Internal problem ID [11013]
Internal file name [OUTPUT/10270_Sunday_December_31_2023_11_33_36_AM_47036669/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 2.1.2-6 Equation of form
\((a_3 x^3+a_2 x^2 x+a_1 x+a_0) y''+f(x)y'+g(x)y=0\)
Problem number: 190.
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.
\[ \boxed {x \left (x^{2}+a \right ) y^{\prime \prime }+\left (x^{2} b +c \right ) y^{\prime }+s x y=0} \]
\[ \begin {array}{lll} & {} & \textrm {Let's solve}\hspace {3pt} \\ {} & {} & x \left (x^{2}+a \right ) \left (\frac {d}{d x}y^{\prime }\right )+\left (x^{2} b +c \right ) y^{\prime }+s x 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 (x^{2} b +c \right ) y^{\prime }}{x \left (x^{2}+a \right )}-\frac {s y}{x^{2}+a} \\ \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^{2} b +c \right ) y^{\prime }}{x \left (x^{2}+a \right )}+\frac {s y}{x^{2}+a}=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^{2} b +c}{x \left (x^{2}+a \right )}, P_{3}\left (x \right )=\frac {s}{x^{2}+a}\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^{2}+a \right ) \left (\frac {d}{d x}y^{\prime }\right )+\left (x^{2} b +c \right ) y^{\prime }+s x 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 \cdot y\hspace {3pt}\textrm {to series expansion}\hspace {3pt} \\ {} & {} & x \cdot y=\moverset {\infty }{\munderset {k =0}{\sum }}a_{k} x^{k +r +1} \\ {} & \circ & \textrm {Shift index using}\hspace {3pt} k \mathrm {->}k -1 \\ {} & {} & x \cdot y=\moverset {\infty }{\munderset {k =1}{\sum }}a_{k -1} x^{k +r} \\ {} & \circ & \textrm {Convert}\hspace {3pt} x^{m}\cdot y^{\prime }\hspace {3pt}\textrm {to series expansion for}\hspace {3pt} m =0..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 =1..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} r \left (a r -a +c \right ) x^{r -1}+a_{1} \left (1+r \right ) \left (a r +c \right ) x^{r}+\left (\moverset {\infty }{\munderset {k =1}{\sum }}\left (a_{k +1} \left (k +r +1\right ) \left (a \left (k +1\right )+a r -a +c \right )+a_{k -1} \left (b \left (k -1\right )+b r +\left (k -1\right )^{2}+2 \left (k -1\right ) r +r^{2}-k +1-r +s \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 must be 0}\hspace {3pt} \\ {} & {} & a_{1} \left (1+r \right ) \left (a r +c \right )=0 \\ \bullet & {} & \textrm {Each term in the series must be 0, giving the recursion relation}\hspace {3pt} \\ {} & {} & \left (k^{2}+\left (b +2 r -3\right ) k +r^{2}+\left (b -3\right ) r -b +s +2\right ) a_{k -1}+a_{k +1} \left (k +r +1\right ) \left (a k +a r +c \right )=0 \\ \bullet & {} & \textrm {Shift index using}\hspace {3pt} k \mathrm {->}k +1 \\ {} & {} & \left (\left (k +1\right )^{2}+\left (b +2 r -3\right ) \left (k +1\right )+r^{2}+\left (b -3\right ) r -b +s +2\right ) a_{k}+a_{k +2} \left (k +2+r \right ) \left (a \left (k +1\right )+a r +c \right )=0 \\ \bullet & {} & \textrm {Recursion relation that defines series solution to ODE}\hspace {3pt} \\ {} & {} & a_{k +2}=-\frac {\left (b k +b r +k^{2}+2 k r +r^{2}-k -r +s \right ) a_{k}}{\left (k +2+r \right ) \left (a k +a r +a +c \right )} \\ \bullet & {} & \textrm {Recursion relation for}\hspace {3pt} r =0 \\ {} & {} & a_{k +2}=-\frac {\left (b k +k^{2}-k +s \right ) a_{k}}{\left (k +2\right ) \left (a k +a +c \right )} \\ \bullet & {} & \textrm {Solution for}\hspace {3pt} r =0 \\ {} & {} & \left [y=\moverset {\infty }{\munderset {k =0}{\sum }}a_{k} x^{k}, a_{k +2}=-\frac {\left (b k +k^{2}-k +s \right ) a_{k}}{\left (k +2\right ) \left (a k +a +c \right )}, a_{1} c =0\right ] \\ \bullet & {} & \textrm {Recursion relation for}\hspace {3pt} r =\frac {a -c}{a} \\ {} & {} & a_{k +2}=-\frac {\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}}-k -\frac {a -c}{a}+s \right ) a_{k}}{\left (k +2+\frac {a -c}{a}\right ) \left (a k +2 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 +2}=-\frac {\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}}-k -\frac {a -c}{a}+s \right ) a_{k}}{\left (k +2+\frac {a -c}{a}\right ) \left (a k +2 a \right )}, a_{1} \left (1+\frac {a -c}{a}\right ) a =0\right ] \\ \bullet & {} & \textrm {Combine solutions and rename parameters}\hspace {3pt} \\ {} & {} & \left [y=\left (\moverset {\infty }{\munderset {k =0}{\sum }}d_{k} x^{k}\right )+\left (\moverset {\infty }{\munderset {k =0}{\sum }}e_{k} x^{k +\frac {a -c}{a}}\right ), d_{k +2}=-\frac {\left (b k +k^{2}-k +s \right ) d_{k}}{\left (k +2\right ) \left (a k +a +c \right )}, d_{1} c =0, e_{k +2}=-\frac {\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}}-k -\frac {a -c}{a}+s \right ) e_{k}}{\left (k +2+\frac {a -c}{a}\right ) \left (a k +2 a \right )}, e_{1} \left (1+\frac {a -c}{a}\right ) a =0\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: 0.157 (sec). Leaf size: 175
dsolve(x*(x^2+a)*diff(y(x),x$2)+(b*x^2+c)*diff(y(x),x)+s*x*y(x)=0,y(x), singsol=all)
\[ y \left (x \right ) = \left (x^{2}+a \right )^{\frac {\left (-b +2\right ) a +c}{2 a}} \left (x^{\frac {a -c}{a}} \operatorname {hypergeom}\left (\left [-\frac {b}{4}+\frac {5}{4}-\frac {\sqrt {b^{2}-2 b -4 s +1}}{4}, -\frac {b}{4}+\frac {5}{4}+\frac {\sqrt {b^{2}-2 b -4 s +1}}{4}\right ], \left [\frac {3 a -c}{2 a}\right ], -\frac {x^{2}}{a}\right ) c_{1} +\operatorname {hypergeom}\left (\left [-\frac {b}{4}+\frac {3}{4}+\frac {c}{2 a}+\frac {\sqrt {b^{2}-2 b -4 s +1}}{4}, -\frac {\sqrt {b^{2}-2 b -4 s +1}}{4}-\frac {b}{4}+\frac {3}{4}+\frac {c}{2 a}\right ], \left [\frac {1}{2}+\frac {c}{2 a}\right ], -\frac {x^{2}}{a}\right ) c_{2} \right ) \]
✓ Solution by Mathematica
Time used: 0.967 (sec). Leaf size: 185
DSolve[x*(x^2+a)*y''[x]+(b*x^2+c)*y'[x]+s*x*y[x]==0,y[x],x,IncludeSingularSolutions -> True]
\[ y(x)\to c_2 a^{\frac {1}{2} \left (\frac {c}{a}-1\right )} x^{1-\frac {c}{a}} \operatorname {Hypergeometric2F1}\left (\frac {a \left (b+\sqrt {b^2-2 b-4 s+1}+1\right )-2 c}{4 a},\frac {b a-\sqrt {b^2-2 b-4 s+1} a+a-2 c}{4 a},\frac {3}{2}-\frac {c}{2 a},-\frac {x^2}{a}\right )+c_1 \operatorname {Hypergeometric2F1}\left (\frac {1}{4} \left (b-\sqrt {b^2-2 b-4 s+1}-1\right ),\frac {1}{4} \left (b+\sqrt {b^2-2 b-4 s+1}-1\right ),\frac {a+c}{2 a},-\frac {x^2}{a}\right ) \]