Internal problem ID [9589]
Internal file name [OUTPUT/8530_Monday_June_06_2022_03_39_45_AM_72135444/index.tex]
Book: Differential Gleichungen, E. Kamke, 3rd ed. Chelsea Pub. NY, 1948
Section: Chapter 2, linear second order
Problem number: 1261.
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.
\[ \boxed {x \left (x +2\right ) y^{\prime \prime }+2 \left (n +1+\left (n +1-2 l \right ) x -l \,x^{2}\right ) y^{\prime }+\left (2 l \left (p -n -1\right ) x +2 p l +m \right ) y=0} \]
\[ \begin {array}{lll} & {} & \textrm {Let's solve}\hspace {3pt} \\ {} & {} & \left (x^{2}+2 x \right ) \left (\frac {d}{d x}y^{\prime }\right )+\left (-2 l \,x^{2}+\left (-4 l +2 n +2\right ) x +2 n +2\right ) y^{\prime }+\left (\left (\left (-2 n +2 p -2\right ) x +2 p \right ) l +m \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 (2 l x n -2 l x p -2 p l +2 l x -m \right ) y}{x \left (x +2\right )}+\frac {2 \left (l \,x^{2}+2 l x -x n -n -x -1\right ) y^{\prime }}{x \left (x +2\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 {2 \left (l \,x^{2}+2 l x -x n -n -x -1\right ) y^{\prime }}{x \left (x +2\right )}-\frac {\left (2 l x n -2 l x p -2 p l +2 l x -m \right ) y}{x \left (x +2\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 \left (l \,x^{2}+2 l x -x n -n -x -1\right )}{x \left (x +2\right )}, P_{3}\left (x \right )=-\frac {2 l x n -2 l x p -2 p l +2 l x -m}{x \left (x +2\right )}\right ] \\ {} & \circ & \left (x +2\right )\cdot P_{2}\left (x \right )\textrm {is analytic at}\hspace {3pt} x =-2 \\ {} & {} & \left (\left (x +2\right )\cdot P_{2}\left (x \right )\right )\bigg | {\mstack {}{_{x \hiderel {=}-2}}}=n +1 \\ {} & \circ & \left (x +2\right )^{2}\cdot P_{3}\left (x \right )\textrm {is analytic at}\hspace {3pt} x =-2 \\ {} & {} & \left (\left (x +2\right )^{2}\cdot P_{3}\left (x \right )\right )\bigg | {\mstack {}{_{x \hiderel {=}-2}}}=0 \\ {} & \circ & x =-2\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}=-2 \\ \bullet & {} & \textrm {Multiply by denominators}\hspace {3pt} \\ {} & {} & x \left (x +2\right ) \left (\frac {d}{d x}y^{\prime }\right )+\left (-2 l \,x^{2}-4 l x +2 x n +2 n +2 x +2\right ) y^{\prime }+\left (-2 l x n +2 l x p +2 p l -2 l x +m \right ) y=0 \\ \bullet & {} & \textrm {Change variables using}\hspace {3pt} x =u -2\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 l \,u^{2}+4 l u +2 n u -2 n +2 u -2\right ) \left (\frac {d}{d u}y \left (u \right )\right )+\left (-2 l n u +2 p l u +4 l n -2 p l -2 l u +4 l +m \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} u^{m}\cdot y \left (u \right )\hspace {3pt}\textrm {to series expansion for}\hspace {3pt} m =0..1 \\ {} & {} & u^{m}\cdot y \left (u \right )=\moverset {\infty }{\munderset {k =0}{\sum }}a_{k} u^{k +r +m} \\ {} & \circ & \textrm {Shift index using}\hspace {3pt} k \mathrm {->}k -m \\ {} & {} & u^{m}\cdot y \left (u \right )=\moverset {\infty }{\munderset {k =m}{\sum }}a_{k -m} u^{k +r} \\ {} & \circ & \textrm {Convert}\hspace {3pt} u^{m}\cdot \left (\frac {d}{d u}y \left (u \right )\right )\hspace {3pt}\textrm {to series expansion for}\hspace {3pt} m =0..2 \\ {} & {} & u^{m}\cdot \left (\frac {d}{d u}y \left (u \right )\right )=\moverset {\infty }{\munderset {k =0}{\sum }}a_{k} \left (k +r \right ) u^{k +r -1+m} \\ {} & \circ & \textrm {Shift index using}\hspace {3pt} k \mathrm {->}k +1-m \\ {} & {} & u^{m}\cdot \left (\frac {d}{d u}y \left (u \right )\right )=\moverset {\infty }{\munderset {k =-1+m}{\sum }}a_{k +1-m} \left (k +1-m +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} \\ {} & {} & -2 a_{0} r \left (r +n \right ) u^{-1+r}+\left (-2 a_{1} \left (1+r \right ) \left (1+r +n \right )+a_{0} \left (4 l n -2 p l +4 l r +2 n r +r^{2}+4 l +m +r \right )\right ) u^{r}+\left (\moverset {\infty }{\munderset {k =1}{\sum }}\left (-2 a_{k +1} \left (k +1+r \right ) \left (k +1+r +n \right )+a_{k} \left (k^{2}+4 k l +2 k n +2 k r +4 l n -2 p l +4 l r +2 n r +r^{2}+k +4 l +m +r \right )-2 a_{k -1} l \left (k +r +n -p \right )\right ) u^{k +r}\right )=0 \\ \bullet & {} & a_{0}\textrm {cannot be 0 by assumption, giving the indicial equation}\hspace {3pt} \\ {} & {} & -2 r \left (r +n \right )=0 \\ \bullet & {} & \textrm {Values of r that satisfy the indicial equation}\hspace {3pt} \\ {} & {} & r \in \left \{0, -n \right \} \\ \bullet & {} & \textrm {Each term must be 0}\hspace {3pt} \\ {} & {} & -2 a_{1} \left (1+r \right ) \left (1+r +n \right )+a_{0} \left (4 l n -2 p l +4 l r +2 n r +r^{2}+4 l +m +r \right )=0 \\ \bullet & {} & \textrm {Each term in the series must be 0, giving the recursion relation}\hspace {3pt} \\ {} & {} & -2 a_{k +1} \left (k +1+r \right ) \left (k +1+r +n \right )+\left (k^{2}+\left (4 l +2 n +2 r +1\right ) k +\left (4 n -2 p +4 r +4\right ) l +r^{2}+\left (1+2 n \right ) r +m \right ) a_{k}-2 a_{k -1} l \left (k +r +n -p \right )=0 \\ \bullet & {} & \textrm {Shift index using}\hspace {3pt} k \mathrm {->}k +1 \\ {} & {} & -2 a_{k +2} \left (k +2+r \right ) \left (k +2+r +n \right )+\left (\left (k +1\right )^{2}+\left (4 l +2 n +2 r +1\right ) \left (k +1\right )+\left (4 n -2 p +4 r +4\right ) l +r^{2}+\left (1+2 n \right ) r +m \right ) a_{k +1}-2 a_{k} l \left (k +r +n -p +1\right )=0 \\ \bullet & {} & \textrm {Recursion relation that defines series solution to ODE}\hspace {3pt} \\ {} & {} & a_{k +2}=\frac {k^{2} a_{k +1}-2 k l a_{k}+4 k l a_{k +1}+2 k n a_{k +1}+2 k r a_{k +1}-2 a_{k} l n +4 l n a_{k +1}+2 a_{k} p l -2 l p a_{k +1}-2 l r a_{k}+4 l r a_{k +1}+2 n r a_{k +1}+r^{2} a_{k +1}+3 k a_{k +1}-2 a_{k} l +8 l a_{k +1}+m a_{k +1}+2 n a_{k +1}+3 r a_{k +1}+2 a_{k +1}}{2 \left (k +2+r \right ) \left (k +2+r +n \right )} \\ \bullet & {} & \textrm {Recursion relation for}\hspace {3pt} r =0 \\ {} & {} & a_{k +2}=\frac {k^{2} a_{k +1}-2 k l a_{k}+4 k l a_{k +1}+2 k n a_{k +1}-2 a_{k} l n +4 l n a_{k +1}+2 a_{k} p l -2 l p a_{k +1}+3 k a_{k +1}-2 a_{k} l +8 l a_{k +1}+m a_{k +1}+2 n a_{k +1}+2 a_{k +1}}{2 \left (k +2\right ) \left (k +2+n \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 +2}=\frac {k^{2} a_{k +1}-2 k l a_{k}+4 k l a_{k +1}+2 k n a_{k +1}-2 a_{k} l n +4 l n a_{k +1}+2 a_{k} p l -2 l p a_{k +1}+3 k a_{k +1}-2 a_{k} l +8 l a_{k +1}+m a_{k +1}+2 n a_{k +1}+2 a_{k +1}}{2 \left (k +2\right ) \left (k +2+n \right )}, -2 a_{1} \left (n +1\right )+a_{0} \left (4 l n -2 p l +4 l +m \right )=0\right ] \\ \bullet & {} & \textrm {Revert the change of variables}\hspace {3pt} u =x +2 \\ {} & {} & \left [y=\moverset {\infty }{\munderset {k =0}{\sum }}a_{k} \left (x +2\right )^{k}, a_{k +2}=\frac {k^{2} a_{k +1}-2 k l a_{k}+4 k l a_{k +1}+2 k n a_{k +1}-2 a_{k} l n +4 l n a_{k +1}+2 a_{k} p l -2 l p a_{k +1}+3 k a_{k +1}-2 a_{k} l +8 l a_{k +1}+m a_{k +1}+2 n a_{k +1}+2 a_{k +1}}{2 \left (k +2\right ) \left (k +2+n \right )}, -2 a_{1} \left (n +1\right )+a_{0} \left (4 l n -2 p l +4 l +m \right )=0\right ] \\ \bullet & {} & \textrm {Recursion relation for}\hspace {3pt} r =-n \\ {} & {} & a_{k +2}=\frac {k^{2} a_{k +1}-2 k l a_{k}+4 k l a_{k +1}+2 a_{k} p l -2 l p a_{k +1}-n^{2} a_{k +1}+3 k a_{k +1}-2 a_{k} l +8 l a_{k +1}+m a_{k +1}-n a_{k +1}+2 a_{k +1}}{2 \left (k +2-n \right ) \left (k +2\right )} \\ \bullet & {} & \textrm {Solution for}\hspace {3pt} r =-n \\ {} & {} & \left [y \left (u \right )=\moverset {\infty }{\munderset {k =0}{\sum }}a_{k} u^{k -n}, a_{k +2}=\frac {k^{2} a_{k +1}-2 k l a_{k}+4 k l a_{k +1}+2 a_{k} p l -2 l p a_{k +1}-n^{2} a_{k +1}+3 k a_{k +1}-2 a_{k} l +8 l a_{k +1}+m a_{k +1}-n a_{k +1}+2 a_{k +1}}{2 \left (k +2-n \right ) \left (k +2\right )}, -2 a_{1} \left (-n +1\right )+a_{0} \left (-2 p l -n^{2}+4 l +m -n \right )=0\right ] \\ \bullet & {} & \textrm {Revert the change of variables}\hspace {3pt} u =x +2 \\ {} & {} & \left [y=\moverset {\infty }{\munderset {k =0}{\sum }}a_{k} \left (x +2\right )^{k -n}, a_{k +2}=\frac {k^{2} a_{k +1}-2 k l a_{k}+4 k l a_{k +1}+2 a_{k} p l -2 l p a_{k +1}-n^{2} a_{k +1}+3 k a_{k +1}-2 a_{k} l +8 l a_{k +1}+m a_{k +1}-n a_{k +1}+2 a_{k +1}}{2 \left (k +2-n \right ) \left (k +2\right )}, -2 a_{1} \left (-n +1\right )+a_{0} \left (-2 p l -n^{2}+4 l +m -n \right )=0\right ] \\ \bullet & {} & \textrm {Combine solutions and rename parameters}\hspace {3pt} \\ {} & {} & \left [y=\left (\moverset {\infty }{\munderset {k =0}{\sum }}a_{k} \left (x +2\right )^{k}\right )+\left (\moverset {\infty }{\munderset {k =0}{\sum }}b_{k} \left (x +2\right )^{k -n}\right ), a_{k +2}=\frac {k^{2} a_{k +1}-2 k l a_{k}+4 k l a_{k +1}+2 k n a_{k +1}-2 l n a_{k}+4 l n a_{k +1}+2 l p a_{k}-2 l p a_{k +1}+3 k a_{k +1}-2 l a_{k}+8 l a_{k +1}+m a_{k +1}+2 n a_{k +1}+2 a_{k +1}}{2 \left (k +2\right ) \left (k +2+n \right )}, -2 a_{1} \left (n +1\right )+a_{0} \left (4 l n -2 p l +4 l +m \right )=0, b_{k +2}=\frac {k^{2} b_{k +1}-2 k l b_{k}+4 k l b_{k +1}+2 l p b_{k}-2 l p b_{k +1}-n^{2} b_{k +1}+3 k b_{k +1}-2 l b_{k}+8 l b_{k +1}+m b_{k +1}-n b_{k +1}+2 b_{k +1}}{2 \left (k +2-n \right ) \left (k +2\right )}, -2 b_{1} \left (-n +1\right )+b_{0} \left (-2 p l -n^{2}+4 l +m -n \right )=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 -> hyper3: Equivalence to 2F1, 1F1 or 0F1 under a power @ Moebius -> Mathieu -> Equivalence to the rational form of Mathieu ODE under a power @ Moebius trying a solution in terms of MeijerG functions -> Heun: Equivalence to the GHE or one of its 4 confluent cases under a power @ Moebius <- Heun successful: received ODE is equivalent to the HeunC ODE, case a <> 0, e <> 0, c = 0 `
✓ Solution by Maple
Time used: 0.5 (sec). Leaf size: 106
dsolve(x*(x+2)*diff(diff(y(x),x),x)+2*(n+1+(n+1-2*l)*x-l*x^2)*diff(y(x),x)+(2*l*(p-n-1)*x+2*p*l+m)*y(x)=0,y(x), singsol=all)
\[ y \left (x \right ) = \frac {\sqrt {2}\, \sqrt {-2-x}\, \left (x +2\right )^{-\frac {n}{2}-\frac {1}{2}} \left (-\frac {x}{2}-1\right )^{\frac {n}{2}} \left (c_{2} x^{-n} \operatorname {HeunC}\left (4 l , -n , n , -4 p l , 2 \left (n +1+p \right ) l -\frac {n^{2}}{2}+m -n , -\frac {x}{2}\right )+c_{1} \operatorname {HeunC}\left (4 l , n , n , -4 p l , 2 \left (n +1+p \right ) l -\frac {n^{2}}{2}+m -n , -\frac {x}{2}\right )\right )}{2} \]
✓ Solution by Mathematica
Time used: 0.661 (sec). Leaf size: 120
DSolve[(m + 2*l*p + 2*l*(-1 - n + p)*x)*y[x] + 2*(1 + n + (1 - 2*l + n)*x - l*x^2)*y'[x] + x*(2 + x)*y''[x] == 0,y[x],x,IncludeSingularSolutions -> True]
\[ y(x)\to \left (-\frac {x}{2}-1\right )^{\frac {n+1}{2}} x^{-n} (x+2)^{-\frac {n}{2}-\frac {1}{2}} \left (c_2 \text {HeunC}\left [-4 l n-2 l p-m+n^2+n,-4 l (p-1),1-n,n+1,4 l,-\frac {x}{2}\right ]+c_1 x^n \text {HeunC}\left [-2 l p-m,4 l (n-p+1),n+1,n+1,4 l,-\frac {x}{2}\right ]\right ) \]