Internal problem ID [9715]
Internal file name [OUTPUT/8657_Monday_June_06_2022_04_40_58_AM_98459127/index.tex]
Book: Differential Gleichungen, E. Kamke, 3rd ed. Chelsea Pub. NY, 1948
Section: Chapter 2, linear second order
Problem number: 1388.
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.
\[ \boxed {y^{\prime \prime }+\frac {\left (3 x -1\right ) y^{\prime }}{2 x \left (x -1\right )}+\frac {\left (v \left (v +1\right ) \left (x -1\right )-a^{2} x \right ) y}{4 x^{2} \left (x -1\right )^{2}}=0} \]
\[ \begin {array}{lll} & {} & \textrm {Let's solve}\hspace {3pt} \\ {} & {} & 4 \left (\frac {d}{d x}y^{\prime }\right ) x^{2} \left (x -1\right )^{2}+\left (6 x^{3}-8 x^{2}+2 x \right ) y^{\prime }+\left (\left (x -1\right ) v^{2}+\left (x -1\right ) v -a^{2} x \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 (a^{2} x -v^{2} x +v^{2}-v x +v \right ) y}{4 x^{2} \left (x -1\right )^{2}}-\frac {\left (3 x -1\right ) y^{\prime }}{2 x \left (x -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 (3 x -1\right ) y^{\prime }}{2 x \left (x -1\right )}-\frac {\left (a^{2} x -v^{2} x +v^{2}-v x +v \right ) y}{4 x^{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 {3 x -1}{2 x \left (x -1\right )}, P_{3}\left (x \right )=-\frac {a^{2} x -v^{2} x +v^{2}-v x +v}{4 x^{2} \left (x -1\right )^{2}}\right ] \\ {} & \circ & x \cdot P_{2}\left (x \right )\textrm {is analytic at}\hspace {3pt} x =0 \\ {} & {} & \left (x \cdot P_{2}\left (x \right )\right )\bigg | {\mstack {}{_{x \hiderel {=}0}}}=\frac {1}{2} \\ {} & \circ & x^{2}\cdot P_{3}\left (x \right )\textrm {is analytic at}\hspace {3pt} x =0 \\ {} & {} & \left (x^{2}\cdot P_{3}\left (x \right )\right )\bigg | {\mstack {}{_{x \hiderel {=}0}}}=-\frac {1}{4} v^{2}-\frac {1}{4} v \\ {} & \circ & x =0\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}=0 \\ \bullet & {} & \textrm {Multiply by denominators}\hspace {3pt} \\ {} & {} & 4 \left (\frac {d}{d x}y^{\prime }\right ) x^{2} \left (x -1\right )^{2}+2 \left (3 x -1\right ) x \left (x -1\right ) y^{\prime }+\left (-a^{2} x +v^{2} x -v^{2}+v x -v \right ) y=0 \\ \bullet & {} & \textrm {Assume series solution for}\hspace {3pt} y \\ {} & {} & y=\moverset {\infty }{\munderset {k =0}{\sum }}a_{k} x^{k +r} \\ \square & {} & \textrm {Rewrite ODE with series expansions}\hspace {3pt} \\ {} & \circ & \textrm {Convert}\hspace {3pt} x^{m}\cdot y\hspace {3pt}\textrm {to series expansion for}\hspace {3pt} m =0..1 \\ {} & {} & x^{m}\cdot y=\moverset {\infty }{\munderset {k =0}{\sum }}a_{k} x^{k +r +m} \\ {} & \circ & \textrm {Shift index using}\hspace {3pt} k \mathrm {->}k -m \\ {} & {} & x^{m}\cdot y=\moverset {\infty }{\munderset {k =m}{\sum }}a_{k -m} x^{k +r} \\ {} & \circ & \textrm {Convert}\hspace {3pt} x^{m}\cdot y^{\prime }\hspace {3pt}\textrm {to series expansion for}\hspace {3pt} m =1..3 \\ {} & {} & x^{m}\cdot y^{\prime }=\moverset {\infty }{\munderset {k =0}{\sum }}a_{k} \left (k +r \right ) x^{k +r -1+m} \\ {} & \circ & \textrm {Shift index using}\hspace {3pt} k \mathrm {->}k +1-m \\ {} & {} & x^{m}\cdot y^{\prime }=\moverset {\infty }{\munderset {k =-1+m}{\sum }}a_{k +1-m} \left (k +1-m +r \right ) x^{k +r} \\ {} & \circ & \textrm {Convert}\hspace {3pt} x^{m}\cdot \left (\frac {d}{d x}y^{\prime }\right )\hspace {3pt}\textrm {to series expansion for}\hspace {3pt} m =2..4 \\ {} & {} & x^{m}\cdot \left (\frac {d}{d x}y^{\prime }\right )=\moverset {\infty }{\munderset {k =0}{\sum }}a_{k} \left (k +r \right ) \left (k +r -1\right ) x^{k +r -2+m} \\ {} & \circ & \textrm {Shift index using}\hspace {3pt} k \mathrm {->}k +2-m \\ {} & {} & x^{m}\cdot \left (\frac {d}{d x}y^{\prime }\right )=\moverset {\infty }{\munderset {k =-2+m}{\sum }}a_{k +2-m} \left (k +2-m +r \right ) \left (k +1-m +r \right ) x^{k +r} \\ & {} & \textrm {Rewrite ODE with series expansions}\hspace {3pt} \\ {} & {} & a_{0} \left (v +2 r \right ) \left (-v -1+2 r \right ) x^{r}+\left (a_{1} \left (v +2 r +2\right ) \left (-v +1+2 r \right )-a_{0} \left (a^{2}+8 r^{2}-v^{2}-v \right )\right ) x^{1+r}+\left (\moverset {\infty }{\munderset {k =2}{\sum }}\left (a_{k} \left (v +2 r +2 k \right ) \left (-v -1+2 r +2 k \right )-a_{k -1} \left (a^{2}+8 \left (k -1\right )^{2}+16 \left (k -1\right ) r +8 r^{2}-v^{2}-v \right )+2 a_{k -2} \left (k -2+r \right ) \left (2 k -3+2 r \right )\right ) x^{k +r}\right )=0 \\ \bullet & {} & a_{0}\textrm {cannot be 0 by assumption, giving the indicial equation}\hspace {3pt} \\ {} & {} & \left (v +2 r \right ) \left (-v -1+2 r \right )=0 \\ \bullet & {} & \textrm {Values of r that satisfy the indicial equation}\hspace {3pt} \\ {} & {} & r \in \left \{-\frac {v}{2}, \frac {v}{2}+\frac {1}{2}\right \} \\ \bullet & {} & \textrm {Each term must be 0}\hspace {3pt} \\ {} & {} & a_{1} \left (v +2 r +2\right ) \left (-v +1+2 r \right )-a_{0} \left (a^{2}+8 r^{2}-v^{2}-v \right )=0 \\ \bullet & {} & \textrm {Solve for the dependent coefficient(s)}\hspace {3pt} \\ {} & {} & a_{1}=\frac {a_{0} \left (a^{2}+8 r^{2}-v^{2}-v \right )}{4 r^{2}-v^{2}+6 r -v +2} \\ \bullet & {} & \textrm {Each term in the series must be 0, giving the recursion relation}\hspace {3pt} \\ {} & {} & a_{k} \left (v +2 r +2 k \right ) \left (-v -1+2 r +2 k \right )-8 \left (r^{2}+\left (2 k -2\right ) r -\frac {v^{2}}{8}+\frac {a^{2}}{8}+k^{2}-\frac {v}{8}-2 k +1\right ) a_{k -1}+4 a_{k -2} \left (k -2+r \right ) \left (k -\frac {3}{2}+r \right )=0 \\ \bullet & {} & \textrm {Shift index using}\hspace {3pt} k \mathrm {->}k +2 \\ {} & {} & a_{k +2} \left (v +2 r +2 k +4\right ) \left (-v +3+2 r +2 k \right )-8 \left (r^{2}+\left (2 k +2\right ) r -\frac {v^{2}}{8}+\frac {a^{2}}{8}+\left (k +2\right )^{2}-\frac {v}{8}-2 k -3\right ) a_{k +1}+4 a_{k} \left (k +r \right ) \left (k +\frac {1}{2}+r \right )=0 \\ \bullet & {} & \textrm {Recursion relation that defines series solution to ODE}\hspace {3pt} \\ {} & {} & a_{k +2}=\frac {a^{2} a_{k +1}-4 k^{2} a_{k}+8 k^{2} a_{k +1}-8 k r a_{k}+16 k r a_{k +1}-4 r^{2} a_{k}+8 r^{2} a_{k +1}-v^{2} a_{k +1}-2 k a_{k}+16 k a_{k +1}-2 r a_{k}+16 r a_{k +1}-v a_{k +1}+8 a_{k +1}}{\left (v +2 r +2 k +4\right ) \left (-v +3+2 r +2 k \right )} \\ \bullet & {} & \textrm {Recursion relation for}\hspace {3pt} r =-\frac {v}{2} \\ {} & {} & a_{k +2}=\frac {a^{2} a_{k +1}-4 k^{2} a_{k}+8 k^{2} a_{k +1}+4 k v a_{k}-8 k v a_{k +1}-a_{k} v^{2}+v^{2} a_{k +1}-2 k a_{k}+16 k a_{k +1}+a_{k} v -9 v a_{k +1}+8 a_{k +1}}{\left (2 k +4\right ) \left (-2 v +3+2 k \right )} \\ \bullet & {} & \textrm {Solution for}\hspace {3pt} r =-\frac {v}{2} \\ {} & {} & \left [y=\moverset {\infty }{\munderset {k =0}{\sum }}a_{k} x^{k -\frac {v}{2}}, a_{k +2}=\frac {a^{2} a_{k +1}-4 k^{2} a_{k}+8 k^{2} a_{k +1}+4 k v a_{k}-8 k v a_{k +1}-a_{k} v^{2}+v^{2} a_{k +1}-2 k a_{k}+16 k a_{k +1}+a_{k} v -9 v a_{k +1}+8 a_{k +1}}{\left (2 k +4\right ) \left (-2 v +3+2 k \right )}, a_{1}=\frac {a_{0} \left (a^{2}+v^{2}-v \right )}{-4 v +2}\right ] \\ \bullet & {} & \textrm {Recursion relation for}\hspace {3pt} r =\frac {v}{2}+\frac {1}{2} \\ {} & {} & a_{k +2}=\frac {a^{2} a_{k +1}-4 k^{2} a_{k}+8 k^{2} a_{k +1}-8 k \left (\frac {v}{2}+\frac {1}{2}\right ) a_{k}+16 k \left (\frac {v}{2}+\frac {1}{2}\right ) a_{k +1}-4 \left (\frac {v}{2}+\frac {1}{2}\right )^{2} a_{k}+8 \left (\frac {v}{2}+\frac {1}{2}\right )^{2} a_{k +1}-v^{2} a_{k +1}-2 k a_{k}+16 k a_{k +1}-2 \left (\frac {v}{2}+\frac {1}{2}\right ) a_{k}+16 \left (\frac {v}{2}+\frac {1}{2}\right ) a_{k +1}-v a_{k +1}+8 a_{k +1}}{\left (2 v +5+2 k \right ) \left (2 k +4\right )} \\ \bullet & {} & \textrm {Solution for}\hspace {3pt} r =\frac {v}{2}+\frac {1}{2} \\ {} & {} & \left [y=\moverset {\infty }{\munderset {k =0}{\sum }}a_{k} x^{k +\frac {v}{2}+\frac {1}{2}}, a_{k +2}=\frac {a^{2} a_{k +1}-4 k^{2} a_{k}+8 k^{2} a_{k +1}-8 k \left (\frac {v}{2}+\frac {1}{2}\right ) a_{k}+16 k \left (\frac {v}{2}+\frac {1}{2}\right ) a_{k +1}-4 \left (\frac {v}{2}+\frac {1}{2}\right )^{2} a_{k}+8 \left (\frac {v}{2}+\frac {1}{2}\right )^{2} a_{k +1}-v^{2} a_{k +1}-2 k a_{k}+16 k a_{k +1}-2 \left (\frac {v}{2}+\frac {1}{2}\right ) a_{k}+16 \left (\frac {v}{2}+\frac {1}{2}\right ) a_{k +1}-v a_{k +1}+8 a_{k +1}}{\left (2 v +5+2 k \right ) \left (2 k +4\right )}, a_{1}=\frac {a_{0} \left (a^{2}+8 \left (\frac {v}{2}+\frac {1}{2}\right )^{2}-v^{2}-v \right )}{4 \left (\frac {v}{2}+\frac {1}{2}\right )^{2}-v^{2}+2 v +5}\right ] \\ \bullet & {} & \textrm {Combine solutions and rename parameters}\hspace {3pt} \\ {} & {} & \left [y=\left (\moverset {\infty }{\munderset {k =0}{\sum }}b_{k} x^{k -\frac {v}{2}}\right )+\left (\moverset {\infty }{\munderset {k =0}{\sum }}c_{k} x^{k +\frac {v}{2}+\frac {1}{2}}\right ), b_{k +2}=\frac {a^{2} b_{k +1}-4 k^{2} b_{k}+8 k^{2} b_{k +1}+4 k v b_{k}-8 k v b_{k +1}-v^{2} b_{k}+v^{2} b_{k +1}-2 k b_{k}+16 k b_{k +1}+v b_{k}-9 v b_{k +1}+8 b_{k +1}}{\left (2 k +4\right ) \left (-2 v +3+2 k \right )}, b_{1}=\frac {b_{0} \left (a^{2}+v^{2}-v \right )}{-4 v +2}, c_{k +2}=\frac {a^{2} c_{k +1}-4 k^{2} c_{k}+8 k^{2} c_{k +1}-8 k \left (\frac {v}{2}+\frac {1}{2}\right ) c_{k}+16 k \left (\frac {v}{2}+\frac {1}{2}\right ) c_{k +1}-4 \left (\frac {v}{2}+\frac {1}{2}\right )^{2} c_{k}+8 \left (\frac {v}{2}+\frac {1}{2}\right )^{2} c_{k +1}-v^{2} c_{k +1}-2 k c_{k}+16 k c_{k +1}-2 \left (\frac {v}{2}+\frac {1}{2}\right ) c_{k}+16 \left (\frac {v}{2}+\frac {1}{2}\right ) c_{k +1}-v c_{k +1}+8 c_{k +1}}{\left (2 v +5+2 k \right ) \left (2 k +4\right )}, c_{1}=\frac {c_{0} \left (a^{2}+8 \left (\frac {v}{2}+\frac {1}{2}\right )^{2}-v^{2}-v \right )}{4 \left (\frac {v}{2}+\frac {1}{2}\right )^{2}-v^{2}+2 v +5}\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 <- heuristic approach successful <- hypergeometric successful <- special function solution successful`
✓ Solution by Maple
Time used: 0.156 (sec). Leaf size: 77
dsolve(diff(diff(y(x),x),x) = -1/2/x*(3*x-1)/(x-1)*diff(y(x),x)-1/4*(v*(v+1)*(x-1)-a^2*x)/x^2/(x-1)^2*y(x),y(x), singsol=all)
\[ y \left (x \right ) = \left (x -1\right )^{-\frac {a}{2}} \left (\operatorname {hypergeom}\left (\left [-\frac {v}{2}-\frac {a}{2}, \frac {1}{2}-\frac {v}{2}-\frac {a}{2}\right ], \left [\frac {1}{2}-v \right ], x\right ) x^{-\frac {v}{2}} c_{1} +c_{2} \sqrt {x}\, x^{\frac {v}{2}} \operatorname {hypergeom}\left (\left [1+\frac {v}{2}-\frac {a}{2}, \frac {1}{2}+\frac {v}{2}-\frac {a}{2}\right ], \left [\frac {3}{2}+v \right ], x\right )\right ) \]
✓ Solution by Mathematica
Time used: 0.342 (sec). Leaf size: 109
DSolve[y''[x] == -1/4*((v*(1 + v)*(-1 + x) - a^2*x)*y[x])/((-1 + x)^2*x^2) - ((-1 + 3*x)*y'[x])/(2*(-1 + x)*x),y[x],x,IncludeSingularSolutions -> True]
\[ y(x)\to \frac {(-1)^{-v} (x-1)^{\frac {a+1}{2}} x^{-v/2} \left (c_1 (-1)^v x^{v+\frac {1}{2}} \operatorname {Hypergeometric2F1}\left (\frac {1}{2} (a+v+1),\frac {1}{2} (a+v+2),v+\frac {3}{2},x\right )-i c_2 \operatorname {Hypergeometric2F1}\left (\frac {a-v}{2},\frac {1}{2} (a-v+1),\frac {1}{2}-v,x\right )\right )}{\sqrt {1-x}} \]