Internal problem ID [9644]
Internal file name [OUTPUT/8586_Monday_June_06_2022_04_16_59_AM_61283486/index.tex]
Book: Differential Gleichungen, E. Kamke, 3rd ed. Chelsea Pub. NY, 1948
Section: Chapter 2, linear second order
Problem number: 1317.
ODE order: 2.
ODE degree: 1.
The type(s) of ODE detected by this program : "unknown"
Maple gives the following as the ode type
[[_elliptic, _class_I]]
Unable to solve or complete the solution.
\[ \boxed {x \left (x^{2}-1\right ) y^{\prime \prime }+\left (3 x^{2}-1\right ) y^{\prime }+x y=0} \]
\[ \begin {array}{lll} & {} & \textrm {Let's solve}\hspace {3pt} \\ {} & {} & \left (x^{3}-x \right ) \left (\frac {d}{d x}y^{\prime }\right )+\left (3 x^{2}-1\right ) y^{\prime }+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 {y}{x^{2}-1}-\frac {\left (3 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 (3 x^{2}-1\right ) y^{\prime }}{x \left (x^{2}-1\right )}+\frac {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 {3 x^{2}-1}{x \left (x^{2}-1\right )}, P_{3}\left (x \right )=\frac {1}{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}}}=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} \\ {} & {} & x \left (x^{2}-1\right ) \left (\frac {d}{d x}y^{\prime }\right )+\left (3 x^{2}-1\right ) y^{\prime }+x 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^{3}-3 u^{2}+2 u \right ) \left (\frac {d}{d u}\frac {d}{d u}y \left (u \right )\right )+\left (3 u^{2}-6 u +2\right ) \left (\frac {d}{d u}y \left (u \right )\right )+\left (u -1\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..3 \\ {} & {} & 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} \left (3 r^{2}+3 r +1\right )\right ) u^{r}+\left (\moverset {\infty }{\munderset {k =1}{\sum }}\left (2 a_{k +1} \left (k +r +1\right )^{2}-a_{k} \left (3 k^{2}+6 k r +3 r^{2}+3 k +3 r +1\right )+a_{k -1} \left (k +r \right )^{2}\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 {Each term must be 0}\hspace {3pt} \\ {} & {} & 2 a_{1} \left (1+r \right )^{2}-a_{0} \left (3 r^{2}+3 r +1\right )=0 \\ \bullet & {} & \textrm {Each term in the series must be 0, giving the recursion relation}\hspace {3pt} \\ {} & {} & \left (-3 a_{k}+a_{k -1}+2 a_{k +1}\right ) k^{2}+\left (-3 a_{k}+4 a_{k +1}\right ) k -a_{k}+2 a_{k +1}=0 \\ \bullet & {} & \textrm {Shift index using}\hspace {3pt} k \mathrm {->}k +1 \\ {} & {} & \left (-3 a_{k +1}+a_{k}+2 a_{k +2}\right ) \left (k +1\right )^{2}+\left (-3 a_{k +1}+4 a_{k +2}\right ) \left (k +1\right )-a_{k +1}+2 a_{k +2}=0 \\ \bullet & {} & \textrm {Recursion relation that defines series solution to ODE}\hspace {3pt} \\ {} & {} & a_{k +2}=-\frac {k^{2} a_{k}-3 k^{2} a_{k +1}+2 k a_{k}-9 k a_{k +1}+a_{k}-7 a_{k +1}}{2 \left (k^{2}+4 k +4\right )} \\ \bullet & {} & \textrm {Recursion relation for}\hspace {3pt} r =0 \\ {} & {} & a_{k +2}=-\frac {k^{2} a_{k}-3 k^{2} a_{k +1}+2 k a_{k}-9 k a_{k +1}+a_{k}-7 a_{k +1}}{2 \left (k^{2}+4 k +4\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}-3 k^{2} a_{k +1}+2 k a_{k}-9 k a_{k +1}+a_{k}-7 a_{k +1}}{2 \left (k^{2}+4 k +4\right )}, 2 a_{1}-a_{0}=0\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 +2}=-\frac {k^{2} a_{k}-3 k^{2} a_{k +1}+2 k a_{k}-9 k a_{k +1}+a_{k}-7 a_{k +1}}{2 \left (k^{2}+4 k +4\right )}, 2 a_{1}-a_{0}=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 <- elliptic successful <- special function solution successful`
✓ Solution by Maple
Time used: 0.078 (sec). Leaf size: 13
dsolve(x*(x^2-1)*diff(diff(y(x),x),x)+(3*x^2-1)*diff(y(x),x)+x*y(x)=0,y(x), singsol=all)
\[ y \left (x \right ) = c_{1} \operatorname {EllipticK}\left (x \right )+c_{2} \operatorname {EllipticCK}\left (x \right ) \]
✓ Solution by Mathematica
Time used: 0.319 (sec). Leaf size: 38
DSolve[x*y[x] + (-1 + 3*x^2)*y'[x] + x*(-1 + x^2)*y''[x] == 0,y[x],x,IncludeSingularSolutions -> True]
\[ y(x)\to c_2 G_{2,2}^{2,0}\left (x^2| \begin {array}{c} \frac {1}{2},\frac {1}{2} \\ 0,0 \\ \end {array} \right )+\frac {2 c_1 \operatorname {EllipticK}\left (x^2\right )}{\pi } \]