Internal problem ID [9683]
Internal file name [OUTPUT/8625_Monday_June_06_2022_04_30_11_AM_69995561/index.tex]
Book: Differential Gleichungen, E. Kamke, 3rd ed. Chelsea Pub. NY, 1948
Section: Chapter 2, linear second order
Problem number: 1356.
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.
\[ \boxed {y^{\prime \prime }+\frac {\left (2 x^{2}+1\right ) y^{\prime }}{x \left (x^{2}+1\right )}+\frac {\left (-v \left (v +1\right ) x^{2}-n^{2}\right ) y}{x^{2} \left (x^{2}+1\right )}=0} \]
\[ \begin {array}{lll} & {} & \textrm {Let's solve}\hspace {3pt} \\ {} & {} & \left (x^{4}+x^{2}\right ) \left (\frac {d}{d x}y^{\prime }\right )+\left (2 x^{3}+x \right ) y^{\prime }+\left (\left (-v^{2}-v \right ) x^{2}-n^{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 {\left (v^{2} x^{2}+v \,x^{2}+n^{2}\right ) y}{x^{2} \left (x^{2}+1\right )}-\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 {\left (v^{2} x^{2}+v \,x^{2}+n^{2}\right ) y}{x^{2} \left (x^{2}+1\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 {2 x^{2}+1}{x \left (x^{2}+1\right )}, P_{3}\left (x \right )=-\frac {v^{2} x^{2}+v \,x^{2}+n^{2}}{x^{2} \left (x^{2}+1\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}}}=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}}}=-n^{2} \\ {} & \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} \\ {} & {} & \left (\frac {d}{d x}y^{\prime }\right ) x^{2} \left (x^{2}+1\right )+\left (2 x^{2}+1\right ) x y^{\prime }+\left (-v^{2} x^{2}-v \,x^{2}-n^{2}\right ) 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\hspace {3pt}\textrm {to series expansion for}\hspace {3pt} m =0..2 \\ {} & {} & x^{m}\cdot y=\moverset {\infty }{\munderset {k =0}{\sum }}a_{k} x^{k +r +m} \\ {} & \circ & \textrm {Shift index using}\hspace {3pt} k \mathrm {->}k -m \\ {} & {} & x^{m}\cdot y=\moverset {\infty }{\munderset {k =m}{\sum }}a_{k -m} x^{k +r} \\ {} & \circ & \textrm {Convert}\hspace {3pt} x^{m}\cdot y^{\prime }\hspace {3pt}\textrm {to series expansion for}\hspace {3pt} m =1..3 \\ {} & {} & 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 =2..4 \\ {} & {} & 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} \left (n +r \right ) \left (-n +r \right ) x^{r}+a_{1} \left (1+n +r \right ) \left (1-n +r \right ) x^{1+r}+\left (\moverset {\infty }{\munderset {k =2}{\sum }}\left (a_{k} \left (k +n +r \right ) \left (k -n +r \right )+a_{k -2} \left (v -1+r +k \right ) \left (-v +r +k -2\right )\right ) x^{k +r}\right )=0 \\ \bullet & {} & a_{0}\textrm {cannot be 0 by assumption, giving the indicial equation}\hspace {3pt} \\ {} & {} & \left (n +r \right ) \left (-n +r \right )=0 \\ \bullet & {} & \textrm {Values of r that satisfy the indicial equation}\hspace {3pt} \\ {} & {} & r \in \left \{n , -n \right \} \\ \bullet & {} & \textrm {Each term must be 0}\hspace {3pt} \\ {} & {} & a_{1} \left (1+n +r \right ) \left (1-n +r \right )=0 \\ \bullet & {} & \textrm {Solve for the dependent coefficient(s)}\hspace {3pt} \\ {} & {} & a_{1}=0 \\ \bullet & {} & \textrm {Each term in