Internal problem ID [9577]
Internal file name [OUTPUT/8518_Monday_June_06_2022_03_36_26_AM_75516705/index.tex]
Book: Differential Gleichungen, E. Kamke, 3rd ed. Chelsea Pub. NY, 1948
Section: Chapter 2, linear second order
Problem number: 1249.
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.
\[ \boxed {\left (x^{2}-1\right ) y^{\prime \prime }+\left (a x +b \right ) y^{\prime }+c y=0} \]
\[ \begin {array}{lll} & {} & \textrm {Let's solve}\hspace {3pt} \\ {} & {} & \left (x^{2}-1\right ) \left (\frac {d}{d x}y^{\prime }\right )+\left (a x +b \right ) y^{\prime }+c 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 +b \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 {\left (a x +b \right ) y^{\prime }}{x^{2}-1}+\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 +b}{x^{2}-1}, 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 {b}{2}+\frac {a}{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} \\ {} & {} & \left (x^{2}-1\right ) \left (\frac {d}{d x}y^{\prime }\right )+\left (a x +b \right ) y^{\prime }+c 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^{2}-2 u \right ) \left (\frac {d}{d u}\frac {d}{d u}y \left (u \right )\right )+\left (a u -a +b \right ) \left (\frac {d}{d u}y \left (u \right )\right )+c 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 \left (\frac {d}{d u}y \left (u \right )\right )\hspace {3pt}\textrm {to series expansion for}\hspace {3pt} m =0..1 \\ {} & {} & 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..2 \\ {} & {} & 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 (\moverset {\infty }{\munderset {k =0}{\sum }}\left (-a_{k +1} \left (k +1+r \right ) \left (2 k +2 r +a -b \right )+a_{k} \left (a k +a r +k^{2}+2 k r +r^{2}+c -k -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 {b}{2}-\frac {a}{2}+1\right \} \\ \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 (k^{2}+\left (a +2 r -1\right ) k +r^{2}+\left (a -1\right ) r +c \right ) a_{k}=0 \\ \bullet & {} & \textrm {Recursion relation that defines series solution to ODE}\hspace {3pt} \\ {} & {} & a_{k +1}=\frac {\left (a k +a r +k^{2}+2 k r +r^{2}+c -k -r \right ) a_{k}}{\left (k +1+r \right ) \left (2 k +2 r +a -b \right )} \\ \bullet & {} & \textrm {Recursion relation for}\hspace {3pt} r =0 \\ {} & {} & a_{k +1}=\frac {\left (a k +k^{2}+c -k \right ) a_{k}}{\left (k +1\right ) \left (2 k +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 +1}=\frac {\left (a k +k^{2}+c -k \right ) a_{k}}{\left (k +1\right ) \left (2 k +a -b \right )}\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 +1}=\frac {\left (a k +k^{2}+c -k \right ) a_{k}}{\left (k +1\right ) \left (2 k +a -b \right )}\right ] \\ \bullet & {} & \textrm {Recursion relation for}\hspace {3pt} r =\frac {b}{2}-\frac {a}{2}+1 \\ {} & {} & a_{k +1}=\frac {\left (a k +a \left (\frac {b}{2}-\frac {a}{2}+1\right )+k^{2}+2 k \left (\frac {b}{2}-\frac {a}{2}+1\right )+\left (\frac {b}{2}-\frac {a}{2}+1\right )^{2}+c -k -\frac {b}{2}+\frac {a}{2}-1\right ) a_{k}}{\left (k +2+\frac {b}{2}-\frac {a}{2}\right ) \left (2 k +2\right )} \\ \bullet & {} & \textrm {Solution for}\hspace {3pt} r =\frac {b}{2}-\frac {a}{2}+1 \\ {} & {} & \left [y \left (u \right )=\moverset {\infty }{\munderset {k =0}{\sum }}a_{k} u^{k +\frac {b}{2}-\frac {a}{2}+1}, a_{k +1}=\frac {\left (a k +a \left (\frac {b}{2}-\frac {a}{2}+1\right )+k^{2}+2 k \left (\frac {b}{2}-\frac {a}{2}+1\right )+\left (\frac {b}{2}-\frac {a}{2}+1\right )^{2}+c -k -\frac {b}{2}+\frac {a}{2}-1\right ) a_{k}}{\left (k +2+\frac {b}{2}-\frac {a}{2}\right ) \left (2 k +2\right )}\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 {b}{2}-\frac {a}{2}+1}, a_{k +1}=\frac {\left (a k +a \left (\frac {b}{2}-\frac {a}{2}+1\right )+k^{2}+2 k \left (\frac {b}{2}-\frac {a}{2}+1\right )+\left (\frac {b}{2}-\frac {a}{2}+1\right )^{2}+c -k -\frac {b}{2}+\frac {a}{2}-1\right ) a_{k}}{\left (k +2+\frac {b}{2}-\frac {a}{2}\right ) \left (2 k +2\right )}\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 {b}{2}-\frac {a}{2}+1}\right ), d_{k +1}=\frac {\left (a k +k^{2}+c -k \right ) d_{k}}{\left (k +1\right ) \left (2 k +a -b \right )}, e_{k +1}=\frac {\left (a k +a \left (\frac {b}{2}-\frac {a}{2}+1\right )+k^{2}+2 k \left (\frac {b}{2}-\frac {a}{2}+1\right )+\left (\frac {b}{2}-\frac {a}{2}+1\right )^{2}+c -k -\frac {b}{2}+\frac {a}{2}-1\right ) e_{k}}{\left (k +2+\frac {b}{2}-\frac {a}{2}\right ) \left (2 k +2\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: 0.125 (sec). Leaf size: 134
dsolve((x^2-1)*diff(diff(y(x),x),x)+(a*x+b)*diff(y(x),x)+c*y(x)=0,y(x), singsol=all)
\[ y \left (x \right ) = c_{1} \operatorname {hypergeom}\left (\left [-\frac {1}{2}-\frac {\sqrt {a^{2}-2 a -4 c +1}}{2}+\frac {a}{2}, -\frac {1}{2}+\frac {\sqrt {a^{2}-2 a -4 c +1}}{2}+\frac {a}{2}\right ], \left [\frac {a}{2}-\frac {b}{2}\right ], \frac {x}{2}+\frac {1}{2}\right )+c_{2} \left (\frac {x}{2}+\frac {1}{2}\right )^{1-\frac {a}{2}+\frac {b}{2}} \operatorname {hypergeom}\left (\left [\frac {1}{2}-\frac {\sqrt {a^{2}-2 a -4 c +1}}{2}+\frac {b}{2}, \frac {1}{2}+\frac {\sqrt {a^{2}-2 a -4 c +1}}{2}+\frac {b}{2}\right ], \left [2-\frac {a}{2}+\frac {b}{2}\right ], \frac {x}{2}+\frac {1}{2}\right ) \]
✓ Solution by Mathematica
Time used: 0.204 (sec). Leaf size: 190
DSolve[c*y[x] + (b + a*x)*y'[x] + (-1 + x^2)*y''[x] == 0,y[x],x,IncludeSingularSolutions -> True]
\[ y(x)\to \frac {1}{2} (x-1)^{\frac {1}{2} (-a-b)} \left (2 c_1 (x-1)^{\frac {a+b}{2}} \operatorname {Hypergeometric2F1}\left (\frac {1}{2} \left (a-\sqrt {a^2-2 a-4 c+1}-1\right ),\frac {1}{2} \left (a+\sqrt {a^2-2 a-4 c+1}-1\right ),\frac {a+b}{2},\frac {1-x}{2}\right )+c_2 (x-1) 2^{\frac {a+b}{2}} \operatorname {Hypergeometric2F1}\left (\frac {1}{2} \left (-b-\sqrt {a^2-2 a-4 c+1}+1\right ),\frac {1}{2} \left (-b+\sqrt {a^2-2 a-4 c+1}+1\right ),\frac {1}{2} (-a-b+4),\frac {1-x}{2}\right )\right ) \]