Internal problem ID [9661]
Internal file name [OUTPUT/8603_Monday_June_06_2022_04_24_08_AM_70934534/index.tex]
Book: Differential Gleichungen, E. Kamke, 3rd ed. Chelsea Pub. NY, 1948
Section: Chapter 2, linear second order
Problem number: 1334.
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.
\[ \boxed {y^{\prime \prime }+\frac {\left (\left (a +1\right ) x -1\right ) y^{\prime }}{x \left (x -1\right )}+\frac {\left (\left (a^{2}-b^{2}\right ) x +c^{2}\right ) y}{4 x^{2} \left (x -1\right )}=0} \]
\[ \begin {array}{lll} & {} & \textrm {Let's solve}\hspace {3pt} \\ {} & {} & \left (4 x^{3}-4 x^{2}\right ) \left (\frac {d}{d x}y^{\prime }\right )+\left (4 a \,x^{2}+4 x^{2}-4 x \right ) y^{\prime }+\left (\left (a^{2}-b^{2}\right ) x +c^{2}\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 -b^{2} x +c^{2}\right ) y}{4 x^{2} \left (x -1\right )}-\frac {\left (x a +x -1\right ) y^{\prime }}{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 (x a +x -1\right ) y^{\prime }}{x \left (x -1\right )}+\frac {\left (a^{2} x -b^{2} x +c^{2}\right ) y}{4 x^{2} \left (x -1\right )}=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 a +x -1}{x \left (x -1\right )}, P_{3}\left (x \right )=\frac {a^{2} x -b^{2} x +c^{2}}{4 x^{2} \left (x -1\right )}\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}}}=1 \\ {} & \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 {c^{2}}{4} \\ {} & \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 )+4 \left (x a +x -1\right ) x y^{\prime }+\left (a^{2} x -b^{2} x +c^{2}\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..2 \\ {} & {} & 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..3 \\ {} & {} & 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 (c +2 r \right ) \left (c -2 r \right ) x^{r}+\left (\moverset {\infty }{\munderset {k =1}{\sum }}\left (a_{k} \left (c +2 k +2 r \right ) \left (c -2 k -2 r \right )+a_{k -1} \left (a +b +2 k -2+2 r \right ) \left (a -b +2 k -2+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 (c +2 r \right ) \left (c -2 r \right )=0 \\ \bullet & {} & \textrm {Values of r that satisfy the indicial equation}\hspace {3pt} \\ {} & {} & r \in \left \{-\frac {c}{2}, \frac {c}{2}\right \} \\ \bullet & {} & \textrm {Each term in the series must be 0, giving the recursion relation}\hspace {3pt} \\ {} & {} & a_{k} \left (c +2 k +2 r \right ) \left (c -2 k -2 r \right )+a_{k -1} \left (a +b +2 k -2+2 r \right ) \left (a -b +2 k -2+2 r \right )=0 \\ \bullet & {} & \textrm {Shift index using}\hspace {3pt} k \mathrm {->}1+k \\ {} & {} & a_{1+k} \left (c +2+2 k +2 r \right ) \left (c -2-2 k -2 r \right )+a_{k} \left (a +b +2 k +2 r \right ) \left (a -b +2 k +2 r \right )=0 \\ \bullet & {} & \textrm {Recursion relation that defines series solution to ODE}\hspace {3pt} \\ {} & {} & a_{1+k}=-\frac {a_{k} \left (a +b +2 k +2 r \right ) \left (a -b +2 k +2 r \right )}{\left (c +2+2 k +2 r \right ) \left (c -2-2 k -2 r \right )} \\ \bullet & {} & \textrm {Recursion relation for}\hspace {3pt} r =-\frac {c}{2} \\ {} & {} & a_{1+k}=-\frac {a_{k} \left (a +b +2 k -c \right ) \left (a -b +2 k -c \right )}{\left (2+2 k \right ) \left (2 c -2-2 k \right )} \\ \bullet & {} & \textrm {Solution for}\hspace {3pt} r =-\frac {c}{2} \\ {} & {} & \left [y=\moverset {\infty }{\munderset {k =0}{\sum }}a_{k} x^{k -\frac {c}{2}}, a_{1+k}=-\frac {a_{k} \left (a +b +2 k -c \right ) \left (a -b +2 k -c \right )}{\left (2+2 k \right ) \left (2 c -2-2 k \right )}\right ] \\ \bullet & {} & \textrm {Recursion relation for}\hspace {3pt} r =\frac {c}{2} \\ {} & {} & a_{1+k}=-\frac {a_{k} \left (a +b +2 k +c \right ) \left (a -b +2 k +c \right )}{\left (2 c +2+2 k \right ) \left (-2-2 k \right )} \\ \bullet & {} & \textrm {Solution for}\hspace {3pt} r =\frac {c}{2} \\ {} & {} & \left [y=\moverset {\infty }{\munderset {k =0}{\sum }}a_{k} x^{k +\frac {c}{2}}, a_{1+k}=-\frac {a_{k} \left (a +b +2 k +c \right ) \left (a -b +2 k +c \right )}{\left (2 c +2+2 k \right ) \left (-2-2 k \right )}\right ] \\ \bullet & {} & \textrm {Combine solutions and rename parameters}\hspace {3pt} \\ {} & {} & \left [y=\left (\moverset {\infty }{\munderset {k =0}{\sum }}d_{k} x^{k -\frac {c}{2}}\right )+\left (\moverset {\infty }{\munderset {k =0}{\sum }}e_{k} x^{k +\frac {c}{2}}\right ), d_{k +1}=-\frac {d_{k} \left (a +b +2 k -c \right ) \left (a -b +2 k -c \right )}{\left (2 k +2\right ) \left (2 c -2-2 k \right )}, e_{k +1}=-\frac {e_{k} \left (a +b +2 k +c \right ) \left (a -b +2 k +c \right )}{\left (2 c +2+2 k \right ) \left (-2-2 k \right )}\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: 0.172 (sec). Leaf size: 89
dsolve(diff(diff(y(x),x),x) = -((a+1)*x-1)/x/(x-1)*diff(y(x),x)-1/4*((a^2-b^2)*x+c^2)/x^2/(x-1)*y(x),y(x), singsol=all)
\[ y \left (x \right ) = \left (\operatorname {hypergeom}\left (\left [-\frac {a}{2}-\frac {b}{2}+\frac {c}{2}+1, -\frac {a}{2}+\frac {b}{2}+\frac {c}{2}+1\right ], \left [c +1\right ], x\right ) x^{\frac {c}{2}} c_{1} +\operatorname {hypergeom}\left (\left [-\frac {a}{2}-\frac {b}{2}-\frac {c}{2}+1, -\frac {a}{2}+\frac {b}{2}-\frac {c}{2}+1\right ], \left [-c +1\right ], x\right ) x^{-\frac {c}{2}} c_{2} \right ) \left (x -1\right )^{-a +1} \]
✓ Solution by Mathematica
Time used: 0.223 (sec). Leaf size: 89
DSolve[y''[x] == -1/4*((c^2 + (a^2 - b^2)*x)*y[x])/((-1 + x)*x^2) - ((-1 + (1 + a)*x)*y'[x])/((-1 + x)*x),y[x],x,IncludeSingularSolutions -> True]
\[ y(x)\to i^{-c} x^{-c/2} \left (i^{2 c} c_2 x^c \operatorname {Hypergeometric2F1}\left (\frac {1}{2} (a-b+c),\frac {1}{2} (a+b+c),c+1,x\right )+c_1 \operatorname {Hypergeometric2F1}\left (\frac {1}{2} (a-b-c),\frac {1}{2} (a+b-c),1-c,x\right )\right ) \]