Internal problem ID [9701]
Internal file name [OUTPUT/8643_Monday_June_06_2022_04_35_42_AM_17557834/index.tex]
Book: Differential Gleichungen, E. Kamke, 3rd ed. Chelsea Pub. NY, 1948
Section: Chapter 2, linear second order
Problem number: 1374.
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.
\[ \boxed {y^{\prime \prime }-\frac {2 x \left (2 a -1\right ) y^{\prime }}{x^{2}-1}+\frac {\left (x^{2} \left (2 a \left (2 a -1\right )-v \left (v +1\right )\right )+2 a +v \left (v +1\right )\right ) y}{\left (x^{2}-1\right )^{2}}=0} \]
\[ \begin {array}{lll} & {} & \textrm {Let's solve}\hspace {3pt} \\ {} & {} & \left (\frac {d}{d x}y^{\prime }\right ) \left (x^{4}-2 x^{2}+1\right )+\left (\left (-4 a +2\right ) x^{3}+\left (-2+4 a \right ) x \right ) y^{\prime }+\left (\left (4 a^{2}-v^{2}-2 a -v \right ) x^{2}+v^{2}+2 a +v \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 (4 a^{2} x^{2}-v^{2} x^{2}-2 a \,x^{2}-v \,x^{2}+v^{2}+2 a +v \right ) y}{x^{4}-2 x^{2}+1}+\frac {2 x \left (2 a -1\right ) y^{\prime }}{x^{2}-1} \\ \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 {2 x \left (2 a -1\right ) y^{\prime }}{x^{2}-1}+\frac {\left (4 a^{2} x^{2}-v^{2} x^{2}-2 a \,x^{2}-v \,x^{2}+v^{2}+2 a +v \right ) y}{x^{4}-2 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 {2 x \left (2 a -1\right )}{x^{2}-1}, P_{3}\left (x \right )=\frac {4 a^{2} x^{2}-v^{2} x^{2}-2 a \,x^{2}-v \,x^{2}+v^{2}+2 a +v}{x^{4}-2 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}}}=-2 a +1 \\ {} & \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}}}=a^{2} \\ {} & \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 ) \left (x^{2}-1\right ) \left (x^{4}-2 x^{2}+1\right )-2 x \left (2 a -1\right ) \left (x^{4}-2 x^{2}+1\right ) y^{\prime }+\left (4 a^{2} x^{2}-v^{2} x^{2}-2 a \,x^{2}-v \,x^{2}+v^{2}+2 a +v \right ) \left (x^{2}-1\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 (u^{6}-6 u^{5}+12 u^{4}-8 u^{3}\right ) \left (\frac {d}{d u}\frac {d}{d u}y \left (u \right )\right )+\left (-4 a \,u^{5}+20 a \,u^{4}+2 u^{5}-32 a \,u^{3}-10 u^{4}+16 a \,u^{2}+16 u^{3}-8 u^{2}\right ) \left (\frac {d}{d u}y \left (u \right )\right )+\left (4 a^{2} u^{4}-u^{4} v^{2}-16 a^{2} u^{3}-2 a \,u^{4}-u^{4} v +4 u^{3} v^{2}+20 a^{2} u^{2}+8 a \,u^{3}+4 u^{3} v -4 u^{2} v^{2}-8 a^{2} u -8 a \,u^{2}-4 u^{2} v \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 =1..4 \\ {} & {} & 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 =2..5 \\ {} & {} & 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 =3..6 \\ {} & {} & 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} \\ {} & {} & -8 a_{0} \left (a -r \right )^{2} u^{1+r}+\left (-8 a_{1} \left (a -1-r \right )^{2}+4 a_{0} \left (5 a^{2}-8 a r +3 r^{2}-v^{2}-2 a +r -v \right )\right ) u^{2+r}+\left (-8 a_{2} \left (a -2-r \right )^{2}+4 a_{1} \left (5 a^{2}-8 a r +3 r^{2}-v^{2}-10 a +7 r -v +4\right )-2 a_{0} \left (8 a^{2}-10 a r +3 r^{2}-2 v^{2}-4 a +2 r -2 v \right )\right ) u^{3+r}+\left (\moverset {\infty }{\munderset {k =4}{\sum }}\left (-8 a_{k -1} \left (a -k +1-r \right )^{2}+4 a_{k -2} \left (5 a^{2}-8 a \left (k -2\right )-8 a r +3 \left (k -2\right )^{2}+6 \left (k -2\right ) r +3 r^{2}-v^{2}-2 a +k -2+r -v \right )-2 a_{k -3} \left (8 a^{2}-10 a \left (k -3\right )-10 a r +3 \left (k -3\right )^{2}+6 \left (k -3\right ) r +3 r^{2}-2 v^{2}-4 a +2 k -6+2 r -2 v \right )+a_{k -4} \left (2 a -k +4-r +v \right ) \left (-v +3-r -k +2 a \right )\right ) u^{k +r}\right )=0 \\ \bullet & {} & a_{0}\textrm {cannot be 0 by assumption, giving the indicial equation}\hspace {3pt} \\ {} & {} & -8 \left (a -r \right )^{2}=0 \\ \bullet & {} & \textrm {Values of r that satisfy the indicial equation}\hspace {3pt} \\ {} & {} & r =a \\ \bullet & {} & \textrm {The coefficients of each power of}\hspace {3pt} u \hspace {3pt}\textrm {must be 0}\hspace {3pt} \\ {} & {} & \left [-8 a_{1} \left (a -1-r \right )^{2}+4 a_{0} \left (5 a^{2}-8 a r +3 r^{2}-v^{2}-2 a +r -v \right )=0, -8 a_{2} \left (a -2-r \right )^{2}+4 a_{1} \left (5 a^{2}-8 a r +3 r^{2}-v^{2}-10 a +7 r -v +4\right )-2 a_{0} \left (8 a^{2}-10 a r +3 r^{2}-2 v^{2}-4 a +2 r -2 v \right )=0\right ] \\ \bullet & {} & \textrm {Solve for the dependent coefficient(s)}\hspace {3pt} \\ {} & {} & \left \{a_{1}=\frac {a_{0} \left (5 a^{2}-8 a r +3 r^{2}-v^{2}-2 a +r -v \right )}{2 \left (a^{2}-2 a r +r^{2}-2 a +2 r +1\right )}, a_{2}=\frac {a_{0} \left (17 a^{4}-54 a^{3} r +63 a^{2} r^{2}-8 v^{2} a^{2}-32 a \,r^{3}+12 a r \,v^{2}+6 r^{4}-4 r^{2} v^{2}+v^{4}-40 a^{3}+90 a^{2} r -8 v \,a^{2}-66 a \,r^{2}+12 a r v +8 a \,v^{2}+16 r^{3}-4 r^{2} v -4 r \,v^{2}+2 v^{3}+24 a^{2}-34 a r +8 a v +12 r^{2}-4 r v -v^{2}-4 a +2 r -2 v \right )}{4 \left (a^{4}-4 a^{3} r +6 a^{2} r^{2}-4 a \,r^{3}+r^{4}-6 a^{3}+18 a^{2} r -18 a \,r^{2}+6 r^{3}+13 a^{2}-26 a r +13 r^{2}-12 a +12 r +4\right )}\right \} \\ \bullet & {} & \textrm {Each term in the series must be 0, giving the recursion relation}\hspace {3pt} \\ {} & {} & -8 a_{k -1} \left (-a +k -1+r \right )^{2}+4 a_{k -2} \left (3 k^{2}+\left (6 r -8 a -11\right ) k +3 r^{2}-8 a r -v^{2}+5 a^{2}-11 r -v +14 a +10\right )+2 a_{k -3} \left (-3 k^{2}-2 \left (3 r -5 a -8\right ) k -3 r^{2}+10 a r +2 v^{2}-8 a^{2}+16 r +2 v -26 a -21\right )+a_{k -4} \left (-2 a +k -4+r -v \right ) \left (v -3+r +k -2 a \right )=0 \\ \bullet & {} & \textrm {Shift index using}\hspace {3pt} k \mathrm {->}k +4 \\ {} & {} & -8 a_{k +3} \left (-a +k +3+r \right )^{2}+4 a_{k +2} \left (3 \left (k +4\right )^{2}+\left (6 r -8 a -11\right ) \left (k +4\right )+3 r^{2}-8 a r -v^{2}+5 a^{2}-11 r -v +14 a +10\right )+2 a_{k +1} \left (-3 \left (k +4\right )^{2}-2 \left (3 r -5 a -8\right ) \left (k +4\right )-3 r^{2}+10 a r +2 v^{2}-8 a^{2}+16 r +2 v -26 a -21\right )+a_{k} \left (-2 a +k +r -v \right ) \left (v +1+r +k -2 a \right )=0 \\ \bullet & {} & \textrm {Recursion relation that defines series solution to ODE}\hspace {3pt} \\ {} & {} & a_{k +3}=\frac {4 