{4 a}+2 c -d \right )}{\frac {\left (2 a -b +\sqrt {4 a^{2}-4 b a -4 a c +4 d a -4 a e +b^{2}}\right )^{2}}{4 a}+2 a +b +\sqrt {4 a^{2}-4 b a -4 a c +4 d a -4 a e +b^{2}}+\frac {b \left (2 a -b +\sqrt {4 a^{2}-4 b a -4 a c +4 d a -4 a e +b^{2}}\right )}{2 a}+c -d +e}\right ] \\ \bullet & {} & \textrm {Revert the change of variables}\hspace {3pt} u =x +1 \\ {} & {} & \left [y=\moverset {\infty }{\munderset {k =0}{\sum }}a_{k} \left (x +1\right )^{k +\frac {2 a -b +\sqrt {4 a^{2}-4 b a -4 a c +4 d a -4 a e +b^{2}}}{4 a}}, a_{k +2}=-\frac {a \,k^{2} a_{k}-4 a \,k^{2} a_{k +1}+\frac {k \left (2 a -b +\sqrt {4 a^{2}-4 b a -4 a c +4 d a -4 a e +b^{2}}\right ) a_{k}}{2}-2 k \left (2 a -b +\sqrt {4 a^{2}-4 b a -4 a c +4 d a -4 a e +b^{2}}\right ) a_{k +1}+\frac {\left (2 a -b +\sqrt {4 a^{2}-4 b a -4 a c +4 d a -4 a e +b^{2}}\right )^{2} a_{k}}{16 a}-\frac {\left (2 a -b +\sqrt {4 a^{2}-4 b a -4 a c +4 d a -4 a e +b^{2}}\right )^{2} a_{k +1}}{4 a}-a k a_{k}-4 a k a_{k +1}-\frac {\left (2 a -b +\sqrt {4 a^{2}-4 b a -4 a c +4 d a -4 a e +b^{2}}\right ) a_{k}}{4}-\left (2 a -b +\sqrt {4 a^{2}-4 b a -4 a c +4 d a -4 a e +b^{2}}\right ) a_{k +1}+a_{k} b k -3 b k a_{k +1}+\frac {a_{k} b \left (2 a -b +\sqrt {4 a^{2}-4 b a -4 a c +4 d a -4 a e +b^{2}}\right )}{4 a}-\frac {3 b \left (2 a -b +\sqrt {4 a^{2}-4 b a -4 a c +4 d a -4 a e +b^{2}}\right ) a_{k +1}}{4 a}-3 b a_{k +1}+a_{k} c -2 c a_{k +1}+d a_{k +1}}{4 a \,k^{2}+2 k \left (2 a -b +\sqrt {4 a^{2}-4 b a -4 a c +4 d a -4 a e +b^{2}}\right )+\frac {\left (2 a -b +\sqrt {4 a^{2}-4 b a -4 a c +4 d a -4 a e +b^{2}}\right )^{2}}{4 a}+12 a k +14 a +b +3 \sqrt {4 a^{2}-4 b a -4 a c +4 d a -4 a e +b^{2}}+2 b k +\frac {b \left (2 a -b +\sqrt {4 a^{2}-4 b a -4 a c +4 d a -4 a e +b^{2}}\right )}{2 a}+c -d +e}, a_{1}=\frac {a_{0} \left (\frac {\left (2 a -b +\sqrt {4 a^{2}-4 b a -4 a c +4 d a -4 a e +b^{2}}\right )^{2}}{4 a}-2 a +b -\sqrt {4 a^{2}-4 b a -4 a c +4 d a -4 a e +b^{2}}+\frac {3 b \left (2 a -b +\sqrt {4 a^{2}-4 b a -4 a c +4 d a -4 a e +b^{2}}\right )}{4 a}+2 c -d \right )}{\frac {\left (2 a -b +\sqrt {4 a^{2}-4 b a -4 a c +4 d a -4 a e +b^{2}}\right )^{2}}{4 a}+2 a +b +\sqrt {4 a^{2}-4 b a -4 a c +4 d a -4 a e +b^{2}}+\frac {b \left (2 a -b +\sqrt {4 a^{2}-4 b a -4 a c +4 d a -4 a e +b^{2}}\right )}{2 a}+c -d +e}\right ] \\ \bullet & {} & \textrm {Combine solutions and rename parameters}\hspace {3pt} \\ {} & {} & \left [y=\left (\moverset {\infty }{\munderset {k =0}{\sum }}f_{k} \left (x +1\right )^{k -\frac {-2 a +b +\sqrt {4 a^{2}-4 b a -4 a c +4 d a -4 a e +b^{2}}}{4 a}}\right )+\left (\moverset {\infty }{\munderset {k =0}{\sum }}g_{k} \left (x +1\right )^{k +\frac {2 a -b +\sqrt {4 a^{2}-4 b a -4 a c +4 d a -4 a e +b^{2}}}{4 a}}\right ), f_{k +2}=-\frac {a \,k^{2} f_{k}-4 a \,k^{2} f_{k +1}-\frac {k \left (-2 a +b +\sqrt {4 a^{2}-4 b a -4 a c +4 d a -4 a