Internal problem ID [10995]
Internal file name [OUTPUT/10252_Sunday_December_31_2023_11_24_13_AM_550928/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: 172.
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.
\[ \boxed {x \left (x +a \right ) y^{\prime \prime }+\left (b x +c \right ) y^{\prime }+d y=0} \]
\[ \begin {array}{lll} & {} & \textrm {Let's solve}\hspace {3pt} \\ {} & {} & x \left (x +a \right ) \left (\frac {d}{d x}y^{\prime }\right )+\left (b x +c \right ) y^{\prime }+d 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 {d y}{x \left (x +a \right )}-\frac {\left (b x +c \right ) y^{\prime }}{x \left (x +a \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 (b x +c \right ) y^{\prime }}{x \left (x +a \right )}+\frac {d y}{x \left (x +a \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 {b x +c}{x \left (x +a \right )}, P_{3}\left (x \right )=\frac {d}{x \left (x +a \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}}}=\frac {c}{a} \\ {} & \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}}}=0 \\ {} & \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} \\ {} & {} & x \left (x +a \right ) \left (\frac {d}{d x}y^{\prime }\right )+\left (b x +c \right ) y^{\prime }+d 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^{\prime }\hspace {3pt}\textrm {to series expansion for}\hspace {3pt} m =0..1 \\ {} & {} & 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 =1..2 \\ {} & {} & 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} r \left (a r -a +c \right ) x^{r -1}+\left (\moverset {\infty }{\munderset {k =0}{\sum }}\left (a_{k +1} \left (k +1+r \right ) \left (a \left (k +1\right )+a r -a +c \right )+a_{k} \left (b k +b r +k^{2}+2 k r +r^{2}+d -k -r \right )\right ) x^{k +r}\right )=0 \\ \bullet & {} & a_{0}\textrm {cannot be 0 by assumption, giving the indicial equation}\hspace {3pt} \\ {} & {} & r \left (a r -a +c \right )=0 \\ \bullet & {} & \textrm {Values of r that satisfy the indicial equation}\hspace {3pt} \\ {} & {} & r \in \left \{0, \frac {a -c}{a}\right \} \\ \bullet & {} & \textrm {Each term in the series must be 0, giving the recursion relation}\hspace {3pt} \\ {} & {} & \left (k +1+r \right ) \left (a k +a r +c \right ) a_{k +1}+a_{k} \left (k^{2}+\left (b +2 r -1\right ) k +r^{2}+\left (b -1\right ) r +d \right )=0 \\ \bullet & {} & \textrm {Recursion relation that defines series solution to ODE}\hspace {3pt} \\ {} & {} & a_{k +1}=-\frac {a_{k} \left (b k +b r +k^{2}+2 k r +r^{2}+d -k -r \right )}{\left (k +1+r \right ) \left (a k +a r +c \right )} \\ \bullet & {} & \textrm {Recursion relation for}\hspace {3pt} r =0 \\ {} & {} & a_{k +1}=-\frac {a_{k} \left (b k +k^{2}+d -k \right )}{\left (k +1\right ) \left (a k +c \right )} \\ \bullet & {} & \textrm {Solution for}\hspace {3pt} r =0 \\ {} & {} & \left [y=\moverset {\infty }{\munderset {k =0}{\sum }}a_{k} x^{k}, a_{k +1}=-\frac {a_{k} \left (b k +k^{2}+d -k \right )}{\left (k +1\right ) \left (a k +c \right )}\right ] \\ \bullet & {} & \textrm {Recursion relation for}\hspace {3pt} r =\frac {a -c}{a} \\ {} & {} & a_{k +1}=-\frac {a_{k} \left (b k +\frac {b \left (a -c \right )}{a}+k^{2}+\frac {2 k \left (a -c \right )}{a}+\frac {\left (a -c \right )^{2}}{a^{2}}+d -k -\frac {a -c}{a}\right )}{\left (k +1+\frac {a -c}{a}\right ) \left (a k +a \right )} \\ \bullet & {} & \textrm {Solution for}\hspace {3pt} r =\frac {a -c}{a} \\ {} & {} & \left [y=\moverset {\infty }{\munderset {k =0}{\sum }}a_{k} x^{k +\frac {a -c}{a}}, a_{k +1}=-\frac {a_{k} \left (b k +\frac {b \left (a -c \right )}{a}+k^{2}+\frac {2 k \left (a -c \right )}{a}+\frac {\left (a -c \right )^{2}}{a^{2}}+d -k -\frac {a -c}{a}\right )}{\left (k +1+\frac {a -c}{a}\right ) \left (a k +a \right )}\right ] \\ \bullet & {} & \textrm {Combine solutions and rename parameters}\hspace {3pt} \\ {} & {} & \left [y=\left (\moverset {\infty }{\munderset {k =0}{\sum }}e_{k} x^{k}\right )+\left (\moverset {\infty }{\munderset {k =0}{\sum }}f_{k} x^{k +\frac {a -c}{a}}\right ), e_{k +1}=-\frac {e_{k} \left (b k +k^{2}+d -k \right )}{\left (k +1\right ) \left (a k +c \right )}, f_{k +1}=-\frac {f_{k} \left (b k +\frac {b \left (a -c \right )}{a}+k^{2}+\frac {2 k \left (a -c \right )}{a}+\frac {\left (a -c \right )^{2}}{a^{2}}+d -k -\frac {a -c}{a}\right )}{\left (k +1+\frac {a -c}{a}\right ) \left (a k +a \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: 0.109 (sec). Leaf size: 230
dsolve(x*(x+a)*diff(y(x),x$2)+(b*x+c)*diff(y(x),x)+d*y(x)=0,y(x), singsol=all)
\[ y \left (x \right ) = c_{2} \left (\operatorname {csgn}\left (a \right ) a +a +2 x \right )^{-\frac {\left (\left (b -2\right ) \operatorname {csgn}\left (a \right ) a +a b -2 c \right ) \operatorname {csgn}\left (a \right )}{2 a}} \operatorname {hypergeom}\left (\left [\frac {\operatorname {csgn}\left (a \right ) \left (\operatorname {csgn}\left (a \right ) a +\sqrt {b^{2}-2 b -4 d +1}\, \operatorname {csgn}\left (a \right ) a -a b +2 c \right )}{2 a}, -\frac {\operatorname {csgn}\left (a \right ) \left (\sqrt {b^{2}-2 b -4 d +1}\, \operatorname {csgn}\left (a \right ) a -\operatorname {csgn}\left (a \right ) a +a b -2 c \right )}{2 a}\right ], \left [-\frac {\operatorname {csgn}\left (a \right ) \left (\left (b -4\right ) \operatorname {csgn}\left (a \right ) a +a b -2 c \right )}{2 a}\right ], \frac {\operatorname {csgn}\left (a \right ) \left (\operatorname {csgn}\left (a \right ) a +a +2 x \right )}{2 a}\right )+c_{1} \operatorname {hypergeom}\left (\left [-\frac {1}{2}+\frac {b}{2}-\frac {\sqrt {b^{2}-2 b -4 d +1}}{2}, -\frac {1}{2}+\frac {b}{2}+\frac {\sqrt {b^{2}-2 b -4 d +1}}{2}\right ], \left [\frac {\left (b \,\operatorname {csgn}\left (a \right ) a +a b -2 c \right ) \operatorname {csgn}\left (a \right )}{2 a}\right ], \frac {\operatorname {csgn}\left (a \right ) \left (\operatorname {csgn}\left (a \right ) a +a +2 x \right )}{2 a}\right ) \]
✓ Solution by Mathematica
Time used: 0.423 (sec). Leaf size: 165
DSolve[x*(x+a)*y''[x]+(b*x+c)*y'[x]+d*y[x]==0,y[x],x,IncludeSingularSolutions -> True]
\[ y(x)\to c_2 a^{\frac {c}{a}-1} x^{1-\frac {c}{a}} \operatorname {Hypergeometric2F1}\left (\frac {1}{2} \left (b-\frac {2 c}{a}+\sqrt {b^2-2 b-4 d+1}+1\right ),\frac {b a-\sqrt {b^2-2 b-4 d+1} a+a-2 c}{2 a},2-\frac {c}{a},-\frac {x}{a}\right )+c_1 \operatorname {Hypergeometric2F1}\left (\frac {1}{2} \left (b-\sqrt {b^2-2 b-4 d+1}-1\right ),\frac {1}{2} \left (b+\sqrt {b^2-2 b-4 d+1}-1\right ),\frac {c}{a},-\frac {x}{a}\right ) \]