a_{k} a^{2}-16 a^{2} a_{k +1}+20 a^{2} a_{k +2}-4 a k a_{k}+20 a k a_{k +1}-32 a k a_{k +2}-4 a r a_{k}+20 a r a_{k +1}-32 a r a_{k +2}+k^{2} a_{k}-6 k^{2} a_{k +1}+12 k^{2} a_{k +2}+2 k r a_{k}-12 k r a_{k +1}+24 k r a_{k +2}+r^{2} a_{k}-6 r^{2} a_{k +1}+12 r^{2} a_{k +2}-a_{k} v^{2}+4 v^{2} a_{k +1}-4 v^{2} a_{k +2}-2 a_{k} a +28 a a_{k +1}-72 a a_{k +2}+k a_{k}-16 k a_{k +1}+52 k a_{k +2}+r a_{k}-16 r a_{k +1}+52 r a_{k +2}-a_{k} v +4 v a_{k +1}-4 v a_{k +2}-10 a_{k +1}+56 a_{k +2}}{8 \left (a -k -3-r \right )^{2}} \\ \bullet & {} & \textrm {Recursion relation for}\hspace {3pt} r =a \\ {} & {} & a_{k +3}=\frac {a_{k} a^{2}-2 a^{2} a_{k +1}-2 a k a_{k}+8 a k a_{k +1}-8 a k a_{k +2}+k^{2} a_{k}-6 k^{2} a_{k +1}+12 k^{2} a_{k +2}-a_{k} v^{2}+4 v^{2} a_{k +1}-4 v^{2} a_{k +2}-a_{k} a +12 a a_{k +1}-20 a a_{k +2}+k a_{k}-16 k a_{k +1}+52 k a_{k +2}-a_{k} v +4 v a_{k +1}-4 v a_{k +2}-10 a_{k +1}+56 a_{k +2}}{8 \left (-k -3\right )^{2}} \\ \bullet & {} & \textrm {Solution for}\hspace {3pt} r =a \\ {} & {} & \left [y \left (u \right )=\moverset {\infty }{\munderset {k =0}{\sum }}a_{k} u^{k +a}, a_{k +3}=\frac {a_{k} a^{2}-2 a^{2} a_{k +1}-2 a k a_{k}+8 a k a_{k +1}-8 a k a_{k +2}+k^{2} a_{k}-6 k^{2} a_{k +1}+12 k^{2} a_{k +2}-a_{k} v^{2}+4 v^{2} a_{k +1}-4 v^{2} a_{k +2}-a_{k} a +12 a a_{k +1}-20 a a_{k +2}+k a_{k}-16 k a_{k +1}+52 k a_{k +2}-a_{k} v +4 v a_{k +1}-4 v a_{k +2}-10 a_{k +1}+56 a_{k +2}}{8 \left (-k -3\right )^{2}}, a_{1}=\frac {a_{0} \left (-v^{2}-a -v \right )}{2}, a_{2}=\frac {a_{0} \left (v^{4}+4 a \,v^{2}+2 v^{3}+2 a^{2}+4 a v -v^{2}-2 a -2 v \right )}{16}\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}, a_{k +3}=\frac {a_{k} a^{2}-2 a^{2} a_{k +1}-2 a k a_{k}+8 a k a_{k +1}-8 a k a_{k +2}+k^{2} a_{k}-6 k^{2} a_{k +1}+12 k^{2} a_{k +2}-a_{k} v^{2}+4 v^{2} a_{k +1}-4 v^{2} a_{k +2}-a_{k} a +12 a a_{k +1}-20 a a_{k +2}+k a_{k}-16 k a_{k +1}+52 k a_{k +2}-a_{k} v +4 v a_{k +1}-4 v a_{k +2}-10 a_{k +1}+56 a_{k +2}}{8 \left (-k -3\right )^{2}}, a_{1}=\frac {a_{0} \left (-v^{2}-a -v \right )}{2}, a_{2}=\frac {a_{0} \left (v^{4}+4 a \,v^{2}+2 v^{3}+2 a^{2}+4 a v -v^{2}-2 a -2 v \right )}{16}\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 <- Legendre successful <- special function solution successful`
✓ Solution by Maple
Time used: 0.094 (sec). Leaf size: 23
dsolve(diff(diff(y(x),x),x) = 2*x*(2*a-1)/(x^2-1)*diff(y(x),x)-(x^2*(2*a*(2*a-1)-v*(v+1))+2*a+v*(v+1))/(x^2-1)^2*y(x),y(x), singsol=all)
\[ y \left (x \right ) = \left (c_{1} \operatorname {LegendreP}\left (v , x\right )+c_{2} \operatorname {LegendreQ}\left (v , x\right )\right ) \left (x^{2}-1\right )^{a} \]
✓ Solution by Mathematica
Time used: 0.047 (sec). Leaf size: 26
DSolve[y''[x] == -(((2*a + v*(1 + v) + (2*a*(-1 + 2*a) - v*(1 + v))*x^2)*y[x])/(-1 + x^2)^2) + (2*(-1 + 2*a)*x*y'[x])/(-1 + x^2),y[x],x,IncludeSingularSolutions -> True]
\[ y(x)\to \left (x^2-1\right )^a (c_1 \operatorname {LegendreP}(v,x)+c_2 \operatorname {LegendreQ}(v,x)) \]