Internal problem ID [10981]
Internal file name [OUTPUT/10238_Sunday_December_31_2023_11_10_39_AM_64591386/index.tex]
Book: Handbook of exact solutions for ordinary differential equations. By Polyanin and Zaitsev.
Second edition
Section: Chapter 2, Second-Order Differential Equations. section 2.1.2-5 Equation of form
\((a x^2+b x+c) y''+f(x)y'+g(x)y=0\)
Problem number: 158.
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.
\[ \boxed {\left (x^{2}-1\right ) y^{\prime \prime }+\left (2 a +1\right ) y^{\prime }-b \left (2 a +b \right ) y=0} \]
\[ \begin {array}{lll} & {} & \textrm {Let's solve}\hspace {3pt} \\ {} & {} & \left (x^{2}-1\right ) \left (\frac {d}{d x}y^{\prime }\right )+\left (2 a +1\right ) y^{\prime }+\left (-2 b a -b^{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 {b \left (2 a +b \right ) y}{x^{2}-1}-\frac {\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 {\left (2 a +1\right ) y^{\prime }}{x^{2}-1}-\frac {b \left (2 a +b \right ) 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 {2 a +1}{x^{2}-1}, P_{3}\left (x \right )=-\frac {b \left (2 a +b \right )}{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}}}=-a -\frac {1}{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 (2 a +1\right ) y^{\prime }-b \left (2 a +b \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^{2}-2 u \right ) \left (\frac {d}{d u}\frac {d}{d u}y \left (u \right )\right )+\left (2 a +1\right ) \left (\frac {d}{d u}y \left (u \right )\right )+\left (-2 b a -b^{2}\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} \frac {d}{d u}y \left (u \right )\hspace {3pt}\textrm {to series expansion}\hspace {3pt} \\ {} & {} & \frac {d}{d u}y \left (u \right )=\moverset {\infty }{\munderset {k =0}{\sum }}a_{k} \left (k +r \right ) u^{k +r -1} \\ {} & \circ & \textrm {Shift index using}\hspace {3pt} k \mathrm {->}k +1 \\ {} & {} & \frac {d}{d u}y \left (u \right )=\moverset {\infty }{\munderset {k =-1}{\sum }}a_{k +1} \left (k +1+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 (3-2 r +2 a \right ) u^{-1+r}+\left (\moverset {\infty }{\munderset {k =0}{\sum }}\left (a_{k +1} \left (k +1+r \right ) \left (-2 k +1-2 r +2 a \right )-a_{k} \left (2 b a +b^{2}-k^{2}-2 k r -r^{2}+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 (3-2 r +2 a \right )=0 \\ \bullet & {} & \textrm {Values of r that satisfy the indicial equation}\hspace {3pt} \\ {} & {} & r \in \left \{0, a +\frac {3}{2}\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 +1-2 r +2 a \right )+a_{k} \left (r^{2}+\left (2 k -1\right ) r -2 b a -b^{2}+k^{2}-k \right )=0 \\ \bullet & {} & \textrm {Recursion relation that defines series solution to ODE}\hspace {3pt} \\ {} & {} & a_{k +1}=\frac {a_{k} \left (2 b a +b^{2}-k^{2}-2 k r -r^{2}+k +r \right )}{\left (k +1+r \right ) \left (-2 k +1-2 r +2 a \right )} \\ \bullet & {} & \textrm {Recursion relation for}\hspace {3pt} r =0 \\ {} & {} & a_{k +1}=\frac {a_{k} \left (2 b a +b^{2}-k^{2}+k \right )}{\left (k +1\right ) \left (-2 k +1+2 a \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 {a_{k} \left (2 b a +b^{2}-k^{2}+k \right )}{\left (k +1\right ) \left (-2 k +1+2 a \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 {a_{k} \left (2 b a +b^{2}-k^{2}+k \right )}{\left (k +1\right ) \left (-2 k +1+2 a \right )}\right ] \\ \bullet & {} & \textrm {Recursion relation for}\hspace {3pt} r =a +\frac {3}{2} \\ {} & {} & a_{k +1}=\frac {a_{k} \left (2 b a +b^{2}-k^{2}-2 k \left (a +\frac {3}{2}\right )-\left (a +\frac {3}{2}\right )^{2}+k +a +\frac {3}{2}\right )}{\left (k +\frac {5}{2}+a \right ) \left (-2 k -2\right )} \\ \bullet & {} & \textrm {Solution for}\hspace {3pt} r =a +\frac {3}{2} \\ {} & {} & \left [y \left (u \right )=\moverset {\infty }{\munderset {k =0}{\sum }}a_{k} u^{k +a +\frac {3}{2}}, a_{k +1}=\frac {a_{k} \left (2 b a +b^{2}-k^{2}-2 k \left (a +\frac {3}{2}\right )-\left (a +\frac {3}{2}\right )^{2}+k +a +\frac {3}{2}\right )}{\left (k +\frac {5}{2}+a \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 +a +\frac {3}{2}}, a_{k +1}=\frac {a_{k} \left (2 b a +b^{2}-k^{2}-2 k \left (a +\frac {3}{2}\right )-\left (a +\frac {3}{2}\right )^{2}+k +a +\frac {3}{2}\right )}{\left (k +\frac {5}{2}+a \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 }}c_{k} \left (x +1\right )^{k}\right )+\left (\moverset {\infty }{\munderset {k =0}{\sum }}d_{k} \left (x +1\right )^{k +a +\frac {3}{2}}\right ), c_{k +1}=\frac {c_{k} \left (2 b a +b^{2}-k^{2}+k \right )}{\left (k +1\right ) \left (-2 k +1+2 a \right )}, d_{k +1}=\frac {d_{k} \left (2 b a +b^{2}-k^{2}-2 k \left (a +\frac {3}{2}\right )-\left (a +\frac {3}{2}\right )^{2}+k +a +\frac {3}{2}\right )}{\left (k +\frac {5}{2}+a \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: 0.078 (sec). Leaf size: 112
dsolve((x^2-1)*diff(y(x),x$2)+(2*a+1)*diff(y(x),x)-b*(2*a+b)*y(x)=0,y(x), singsol=all)
\[ y \left (x \right ) = c_{1} \operatorname {hypergeom}\left (\left [-\frac {1}{2}-\frac {\sqrt {8 a b +4 b^{2}+1}}{2}, \frac {\sqrt {8 a b +4 b^{2}+1}}{2}-\frac {1}{2}\right ], \left [-a -\frac {1}{2}\right ], \frac {1}{2}+\frac {x}{2}\right )+c_{2} \left (\frac {1}{2}+\frac {x}{2}\right )^{a +\frac {3}{2}} \operatorname {hypergeom}\left (\left [1-\frac {\sqrt {8 a b +4 b^{2}+1}}{2}+a , \frac {\sqrt {8 a b +4 b^{2}+1}}{2}+1+a \right ], \left [\frac {5}{2}+a \right ], \frac {1}{2}+\frac {x}{2}\right ) \]
✓ Solution by Mathematica
Time used: 0.304 (sec). Leaf size: 152
DSolve[(x^2-1)*y''[x]+(2*a+1)*y'[x]-b*(2*a+b)*y[x]==0,y[x],x,IncludeSingularSolutions -> True]
\[ y(x)\to 2^{a-\frac {1}{2}} c_2 (x-1)^{\frac {1}{2}-a} \operatorname {Hypergeometric2F1}\left (-a-\frac {1}{2} \sqrt {4 b^2+8 a b+1},\frac {1}{2} \sqrt {4 b^2+8 a b+1}-a,\frac {3}{2}-a,\frac {1}{2}-\frac {x}{2}\right )+c_1 \operatorname {Hypergeometric2F1}\left (\frac {1}{2} \left (-\sqrt {4 b^2+8 a b+1}-1\right ),\frac {1}{2} \left (\sqrt {4 b^2+8 a b+1}-1\right ),a+\frac {1}{2},\frac {1-x}{2}\right ) \]