Internal problem ID [9687]
Internal file name [OUTPUT/8629_Monday_June_06_2022_04_31_36_AM_30355396/index.tex]
Book: Differential Gleichungen, E. Kamke, 3rd ed. Chelsea Pub. NY, 1948
Section: Chapter 2, linear second order
Problem number: 1360.
ODE order: 2.
ODE degree: 1.
The type(s) of ODE detected by this program : "second_order_bessel_ode"
Maple gives the following as the ode type
[[_2nd_order, _with_linear_symmetries]]
\[ \boxed {y^{\prime \prime }+\frac {2 x y^{\prime }}{x^{2}-1}-\frac {v \left (v +1\right ) y}{x^{2}}=0} \]
Writing the ode as \begin {align*} y^{\prime \prime } x^{2}+2 y^{\prime } x +\left (-v^{2}-v +\frac {v^{2}}{x^{2}}+\frac {v}{x^{2}}\right ) y = 0\tag {1} \end {align*}
Bessel ode has the form \begin {align*} y^{\prime \prime } x^{2}+y^{\prime } x +\left (-n^{2}+x^{2}\right ) y = 0\tag {2} \end {align*}
The generalized form of Bessel ode is given by Bowman (1958) as the following \begin {align*} y^{\prime \prime } x^{2}+\left (1-2 \alpha \right ) x y^{\prime }+\left (\beta ^{2} \gamma ^{2} x^{2 \gamma }-n^{2} \gamma ^{2}+\alpha ^{2}\right ) y = 0\tag {3} \end {align*}
With the standard solution \begin {align*} y&=x^{\alpha } \left (c_{1} \operatorname {BesselJ}\left (n , \beta \,x^{\gamma }\right )+c_{2} \operatorname {BesselY}\left (n , \beta \,x^{\gamma }\right )\right )\tag {4} \end {align*}
Comparing (3) to (1) and solving for \(\alpha ,\beta ,n,\gamma \) gives \begin {align*} \alpha &= -{\frac {1}{2}}\\ \beta &= \sqrt {v^{2}+v}\\ n &= -v -\frac {1}{2}\\ \gamma &= -1 \end {align*}
Substituting all the above into (4) gives the solution as \begin {align*} y = \frac {c_{1} \operatorname {BesselJ}\left (-v -\frac {1}{2}, \frac {\sqrt {v^{2}+v}}{x}\right )}{\sqrt {x}}+\frac {c_{2} \operatorname {BesselY}\left (-v -\frac {1}{2}, \frac {\sqrt {v^{2}+v}}{x}\right )}{\sqrt {x}} \end {align*}
Summary
The solution(s) found are the following \begin{align*} \tag{1} y &= \frac {c_{1} \operatorname {BesselJ}\left (-v -\frac {1}{2}, \frac {\sqrt {v^{2}+v}}{x}\right )}{\sqrt {x}}+\frac {c_{2} \operatorname {BesselY}\left (-v -\frac {1}{2}, \frac {\sqrt {v^{2}+v}}{x}\right )}{\sqrt {x}} \\ \end{align*}
Verification of solutions
\[ y = \frac {c_{1} \operatorname {BesselJ}\left (-v -\frac {1}{2}, \frac {\sqrt {v^{2}+v}}{x}\right )}{\sqrt {x}}+\frac {c_{2} \operatorname {BesselY}\left (-v -\frac {1}{2}, \frac {\sqrt {v^{2}+v}}{x}\right )}{\sqrt {x}} \] Verified OK.
