Internal problem ID [9573]
Internal file name [OUTPUT/8514_Monday_June_06_2022_03_35_27_AM_81681120/index.tex]
Book: Differential Gleichungen, E. Kamke, 3rd ed. Chelsea Pub. NY, 1948
Section: Chapter 2, linear second order
Problem number: 1245.
ODE order: 2.
ODE degree: 1.
The type(s) of ODE detected by this program : "unknown"
Maple gives the following as the ode type
[_Gegenbauer]
Unable to solve or complete the solution.
\[ \boxed {\left (x^{2}-1\right ) y^{\prime \prime }-2 \left (n -1\right ) x y^{\prime }-\left (v -n +1\right ) \left (v +n \right ) y=0} \]
\[ \begin {array}{lll} & {} & \textrm {Let's solve}\hspace {3pt} \\ {} & {} & \left (x^{2}-1\right ) \left (\frac {d}{d x}y^{\prime }\right )+\left (-2 n x +2 x \right ) y^{\prime }+\left (n^{2}-v^{2}-n -v \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 (n^{2}-v^{2}-n -v \right ) y}{x^{2}-1}+\frac {2 \left (n -1\right ) x y^{\prime }}{x^{2}-1} \\ \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 (n -1\right ) x y^{\prime }}{x^{2}-1}+\frac {\left (n^{2}-v^{2}-n -v \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 \left (n -1\right )}{x^{2}-1}, P_{3}\left (x \right )=\frac {n^{2}-v^{2}-n -v}{x^{2}-1}\right ] \\ {} & \circ & \left (x +1\right )\cdot P_{2}\left (x \right )\textrm {is analytic at}\hspace {3pt} x =-1 \\ {} & {} & \left (\left (x +1\right )\cdot P_{2}\left (x \right )\right )\bigg | {\mstack {}{_{x \hiderel {=}-1}}}=-n +1 \\ {} & \circ & \left (x +1\right )^{2}\cdot P_{3}\left (x \right )\textrm {is analytic at}\hspace {3pt} x =-1 \\ {} & {} & \left (\left (x +1\right )^{2}\cdot P_{3}\left (x \right )\right )\bigg | {\mstack {}{_{x \hiderel {=}-1}}}=0 \\ {} & \circ & x =-1\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}=-1 \\ \bullet & {} & \textrm {Multiply by denominators}\hspace {3pt} \\ {} & {} & \left (x^{2}-1\right ) \left (\frac {d}{d x}y^{\prime }\right )-2 \left (n -1\right ) x y^{\prime }+\left (n^{2}-v^{2}-n -v \right ) y=0 \\ \bullet & {} & \textrm {Change variables using}\hspace {3pt} x =u -1\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 n u +2 n +2 u -2\right ) \left (\frac {d}{d u}y \left (u \right )\right )+\left (n^{2}-v^{2}-n -v \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 \left (\frac {d}{d u}y \left (u \right )\right )\hspace {3pt}\textrm {to series expansion for}\hspace {3pt} m =0..1 \\ {} & {} & 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 (\moverset {\infty }{\munderset {k =0}{\sum }}\left (-2 a_{k +1} \left (k +1+r \right ) \left (k +1+r -n \right )+a_{k} \left (k -n +r +v +1\right ) \left (-v +r -n +k \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 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 )+a_{k} \left (k -n +r +v +1\right ) \left (-v +r -n +k \right )=0 \\ \bullet & {} & \textrm {Recursion relation that defines series solution to ODE}\hspace {3pt} \\ {} & {} & a_{k +1}=\frac {a_{k} \left (k -n +r +v +1\right ) \left (-v +r -n +k \right )}{2 \left (k +1+r \right ) \left (k +1+r -n \right )} \\ \bullet & {} & \textrm {Recursion relation for}\hspace {3pt} r =0 \\ {} & {} & a_{k +1}=\frac {a_{k} \left (k -n +1+v \right ) \left (-v -n +k \right )}{2 \left (k +1\right ) \left (k +1-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 +1}=\frac {a_{k} \left (k -n +1+v \right ) \left (-v -n +k \right )}{2 \left (k +1\right ) \left (k +1-n \right )}\right ] \\ \bullet & {} & \textrm {Revert the change of variables}\hspace {3pt} u =x +1 \\ {} & {} & \left [y=\moverset {\infty }{\munderset {k =0}{\sum }}a_{k} \left (x +1\right )^{k}, a_{k +1}=\frac {a_{k} \left (k -n +1+v \right ) \left (-v -n +k \right )}{2 \left (k +1\right ) \left (k +1-n \right )}\right ] \\ \bullet & {} & \textrm {Recursion relation for}\hspace {3pt} r =n \\ {} & {} & a_{k +1}=\frac {a_{k} \left (k +v +1\right ) \left (-v +k \right )}{2 \left (k +1+n \right ) \left (k +1\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 +1}=\frac {a_{k} \left (k +v +1\right ) \left (-v +k \right )}{2 \left (k +1+n \right ) \left (k +1\right )}\right ] \\ \bullet & {} & \textrm {Revert the change of variables}\hspace {3pt} u =x +1 \\ {} & {} & \left [y=\moverset {\infty }{\munderset {k =0}{\sum }}a_{k} \left (x +1\right )^{k +n}, a_{k +1}=\frac {a_{k} \left (k +v +1\right ) \left (-v +k \right )}{2 \left (k +1+n \right ) \left (k +1\right )}\right ] \\ \bullet & {} & \textrm {Combine solutions and rename parameters}\hspace {3pt} \\ {} & {} & \left [y=\left (\moverset {\infty }{\munderset {k =0}{\sum }}a_{k} \left (x +1\right )^{k}\right )+\left (\moverset {\infty }{\munderset {k =0}{\sum }}b_{k} \left (x +1\right )^{k +n}\right ), a_{k +1}=\frac {a_{k} \left (k -n +1+v \right ) \left (-v -n +k \right )}{2 \left (k +1\right ) \left (k +1-n \right )}, b_{k +1}=\frac {b_{k} \left (k +v +1\right ) \left (-v +k \right )}{2 \left (k +1+n \right ) \left (k +1\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 <- Legendre successful <- special function solution successful`
✓ Solution by Maple
Time used: 0.078 (sec). Leaf size: 27
dsolve((x^2-1)*diff(diff(y(x),x),x)-2*(n-1)*x*diff(y(x),x)-(v-n+1)*(v+n)*y(x)=0,y(x), singsol=all)
\[ y \left (x \right ) = \left (\operatorname {LegendreP}\left (v , n , x\right ) c_{1} +\operatorname {LegendreQ}\left (v , n , x\right ) c_{2} \right ) \left (x^{2}-1\right )^{\frac {n}{2}} \]
✓ Solution by Mathematica
Time used: 0.043 (sec). Leaf size: 32
DSolve[(-1 + n - v)*(n + v)*y[x] - 2*(-1 + n)*x*y'[x] + (-1 + x^2)*y''[x] == 0,y[x],x,IncludeSingularSolutions -> True]
\[ y(x)\to \left (x^2-1\right )^{n/2} (c_1 P_v^n(x)+c_2 Q_v^n(x)) \]