Internal problem ID [9645]
Internal file name [OUTPUT/8587_Monday_June_06_2022_04_17_12_AM_65841309/index.tex]
Book: Differential Gleichungen, E. Kamke, 3rd ed. Chelsea Pub. NY, 1948
Section: Chapter 2, linear second order
Problem number: 1318.
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.
\[ \boxed {x \left (x^{2}-1\right ) y^{\prime \prime }+\left (a \,x^{2}+b \right ) y^{\prime }+c x y=0} \]
\[ \begin {array}{lll} & {} & \textrm {Let's solve}\hspace {3pt} \\ {} & {} & \left (x^{3}-x \right ) \left (\frac {d}{d x}y^{\prime }\right )+\left (a \,x^{2}+b \right ) y^{\prime }+c x 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 {c y}{x^{2}-1}-\frac {\left (a \,x^{2}+b \right ) y^{\prime }}{x \left (x^{2}-1\right )} \\ \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 {\left (a \,x^{2}+b \right ) y^{\prime }}{x \left (x^{2}-1\right )}+\frac {c y}{x^{2}-1}=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 {a \,x^{2}+b}{x \left (x^{2}-1\right )}, P_{3}\left (x \right )=\frac {c}{x^{2}-1}\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 {a}{2}+\frac {b}{2} \\ {} & \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}}}=0 \\ {} & \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} \\ {} & {} & x \left (x^{2}-1\right ) \left (\frac {d}{d x}y^{\prime }\right )+\left (a \,x^{2}+b \right ) y^{\prime }+c x 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 (u^{3}-3 u^{2}+2 u \right ) \left (\frac {d}{d u}\frac {d}{d u}y \left (u \right )\right )+\left (a \,u^{2}-2 a u +a +b \right ) \left (\frac {d}{d u}y \left (u \right )\right )+\left (c u -c \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..1 \\ {} & {} & 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 =0..2 \\ {} & {} & 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 =1..3 \\ {} & {} & 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} r \left (-2+2 r +a +b \right ) u^{-1+r}+\left (a_{1} \left (1+r \right ) \left (2 r +a +b \right )-a_{0} \left (2 a r +3 r^{2}+c -3 r \right )\right ) u^{r}+\left (\moverset {\infty }{\munderset {k =1}{\sum }}\left (a_{k +1} \left (k +1+r \right ) \left (2 k +2 r +a +b \right )-a_{k} \left (2 a k +2 a r +3 k^{2}+6 k r +3 r^{2}+c -3 k -3 r \right )+a_{k -1} \left (a \left (k -1\right )+a r +\left (k -1\right )^{2}+2 \left (k -1\right ) r +r^{2}+c -k +1-r \right )\right ) u^{k +r}\right )=0 \\ \bullet & {} & a_{0}\textrm {cannot be 0 by assumption, giving the indicial equation}\hspace {3pt} \\ {} & {} & r \left (-2+2 r +a +b \right )=0 \\ \bullet & {} & \textrm {Values of r that satisfy the indicial equation}\hspace {3pt} \\ {} & {} & r \in \left \{0, -\frac {a}{2}-\frac {b}{2}+1\right \} \\ \bullet & {} & \textrm {Each term must be 0}\hspace {3pt} \\ {} & {} & a_{1} \left (1+r \right ) \left (2 r +a +b \right )-a_{0} \left (2 a r +3 r^{2}+c -3 r \right )=0 \\ \bullet & {} & \textrm {Each term in the series must be 0, giving the recursion relation}\hspace {3pt} \\ {} & {} & a_{k +1} \left (k +1+r \right ) \left (2 k +2 r +a +b \right )-\left (3 r^{2}+\left (6 k +2 a -3\right ) r +3 k^{2}+2 a k +c -3 k \right ) a_{k}+\left (k^{2}+\left (2 r +a -3\right ) k +r^{2}+\left (a -3\right ) r +c -a +2\right ) a_{k -1}=0 \\ \bullet & {} & \textrm {Shift index using}\hspace {3pt} k \mathrm {->}k +1 \\ {} & {} & a_{k +2} \left (k +2+r \right ) \left (2 k +2+2 r +a +b \right )-\left (3 r^{2}+\left (6 k +3+2 a \right ) r +3 \left (k +1\right )^{2}+2 a \left (k +1\right )+c -3 k -3\right ) a_{k +1}+\left (\left (k +1\right )^{2}+\left (2 r +a -3\right ) \left (k +1\right )+r^{2}+\left (a -3\right ) r +c -a +2\right ) a_{k}=0 \\ \bullet & {} & \textrm {Recursion relation that defines series solution to ODE}\hspace {3pt} \\ {} & {} & a_{k +2}=-\frac {a k a_{k}-2 a k a_{k +1}+a r a_{k}-2 a r a_{k +1}+k^{2} a_{k}-3 k^{2} a_{k +1}+2 k r a_{k}-6 k r a_{k +1}+r^{2} a_{k}-3 r^{2} a_{k +1}-2 a a_{k +1}+a_{k} c -c a_{k +1}-k a_{k}-3 k a_{k +1}-r a_{k}-3 r a_{k +1}}{\left (k +2+r \right ) \left (2 k +2+2 r +a +b \right )} \\ \bullet & {} & \textrm {Recursion relation for}\hspace {3pt} r =0 \\ {} & {} & a_{k +2}=-\frac {a k a_{k}-2 a k a_{k +1}+k^{2} a_{k}-3 k^{2} a_{k +1}-2 a a_{k +1}+a_{k} c -c a_{k +1}-k a_{k}-3 k a_{k +1}}{\left (k +2\right ) \left (2 k +2+a +b \right )} \\ \bullet & {} & \textrm {Solution for}\hspace {3pt} r =0 \\ {} & {} & \left [y \left (u \right )=\moverset {\infty }{\munderset {k =0}{\sum }}a_{k} u^{k}, a_{k +2}=-\frac {a k a_{k}-2 a k a_{k +1}+k^{2} a_{k}-3 k^{2} a_{k +1}-2 a a_{k +1}+a_{k} c -c a_{k +1}-k a_{k}-3 k a_{k +1}}{\left (k +2\right ) \left (2 k +2+a +b \right )}, a_{1} \left (a +b \right )-a_{0} c =0\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}, a_{k +2}=-\frac {a k a_{k}-2 a k a_{k +1}+k^{2} a_{k}-3 k^{2} a_{k +1}-2 a a_{k +1}+a_{k} c -c a_{k +1}-k a_{k}-3 k a_{k +1}}{\left (k +2\right ) \left (2 k +2+a +b \right )}, a_{1} \left (a +b \right )-a_{0} c =0\right ] \\ \bullet & {} & \textrm {Recursion relation for}\hspace {3pt} r =-\frac {a}{2}-\frac {b}{2}+1 \\ {} & {} & a_{k +2}=-\frac {a k a_{k}-2 a k a_{k +1}+a \left (-\frac {a}{2}-\frac {b}{2}+1\right ) a_{k}-2 a \left (-\frac {a}{2}-\frac {b}{2}+1\right ) a_{k +1}+k^{2} a_{k}-3 k^{2} a_{k +1}+2 k \left (-\frac {a}{2}-\frac {b}{2}+1\right ) a_{k}-6 k \left (-\frac {a}{2}-\frac {b}{2}+1\right ) a_{k +1}+\left (-\frac {a}{2}-\frac {b}{2}+1\right )^{2} a_{k}-3 \left (-\frac {a}{2}-\frac {b}{2}+1\right )^{2} a_{k +1}-2 a a_{k +1}+a_{k} c -c a_{k +1}-k a_{k}-3 k a_{k +1}-\left (-\frac {a}{2}-\frac {b}{2}+1\right ) a_{k}-3 \left (-\frac {a}{2}-\frac {b}{2}+1\right ) a_{k +1}}{\left (k +3-\frac {a}{2}-\frac {b}{2}\right ) \left (2 k +4\right )} \\ \bullet & {} & \textrm {Solution for}\hspace {3pt} r =-\frac {a}{2}-\frac {b}{2}+1 \\ {} & {} & \left [y \left (u \right )=\moverset {\infty }{\munderset {k =0}{\sum }}a_{k} u^{k -\frac {a}{2}-\frac {b}{2}+1}, a_{k +2}=-\frac {a k a_{k}-2 a k a_{k +1}+a \left (-\frac {a}{2}-\frac {b}{2}+1\right ) a_{k}-2 a \left (-\frac {a}{2}-\frac {b}{2}+1\right ) a_{k +1}+k^{2} a_{k}-3 k^{2} a_{k +1}+2 k \left (-\frac {a}{2}-\frac {b}{2}+1\right ) a_{k}-6 k \left (-\frac {a}{2}-\frac {b}{2}+1\right ) a_{k +1}+\left (-\frac {a}{2}-\frac {b}{2}+1\right )^{2} a_{k}-3 \left (-\frac {a}{2}-\frac {b}{2}+1\right )^{2} a_{k +1}-2 a a_{k +1}+a_{k} c -c a_{k +1}-k a_{k}-3 k a_{k +1}-\left (-\frac {a}{2}-\frac {b}{2}+1\right ) a_{k}-3 \left (-\frac {a}{2}-\frac {b}{2}+1\right ) a_{k +1}}{\left (k +3-\frac {a}{2}-\frac {b}{2}\right ) \left (2 k +4\right )}, 2 a_{1} \left (2-\frac {a}{2}-\frac {b}{2}\right )-a_{0} \left (2 a \left (-\frac {a}{2}-\frac {b}{2}+1\right )+3 \left (-\frac {a}{2}-\frac {b}{2}+1\right )^{2}+c +\frac {3 a}{2}+\frac {3 b}{2}-3\right )=0\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 {a}{2}-\frac {b}{2}+1}, a_{k +2}=-\frac {a k a_{k}-2 a k a_{k +1}+a \left (-\frac {a}{2}-\frac {b}{2}+1\right ) a_{k}-2 a \left (-\frac {a}{2}-\frac {b}{2}+1\right ) a_{k +1}+k^{2} a_{k}-3 k^{2} a_{k +1}+2 k \left (-\frac {a}{2}-\frac {b}{2}+1\right ) a_{k}-6 k \left (-\frac {a}{2}-\frac {b}{2}+1\right ) a_{k +1}+\left (-\frac {a}{2}-\frac {b}{2}+1\right )^{2} a_{k}-3 \left (-\frac {a}{2}-\frac {b}{2}+1\right )^{2} a_{k +1}-2 a a_{k +1}+a_{k} c -c a_{k +1}-k a_{k}-3 k a_{k +1}-\left (-\frac {a}{2}-\frac {b}{2}+1\right ) a_{k}-3 \left (-\frac {a}{2}-\frac {b}{2}+1\right ) a_{k +1}}{\left (k +3-\frac {a}{2}-\frac {b}{2}\right ) \left (2 k +4\right )}, 2 a_{1} \left (2-\frac {a}{2}-\frac {b}{2}\right )-a_{0} \left (2 a \left (-\frac {a}{2}-\frac {b}{2}+1\right )+3 \left (-\frac {a}{2}-\frac {b}{2}+1\right )^{2}+c +\frac {3 a}{2}+\frac {3 b}{2}-3\right )=0\right ] \\ \bullet & {} & \textrm {Combine solutions and rename parameters}\hspace {3pt} \\ {} & {} & \left [y=\left (\moverset {\infty }{\munderset {k =0}{\sum }}d_{k} \left (x +1\right )^{k}\right )+\left (\moverset {\infty }{\munderset {k =0}{\sum }}e_{k} \left (x +1\right )^{k -\frac {a}{2}-\frac {b}{2}+1}\right ), d_{k +2}=-\frac {a k d_{k}-2 a k d_{k +1}+k^{2} d_{k}-3 k^{2} d_{k +1}-2 a d_{k +1}+c d_{k}-c d_{k +1}-k d_{k}-3 k d_{k +1}}{\left (k +2\right ) \left (2 k +2+a +b \right )}, d_{1} \left (a +b \right )-d_{0} c =0, e_{k +2}=-\frac {a k e_{k}-2 a k e_{k +1}+a \left (-\frac {a}{2}-\frac {b}{2}+1\right ) e_{k}-2 a \left (-\frac {a}{2}-\frac {b}{2}+1\right ) e_{k +1}+k^{2} e_{k}-3 k^{2} e_{k +1}+2 k \left (-\frac {a}{2}-\frac {b}{2}+1\right ) e_{k}-6 k \left (-\frac {a}{2}-\frac {b}{2}+1\right ) e_{k +1}+\left (-\frac {a}{2}-\frac {b}{2}+1\right )^{2} e_{k}-3 \left (-\frac {a}{2}-\frac {b}{2}+1\right )^{2} e_{k +1}-2 a e_{k +1}+e_{k} c -c e_{k +1}-k e_{k}-3 k e_{k +1}-\left (-\frac {a}{2}-\frac {b}{2}+1\right ) e_{k}-3 \left (-\frac {a}{2}-\frac {b}{2}+1\right ) e_{k +1}}{\left (k +3-\frac {a}{2}-\frac {b}{2}\right ) \left (2 k +4\right )}, 2 e_{1} \left (2-\frac {a}{2}-\frac {b}{2}\right )-e_{0} \left (2 a \left (-\frac {a}{2}-\frac {b}{2}+1\right )+3 \left (-\frac {a}{2}-\frac {b}{2}+1\right )^{2}+c +\frac {3 a}{2}+\frac {3 b}{2}-3\right )=0\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 -> Kummer -> hyper3: Equivalence to 1F1 under a power @ Moebius -> hypergeometric -> heuristic approach <- heuristic approach successful <- hypergeometric successful <- special function solution successful`
✓ Solution by Maple
Time used: 0.188 (sec). Leaf size: 122
dsolve(x*(x^2-1)*diff(diff(y(x),x),x)+(a*x^2+b)*diff(y(x),x)+c*x*y(x)=0,y(x), singsol=all)
\[ y \left (x \right ) = c_{1} \operatorname {hypergeom}\left (\left [-\frac {1}{4}+\frac {a}{4}+\frac {\sqrt {a^{2}-2 a -4 c +1}}{4}, -\frac {1}{4}+\frac {a}{4}-\frac {\sqrt {a^{2}-2 a -4 c +1}}{4}\right ], \left [\frac {1}{2}-\frac {b}{2}\right ], x^{2}\right )+c_{2} x^{b +1} \operatorname {hypergeom}\left (\left [\frac {1}{4}+\frac {a}{4}+\frac {b}{2}+\frac {\sqrt {a^{2}-2 a -4 c +1}}{4}, \frac {1}{4}+\frac {a}{4}+\frac {b}{2}-\frac {\sqrt {a^{2}-2 a -4 c +1}}{4}\right ], \left [\frac {3}{2}+\frac {b}{2}\right ], x^{2}\right ) \]
✓ Solution by Mathematica
Time used: 0.349 (sec). Leaf size: 146
DSolve[c*x*y[x] + (b + a*x^2)*y'[x] + x*(-1 + x^2)*y''[x] == 0,y[x],x,IncludeSingularSolutions -> True]
\[ y(x)\to c_1 \operatorname {Hypergeometric2F1}\left (\frac {1}{4} \left (a-\sqrt {a^2-2 a-4 c+1}-1\right ),\frac {1}{4} \left (a+\sqrt {a^2-2 a-4 c+1}-1\right ),\frac {1-b}{2},x^2\right )+i^{b+1} c_2 x^{b+1} \operatorname {Hypergeometric2F1}\left (\frac {1}{4} \left (a+2 b-\sqrt {a^2-2 a-4 c+1}+1\right ),\frac {1}{4} \left (a+2 b+\sqrt {a^2-2 a-4 c+1}+1\right ),\frac {b+3}{2},x^2\right ) \]