\[ \begin {array}{lll} & {} & \textrm {Let's solve}\hspace {3pt} \\ {} & {} & \left (x^{4}-x^{2}\right ) \left (\frac {d}{d x}y^{\prime }\right )+2 y^{\prime } x^{3}-v \left (v +1\right ) \left (x -1\right ) \left (x +1\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 {2 x y^{\prime }}{x^{2}-1}+\frac {v \left (v +1\right ) y}{x^{2}} \\ \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 x y^{\prime }}{x^{2}-1}-\frac {v \left (v +1\right ) y}{x^{2}}=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}{x^{2}-1}, P_{3}\left (x \right )=-\frac {v \left (v +1\right )}{x^{2}}\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}}}=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 (\frac {d}{d x}y^{\prime }\right ) x^{2} \left (x^{2}-1\right )+2 y^{\prime } x^{3}-\left (x^{2}-1\right ) v \left (v +1\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^{4}-4 u^{3}+5 u^{2}-2 u \right ) \left (\frac {d}{d u}\frac {d}{d u}y \left (u \right )\right )+\left (2 u^{3}-6 u^{2}+6 u -2\right ) \left (\frac {d}{d u}y \left (u \right )\right )+\left (-u^{2} v^{2}-u^{2} v +2 u \,v^{2}+2 u 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 y \left (u \right )\hspace {3pt}\textrm {to series expansion for}\hspace {3pt} m =1..2 \\ {} & {} & 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..3 \\ {} & {} & 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..4 \\ {} & {} & 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^{2} u^{-1+r}+\left (-2 a_{1} \left (1+r \right )^{2}+a_{0} r \left (1+5 r \right )\right ) u^{r}+\left (-2 a_{2} \left (2+r \right )^{2}+a_{1} \left (1+r \right ) \left (6+5 r \right )-2 a_{0} \left (2 r^{2}-v^{2}+r -v \right )\right ) u^{1+r}+\left (\moverset {\infty }{\munderset {k =2}{\sum }}\left (-2 a_{k +1} \left (k +1+r \right )^{2}+a_{k} \left (k +r \right ) \left (5 k +5 r +1\right )-2 a_{k -1} \left (2 \left (k -1\right )^{2}+4 \left (k -1\right ) r +2 r^{2}-v^{2}+k -1+r -v \right )+a_{k -2} \left (v -1+r +k \right ) \left (-v +r +k -2\right )\right ) u^{k +r}\right )=0 \\ \bullet & {} & a_{0}\textrm {cannot be 0 by assumption, giving the indicial equation}\hspace {3pt} \\ {} & {} & -2 r^{2}=0 \\ \bullet & {} & \textrm {Values of r that satisfy the indicial equation}\hspace {3pt} \\ {} & {} & r =0 \\ \bullet & {} & \textrm {The coefficients of each power of}\hspace {3pt} u \hspace {3pt}\textrm {must be 0}\hspace {3pt} \\ {} & {} & \left [-2 a_{1} \left (1+r \right )^{2}+a_{0} r \left (1+5 r \right )=0, -2 a_{2} \left (2+r \right )^{2}+a_{1} \left (1+r \right ) \left (6+5 r \right )-2 a_{0} \left (2 r^{2}-v^{2}+r -v \right )=0\right ] \\ \bullet & {} & \textrm {Solve for the dependent coefficient(s)}\hspace {3pt} \\ {} & {} & \left \{a_{1}=\frac {a_{0} r \left (1+5 r \right )}{2 \left (r^{2}+2 r +1\right )}, a_{2}=\frac {a_{0} \left (17 r^{3}+4 r \,v^{2}+23 r^{2}+4 v r +4 v^{2}+2 r +4 v \right )}{4 \left (r^{3}+5 r^{2}+8 r +4\right )}\right \} \\ \bullet & {} & \textrm {Each term in the series must be 0, giving the recursion relation}\hspace {3pt} \\ {} & {} & -2 a_{k +1} \left (k +1\right )^{2}+\left (5 a_{k}+a_{k -2}-4 a_{k -1}\right ) k^{2}+\left (a_{k}-3 a_{k -2}+6 a_{k -1}\right ) k +\left (-v^{2}-v +2\right ) a_{k -2}+2 a_{k -1} \left (v^{2}+v -1\right )=0 \\ \bullet & {} & \textrm {Shift index using}\hspace {3pt} k \mathrm {->}k +2 \\ {} & {} & -2 a_{k +3} \left (k +3\right )^{2}+\left (5 a_{k +2}+a_{k}-4 a_{k +1}\right ) \left (k +2\right )^{2}+\left (a_{k +2}-3 a_{k}+6 a_{k +1}\right ) \left (k +2\right )+\left (-v^{2}-v +2\right ) a_{k}+2 a_{k +1} \left (v^{2}+v -1\right )=0 \\ \bullet & {} & \textrm {Recursion relation that defines series solution to ODE}\hspace {3pt} \\ {} & {} & a_{k +3}=\frac {a_{k} k^{2}-4 k^{2} a_{k +1}+5 k^{2} a_{k +2}-a_{k} v^{2}+2 v^{2} a_{k +1}+a_{k} k -10 k a_{k +1}+21 k a_{k +2}-a_{k} v +2 v a_{k +1}-6 a_{k +1}+22 a_{k +2}}{2 \left (k +3\right )^{2}} \\ \bullet & {} & \textrm {Recursion relation for}\hspace {3pt} r =0 \\ {} & {} & a_{k +3}=\frac {a_{k} k^{2}-4 k^{2} a_{k +1}+5 k^{2} a_{k +2}-a_{k} v^{2}+2 v^{2} a_{k +1}+a_{k} k -10 k a_{k +1}+21 k a_{k +2}-a_{k} v +2 v a_{k +1}-6 a_{k +1}+22 a_{k +2}}{2 \left (k +3\right )^{2}} \\ \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 +3}=\frac {a_{k} k^{2}-4 k^{2} a_{k +1}+5 k^{2} a_{k +2}-a_{k} v^{2}+2 v^{2} a_{k +1}+a_{k} k -10 k a_{k +1}+21 k a_{k +2}-a_{k} v +2 v a_{k +1}-6 a_{k +1}+22 a_{k +2}}{2 \left (k +3\right )^{2}}, a_{1}=0, a_{2}=\frac {a_{0} \left (4 v^{2}+4 v \right )}{16}\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 +3}=\frac {a_{k} k^{2}-4 k^{2} a_{k +1}+5 k^{2} a_{k +2}-a_{k} v^{2}+2 v^{2} a_{k +1}+a_{k} k -10 k a_{k +1}+21 k a_{k +2}-a_{k} v +2 v a_{k +1}-6 a_{k +1}+22 a_{k +2}}{2 \left (k +3\right )^{2}}, a_{1}=0, a_{2}=\frac {a_{0} \left (4 v^{2}+4 v \right )}{16}\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 <- hyper3 successful: received ODE is equivalent to the 2F1 ODE <- hypergeometric successful <- special function solution successful`
✓ Solution by Maple
Time used: 0.282 (sec). Leaf size: 109
dsolve(diff(diff(y(x),x),x) = -2*x/(x^2-1)*diff(y(x),x)+v*(v+1)/x^2*y(x),y(x), singsol=all)
\[ y \left (x \right ) = \frac {c_{2} x^{v +1} \left (x^{2}\right )^{-\frac {v}{2}-\frac {1}{4}} \Gamma \left (v +\frac {1}{2}\right )^{2} \left (v +\frac {1}{2}\right ) \operatorname {LegendreP}\left (-\frac {1}{2}, -v -\frac {1}{2}, \frac {-x^{2}-1}{x^{2}-1}\right )+\sec \left (\pi v \right ) \pi \operatorname {LegendreP}\left (-\frac {1}{2}, v +\frac {1}{2}, \frac {-x^{2}-1}{x^{2}-1}\right ) \left (x^{2}\right )^{\frac {1}{4}+\frac {v}{2}} x^{-v} c_{1}}{\sqrt {-x^{2}+1}\, \Gamma \left (v +\frac {1}{2}\right )} \]
✓ Solution by Mathematica
Time used: 0.119 (sec). Leaf size: 68
DSolve[y''[x] == (v*(1 + v)*y[x])/x^2 - (2*x*y'[x])/(-1 + x^2),y[x],x,IncludeSingularSolutions -> True]
\[ y(x)\to c_1 i^{-v} x^{-v} \operatorname {Hypergeometric2F1}\left (\frac {1}{2},-v,\frac {1}{2}-v,x^2\right )+c_2 i^{v+1} x^{v+1} \operatorname {Hypergeometric2F1}\left (\frac {1}{2},v+1,v+\frac {3}{2},x^2\right ) \]