e +b^{2}}\right ) f_{k}}{2}+2 k \left (-2 a +b +\sqrt {4 a^{2}-4 b a -4 a c +4 d a -4 a e +b^{2}}\right ) f_{k +1}+\frac {\left (-2 a +b +\sqrt {4 a^{2}-4 b a -4 a c +4 d a -4 a e +b^{2}}\right )^{2} f_{k}}{16 a}-\frac {\left (-2 a +b +\sqrt {4 a^{2}-4 b a -4 a c +4 d a -4 a e +b^{2}}\right )^{2} f_{k +1}}{4 a}-a k f_{k}-4 a k f_{k +1}+\frac {\left (-2 a +b +\sqrt {4 a^{2}-4 b a -4 a c +4 d a -4 a e +b^{2}}\right ) f_{k}}{4}+\left (-2 a +b +\sqrt {4 a^{2}-4 b a -4 a c +4 d a -4 a e +b^{2}}\right ) f_{k +1}+f_{k} b k -3 b k f_{k +1}-\frac {f_{k} b \left (-2 a +b +\sqrt {4 a^{2}-4 b a -4 a c +4 d a -4 a e +b^{2}}\right )}{4 a}+\frac {3 b \left (-2 a +b +\sqrt {4 a^{2}-4 b a -4 a c +4 d a -4 a e +b^{2}}\right ) f_{k +1}}{4 a}-3 b f_{k +1}+f_{k} c -2 c f_{k +1}+d f_{k +1}}{4 a \,k^{2}-2 k \left (-2 a +b +\sqrt {4 a^{2}-4 b a -4 a c +4 d a -4 a e +b^{2}}\right )+\frac {\left (-2 a +b +\sqrt {4 a^{2}-4 b a -4 a c +4 d a -4 a e +b^{2}}\right )^{2}}{4 a}+12 a k +14 a +b -3 \sqrt {4 a^{2}-4 b a -4 a c +4 d a -4 a e +b^{2}}+2 b k -\frac {b \left (-2 a +b +\sqrt {4 a^{2}-4 b a -4 a c +4 d a -4 a e +b^{2}}\right )}{2 a}+c -d +e}, f_{1}=\frac {f_{0} \left (\frac {\left (-2 a +b +\sqrt {4 a^{2}-4 b a -4 a c +4 d a -4 a e +b^{2}}\right )^{2}}{4 a}-2 a +b +\sqrt {4 a^{2}-4 b a -4 a c +4 d a -4 a e +b^{2}}-\frac {3 b \left (-2 a +b +\sqrt {4 a^{2}-4 b a -4 a c +4 d a -4 a e +b^{2}}\right )}{4 a}+2 c -d \right )}{\frac {\left (-2 a +b +\sqrt {4 a^{2}-4 b a -4 a c +4 d a -4 a e +b^{2}}\right )^{2}}{4 a}+2 a +b -\sqrt {4 a^{2}-4 b a -4 a c +4 d a -4 a e +b^{2}}-\frac {b \left (-2 a +b +\sqrt {4 a^{2}-4 b a -4 a c +4 d a -4 a e +b^{2}}\right )}{2 a}+c -d +e}, g_{k +2}=-\frac {a \,k^{2} g_{k}-4 a \,k^{2} g_{k +1}+\frac {k \left (2 a -b +\sqrt {4 a^{2}-4 b a -4 a c +4 d a -4 a e +b^{2}}\right ) g_{k}}{2}-2 k \left (2 a -b +\sqrt {4 a^{2}-4 b a -4 a c +4 d a -4 a e +b^{2}}\right ) g_{k +1}+\frac {\left (2 a -b +\sqrt {4 a^{2}-4 b a -4 a c +4 d a -4 a e +b^{2}}\right )^{2} g_{k}}{16 a}-\frac {\left (2 a -b +\sqrt {4 a^{2}-4 b a -4 a c +4 d a -4 a e +b^{2}}\right )^{2} g_{k +1}}{4 a}-a k g_{k}-4 a k g_{k +1}-\frac {\left (2 a -b +\sqrt {4 a^{2}-4 b a -4 a c +4 d a -4 a e +b^{2}}\right ) g_{k}}{4}-\left (2 a -b +\sqrt {4 a^{2}-4 b a -4 a c +4 d a -4 a e +b^{2}}\right ) g_{k +1}+g_{k} b k -3 b k g_{k +1}+\frac {g_{k} b \left (2 a -b +\sqrt {4 a^{2}-4 b a -4 a c +4 d a -4 a e +b^{2}}\right )}{4 a}-\frac {3 b \left (2 a -b +\sqrt {4 a^{2}-4 b a -4 a c +4 d a -4 a e +b^{2}}\right ) g_{k +1}}{4 a}-3 b g_{k +1}+g_{k} c -2 c g_{k +1}+d g_{k +1}}{4 a \,k^{2}+2 k \left (2 a -b +\sqrt {4 a^{2}-4 b a -4 a c +4 d a -4 a e +b^{2}}\right )+\frac {\left (2 a -b +\sqrt {4 a^{2}-4 b a -4 a c +4 d a -4 a e +b^{2}}\right )^{2}}{4 a}+12 a k +14 a +b +3 \sqrt {4 a^{2}-4 b a -4 a c +4 d a -4 a e +b^{2}}+2 b k +\frac {b \left (2 a -b +\sqrt {4 a^{2}-4 