the series must be 0, giving the recursion relation}\hspace {3pt} \\ {} & {} & a_{k} \left (k +n +r \right ) \left (k -n +r \right )+a_{k -2} \left (v -1+r +k \right ) \left (-v +r +k -2\right )=0 \\ \bullet & {} & \textrm {Shift index using}\hspace {3pt} k \mathrm {->}k +2 \\ {} & {} & a_{k +2} \left (k +2+n +r \right ) \left (k +2-n +r \right )+a_{k} \left (v +1+r +k \right ) \left (-v +r +k \right )=0 \\ \bullet & {} & \textrm {Recursion relation that defines series solution to ODE}\hspace {3pt} \\ {} & {} & a_{k +2}=-\frac {a_{k} \left (v +1+r +k \right ) \left (-v +r +k \right )}{\left (k +2+n +r \right ) \left (k +2-n +r \right )} \\ \bullet & {} & \textrm {Recursion relation for}\hspace {3pt} r =n \\ {} & {} & a_{k +2}=-\frac {a_{k} \left (v +1+n +k \right ) \left (-v +n +k \right )}{\left (k +2+2 n \right ) \left (k +2\right )} \\ \bullet & {} & \textrm {Solution for}\hspace {3pt} r =n \\ {} & {} & \left [y=\moverset {\infty }{\munderset {k =0}{\sum }}a_{k} x^{k +n}, a_{k +2}=-\frac {a_{k} \left (v +1+n +k \right ) \left (-v +n +k \right )}{\left (k +2+2 n \right ) \left (k +2\right )}, a_{1}=0\right ] \\ \bullet & {} & \textrm {Recursion relation for}\hspace {3pt} r =-n \\ {} & {} & a_{k +2}=-\frac {a_{k} \left (v +1-n +k \right ) \left (-v -n +k \right )}{\left (k +2\right ) \left (k +2-2 n \right )} \\ \bullet & {} & \textrm {Solution for}\hspace {3pt} r =-n \\ {} & {} & \left [y=\moverset {\infty }{\munderset {k =0}{\sum }}a_{k} x^{k -n}, a_{k +2}=-\frac {a_{k} \left (v +1-n +k \right ) \left (-v -n +k \right )}{\left (k +2\right ) \left (k +2-2 n \right )}, a_{1}=0\right ] \\ \bullet & {} & \textrm {Combine solutions and rename parameters}\hspace {3pt} \\ {} & {} & \left [y=\left (\moverset {\infty }{\munderset {k =0}{\sum }}a_{k} x^{k +n}\right )+\left (\moverset {\infty }{\munderset {k =0}{\sum }}b_{k} x^{k -n}\right ), a_{k +2}=-\frac {a_{k} \left (v +1+n +k \right ) \left (-v +n +k \right )}{\left (k +2+2 n \right ) \left (k +2\right )}, a_{1}=0, b_{k +2}=-\frac {b_{k} \left (v +1-n +k \right ) \left (-v -n +k \right )}{\left (k +2\right ) \left (k +2-2 n \right )}, b_{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 <- hypergeometric successful <- special function solution successful`
✓ Solution by Maple
Time used: 0.188 (sec). Leaf size: 29
dsolve(diff(diff(y(x),x),x) = -(2*x^2+1)/x/(x^2+1)*diff(y(x),x)-(-v*(v+1)*x^2-n^2)/x^2/(x^2+1)*y(x),y(x), singsol=all)
\[ y \left (x \right ) = \operatorname {LegendreQ}\left (v , n , \sqrt {x^{2}+1}\right ) c_{2} +\operatorname {LegendreP}\left (v , n , \sqrt {x^{2}+1}\right ) c_{1} \]
✓ Solution by Mathematica
Time used: 0.351 (sec). Leaf size: 78
DSolve[y''[x] == -(((-n^2 - v*(1 + v)*x^2)*y[x])/(x^2*(1 + x^2))) - ((1 + 2*x^2)*y'[x])/(x*(1 + x^2)),y[x],x,IncludeSingularSolutions -> True]
\[ y(x)\to c_1 x^{-n} \operatorname {Hypergeometric2F1}\left (\frac {1}{2} (-n-v),\frac {1}{2} (-n+v+1),1-n,-x^2\right )+c_2 x^n \operatorname {Hypergeometric2F1}\left (\frac {n-v}{2},\frac {1}{2} (n+v+1),n+1,-x^2\right ) \]