Internal problem ID [9638]
Internal file name [OUTPUT/8580_Monday_June_06_2022_04_14_32_AM_28549211/index.tex]
Book: Differential Gleichungen, E. Kamke, 3rd ed. Chelsea Pub. NY, 1948
Section: Chapter 2, linear second order
Problem number: 1311.
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.
\[ \boxed {x \left (x^{2}+1\right ) y^{\prime \prime }+\left (2 x^{2}+1\right ) y^{\prime }-v \left (v +1\right ) x y=0} \]
\[ \begin {array}{lll} & {} & \textrm {Let's solve}\hspace {3pt} \\ {} & {} & \left (x^{3}+x \right ) \left (\frac {d}{d x}y^{\prime }\right )+\left (2 x^{2}+1\right ) y^{\prime }-v \left (v +1\right ) 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 {v \left (v +1\right ) y}{x^{2}+1}-\frac {\left (2 x^{2}+1\right ) y^{\prime }}{x \left (x^{2}+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 (2 x^{2}+1\right ) y^{\prime }}{x \left (x^{2}+1\right )}-\frac {v \left (v +1\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 x^{2}+1}{x \left (x^{2}+1\right )}, P_{3}\left (x \right )=-\frac {v \left (v +1\right )}{x^{2}+1}\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}}}=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}+1\right ) \left (\frac {d}{d x}y^{\prime }\right )+\left (2 x^{2}+1\right ) y^{\prime }-v \left (v +1\right ) 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^{2} x^{-1+r}+a_{1} \left (1+r \right )^{2} x^{r}+\left (\moverset {\infty }{\munderset {k =1}{\sum }}\left (a_{k +1} \left (k +r +1\right )^{2}+a_{k -1} \left (v +r +k \right ) \left (-v +r +k -1\right )\right ) x^{k +r}\right )=0 \\ \bullet & {} & a_{0}\textrm {cannot be 0 by assumption, giving the indicial equation}\hspace {3pt} \\ {} & {} & r^{2}=0 \\ \bullet & {} & \textrm {Values of r that satisfy the indicial equation}\hspace {3pt} \\ {} & {} & r =0 \\ \bullet & {} & \textrm {Each term must be 0}\hspace {3pt} \\ {} & {} & a_{1} \left (1+r \right )^{2}=0 \\ \bullet & {} & \textrm {Each term in the series must be 0, giving the recursion relation}\hspace {3pt} \\ {} & {} & a_{k +1} \left (k +1\right )^{2}+a_{k -1} \left (v +k \right ) \left (-v -1+k \right )=0 \\ \bullet & {} & \textrm {Shift index using}\hspace {3pt} k \mathrm {->}k +1 \\ {} & {} & a_{k +2} \left (k +2\right )^{2}+a_{k} \left (v +k +1\right ) \left (-v +k \right )=0 \\ \bullet & {} & \textrm {Recursion relation that defines series solution to ODE}\hspace {3pt} \\ {} & {} & a_{k +2}=-\frac {a_{k} \left (v +k +1\right ) \left (-v +k \right )}{\left (k +2\right )^{2}} \\ \bullet & {} & \textrm {Recursion relation for}\hspace {3pt} r =0 \\ {} & {} & a_{k +2}=-\frac {a_{k} \left (v +k +1\right ) \left (-v +k \right )}{\left (k +2\right )^{2}} \\ \bullet & {} & \textrm {Solution for}\hspace {3pt} r =0 \\ {} & {} & \left [y=\moverset {\infty }{\munderset {k =0}{\sum }}a_{k} x^{k}, a_{k +2}=-\frac {a_{k} \left (v +k +1\right ) \left (-v +k \right )}{\left (k +2\right )^{2}}, a_{1}=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 -> solution has integrals; searching for one without integrals... -> hyper3: Equivalence to 2F1, 1F1 or 0F1 under a power @ Moebius <- hyper3 successful: received ODE is equivalent to the 2F1 ODE <- hypergeometric solution without integrals succesful <- hypergeometric successful <- special function solution successful`
✓ Solution by Maple
Time used: 0.312 (sec). Leaf size: 52
dsolve(x*(x^2+1)*diff(diff(y(x),x),x)+(2*x^2+1)*diff(y(x),x)-v*(v+1)*x*y(x)=0,y(x), singsol=all)
\[ y \left (x \right ) = c_{1} \operatorname {hypergeom}\left (\left [-\frac {v}{2}, \frac {1}{2}+\frac {v}{2}\right ], \left [\frac {1}{2}\right ], x^{2}+1\right )+c_{2} \sqrt {x^{2}+1}\, \operatorname {hypergeom}\left (\left [1+\frac {v}{2}, \frac {1}{2}-\frac {v}{2}\right ], \left [\frac {3}{2}\right ], x^{2}+1\right ) \]
✓ Solution by Mathematica
Time used: 0.59 (sec). Leaf size: 61
DSolve[-(v*(1 + v)*x*y[x]) + (1 + 2*x^2)*y'[x] + x*(1 + x^2)*y''[x] == 0,y[x],x,IncludeSingularSolutions -> True]
\[ y(x)\to c_2 G_{2,2}^{2,0}\left (-x^2| \begin {array}{c} \frac {1-v}{2},\frac {v+2}{2} \\ 0,0 \\ \end {array} \right )+c_1 \operatorname {Hypergeometric2F1}\left (-\frac {v}{2},\frac {v+1}{2},1,-x^2\right ) \]