b a -4 a c +4 d a -4 a e +b^{2}}\right )}{2 a}+c -d +e}, g_{1}=\frac {g_{0} \left (\frac {\left (2 a -b +\sqrt {4 a^{2}-4 b a -4 a c +4 d a -4 a e +b^{2}}\right )^{2}}{4 a}-2 a +b -\sqrt {4 a^{2}-4 b a -4 a c +4 d a -4 a e +b^{2}}+\frac {3 b \left (2 a -b +\sqrt {4 a^{2}-4 b a -4 a c +4 d a -4 a e +b^{2}}\right )}{4 a}+2 c -d \right )}{\frac {\left (2 a -b +\sqrt {4 a^{2}-4 b a -4 a c +4 d a -4 a e +b^{2}}\right )^{2}}{4 a}+2 a +b +\sqrt {4 a^{2}-4 b a -4 a c +4 d a -4 a e +b^{2}}+\frac {b \left (2 a -b +\sqrt {4 a^{2}-4 b a -4 a c +4 d a -4 a e +b^{2}}\right )}{2 a}+c -d +e}\right ] \end {array} \]
Maple trace
`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 <- No Liouvillian solutions exists -> 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 <- hyper3 successful: received ODE is equivalent to the 2F1 ODE <- hypergeometric successful <- special function solution successful`
✓ Solution by Maple
Time used: 0.141 (sec). Leaf size: 562
dsolve(a*(x^2-1)^2*diff(y(x),x$2)+b*x*(x^2-1)*diff(y(x),x)+(c*x^2+d*x+e)*y(x)=0,y(x), singsol=all)
\[ y \left (x \right ) = \frac {\left (x^{2}-1\right )^{-\frac {b}{4 a}} \sqrt {2+2 x}\, \left (c_{1} \operatorname {hypergeom}\left (\left [-\frac {-\sqrt {4 a^{2}+\left (-4 b -4 c -4 d -4 e \right ) a +b^{2}}+2 \sqrt {a^{2}+\left (-2 b -4 c \right ) a +b^{2}}+\sqrt {4 a^{2}+\left (-4 b -4 c +4 d -4 e \right ) a +b^{2}}-2 a}{4 a}, \frac {\sqrt {4 a^{2}+\left (-4 b -4 c -4 d -4 e \right ) a +b^{2}}+2 \sqrt {a^{2}+\left (-2 b -4 c \right ) a +b^{2}}-\sqrt {4 a^{2}+\left (-4 b -4 c +4 d -4 e \right ) a +b^{2}}+2 a}{4 a}\right ], \left [-\frac {-2 a +\sqrt {4 a^{2}+\left (-4 b -4 c +4 d -4 e \right ) a +b^{2}}}{2 a}\right ], \frac {1}{2}+\frac {x}{2}\right ) \left (\frac {1}{2}+\frac {x}{2}\right )^{-\frac {\sqrt {4 a^{2}+\left (-4 b -4 c +4 d -4 e \right ) a +b^{2}}}{4 a}}+c_{2} \operatorname {hypergeom}\left (\left [\frac {\sqrt {4 a^{2}+\left (-4 b -4 c -4 d -4 e \right ) a +b^{2}}+2 \sqrt {a^{2}+\left (-2 b -4 c \right ) a +b^{2}}+\sqrt {4 a^{2}+\left (-4 b -4 c +4 d -4 e \right ) a +b^{2}}+2 a}{4 a}, \frac {\sqrt {4 a^{2}+\left (-4 b -4 c -4 d -4 e \right ) a +b^{2}}-2 \sqrt {a^{2}+\left (-2 b -4 c \right ) a +b^{2}}+\sqrt {4 a^{2}+\left (-4 b -4 c +4 d -4 e \right ) a +b^{2}}+2 a}{4 a}\right ], \left [\frac {2 a +\sqrt {4 a^{2}+\left (-4 b -4 c +4 d -4 e \right ) a +b^{2}}}{2 a}\right ], \frac {1}{2}+\frac {x}{2}\right ) \left (\frac {1}{2}+\frac {x}{2}\right )^{\frac {\sqrt {4 a^{2}+\left (-4 b -4 c +4 d -4 e \right ) a +b^{2}}}{4 a}}\right ) \sqrt {2 x -2}\, \left (-\frac {1}{2}+\frac {x}{2}\right )^{\frac {\sqrt {4 a^{2}+\left (-4 b -4 c -4 d -4 e \right ) a +b^{2}}}{4 a}}}{4} \]
✗ Solution by Mathematica
Time used: 0.0 (sec). Leaf size: 0
DSolve[a*(x^2-1)^2*y''[x]+b*x*(x^2-1)*y'[x]+(c*x^2+d*x+e)*y[x]==0,y[x],x,IncludeSingularSolutions -> True]
Timed out