Internal problem ID [9806]
Internal file name [OUTPUT/8749_Monday_June_06_2022_05_23_26_AM_35682667/index.tex]
Book: Differential Gleichungen, E. Kamke, 3rd ed. Chelsea Pub. NY, 1948
Section: Chapter 3, linear third order
Problem number: 1480.
ODE order: 3.
ODE degree: 1.
The type(s) of ODE detected by this program : "unknown"
Maple gives the following as the ode type
[[_3rd_order, _with_linear_symmetries]]
Unable to solve or complete the solution.
\[ \boxed {x y^{\prime \prime \prime }-\left (x +2 v \right ) y^{\prime \prime }-\left (x -2 v -1\right ) y^{\prime }+\left (x -1\right ) y=0} \] Unable to solve this ODE.
\[ \begin {array}{lll} & {} & \textrm {Let's solve}\hspace {3pt} \\ {} & {} & x \left (\frac {d}{d x}y^{\prime \prime }\right )-\left (x +2 v \right ) \left (\frac {d}{d x}y^{\prime }\right )-\left (x -2 v -1\right ) y^{\prime }+\left (x -1\right ) y=0 \\ \bullet & {} & \textrm {Highest derivative means the order of the ODE is}\hspace {3pt} 3 \\ {} & {} & \frac {d}{d x}y^{\prime \prime } \\ \square & {} & \textrm {Check to see if}\hspace {3pt} x_{0}=0\hspace {3pt}\textrm {is a regular singular point}\hspace {3pt} \\ {} & \circ & \textrm {Define functions}\hspace {3pt} \\ {} & {} & \left [P_{2}\left (x \right )=-\frac {x +2 v}{x}, P_{3}\left (x \right )=\frac {2 v -x +1}{x}, P_{4}\left (x \right )=\frac {x -1}{x}\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}}}=-2 v \\ {} & \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}}}=0 \\ {} & \circ & x^{3}\cdot P_{4}\left (x \right )\textrm {is analytic at}\hspace {3pt} x =0 \\ {} & {} & \left (x^{3}\cdot P_{4}\left (x \right )\right )\bigg | {\mstack {}{_{x \hiderel {=}0}}}=0 \\ {} & \circ & x =0\textrm {is a regular singular point}\hspace {3pt} \\ & {} & \textrm {Check to see if}\hspace {3pt} x_{0}=0\hspace {3pt}\textrm {is a regular singular point}\hspace {3pt} \\ {} & {} & x_{0}=0 \\ \bullet & {} & \textrm {Multiply by denominators}\hspace {3pt} \\ {} & {} & \left (2 v -x +1\right ) y^{\prime }+\left (-2 v -x \right ) \left (\frac {d}{d x}y^{\prime }\right )+x \left (\frac {d}{d x}y^{\prime \prime }\right )+\left (x -1\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..1 \\ {} & {} & 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 =0..1 \\ {} & {} & 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 =0..1 \\ {} & {} & 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} \\ {} & \circ & \textrm {Convert}\hspace {3pt} x \cdot \left (\frac {d}{d x}y^{\prime \prime }\right )\hspace {3pt}\textrm {to series expansion}\hspace {3pt} \\ {} & {} & x \cdot \left (\frac {d}{d x}y^{\prime \prime }\right )=\moverset {\infty }{\munderset {k =0}{\sum }}a_{k} \left (k +r \right ) \left (k +r -1\right ) \left (k +r -2\right ) x^{k +r -2} \\ {} & \circ & \textrm {Shift index using}\hspace {3pt} k \mathrm {->}k +2 \\ {} & {} & x \cdot \left (\frac {d}{d x}y^{\prime \prime }\right )=\moverset {\infty }{\munderset {k =-2}{\sum }}a_{k +2} \left (k +2+r \right ) \left (k +1+r \right ) \left (k +r \right ) x^{k +r} \\ & {} & \textrm {Rewrite ODE with series expansions}\hspace {3pt} \\ {} & {} & a_{0} r \left (-1+r \right ) \left (-2+r -2 v \right ) x^{-2+r}+\left (a_{1} \left (1+r \right ) r \left (-1+r -2 v \right )-a_{0} r \left (-2+r -2 v \right )\right ) x^{-1+r}+\left (a_{2} \left (2+r \right ) \left (1+r \right ) \left (r -2 v \right )-a_{1} \left (1+r \right ) \left (-1+r -2 v \right )-a_{0} \left (1+r \right )\right ) x^{r}+\left (\moverset {\infty }{\munderset {k =1}{\sum }}\left (a_{k +2} \left (k +2+r \right ) \left (k +1+r \right ) \left (k +r -2 v \right )-a_{k +1} \left (k +1+r \right ) \left (k -1+r -2 v \right )-a_{k} \left (k +1+r \right )+a_{k -1}\right ) x^{k +r}\right )=0 \\ \bullet & {} & a_{0}\textrm {cannot be 0 by assumption, giving the indicial equation}\hspace {3pt} \\ {} & {} & r \left (-1+r \right ) \left (-2+r -2 v \right )=0 \\ \bullet & {} & \textrm {Values of r that satisfy the indicial equation}\hspace {3pt} \\ {} & {} & r \in \left \{0, 1, 2+2 v \right \} \\ \bullet & {} & \textrm {The coefficients of each power of}\hspace {3pt} x \hspace {3pt}\textrm {must be 0}\hspace {3pt} \\ {} & {} & \left [a_{1} \left (1+r \right ) r \left (-1+r -2 v \right )-a_{0} r \left (-2+r -2 v \right )=0, a_{2} \left (2+r \right ) \left (1+r \right ) \left (r -2 v \right )-a_{1} \left (1+r \right ) \left (-1+r -2 v \right )-a_{0} \left (1+r \right )=0\right ] \\ \bullet & {} & \textrm {Each term in the series must be 0, giving the recursion relation}\hspace {3pt} \\ {} & {} & a_{k +2} \left (k +2+r \right ) \left (k +1+r \right ) \left (k +r -2 v \right )+a_{k +1} \left (k +1+r \right ) \left (-k +1-r +2 v \right )-a_{k} \left (k +1+r \right )+a_{k -1}=0 \\ \bullet & {} & \textrm {Shift index using}\hspace {3pt} k \mathrm {->}k +1 \\ {} & {} & a_{k +3} \left (k +3+r \right ) \left (k +2+r \right ) \left (k +1+r -2 v \right )+a_{k +2} \left (k +2+r \right ) \left (-k -r +2 v \right )-a_{k +1} \left (k +2+r \right )+a_{k}=0 \\ \bullet & {} & \textrm {Recursion relation that defines series solution to ODE}\hspace {3pt} \\ {} & {} & a_{k +3}=\frac {k^{2} a_{k +2}+2 k r a_{k +2}-2 k v a_{k +2}+r^{2} a_{k +2}-2 r v a_{k +2}+k a_{k +1}+2 k a_{k +2}+r a_{k +1}+2 r a_{k +2}-4 v a_{k +2}-a_{k}+2 a_{k +1}}{\left (k +3+r \right ) \left (k +2+r \right ) \left (k +1+r -2 v \right )} \\ \bullet & {} & \textrm {Recursion relation for}\hspace {3pt} r =0 \\ {} & {} & a_{k +3}=\frac {k^{2} a_{k +2}-2 k v a_{k +2}+k a_{k +1}+2 k a_{k +2}-4 v a_{k +2}-a_{k}+2 a_{k +1}}{\left (k +3\right ) \left (k +2\right ) \left (k +1-2 v \right )} \\ \bullet & {} & \textrm {Solution for}\hspace {3pt} r =0 \\ {} & {} & \left [y=\moverset {\infty }{\munderset {k =0}{\sum }}a_{k} x^{k}, a_{k +3}=\frac {k^{2} a_{k +2}-2 k v a_{k +2}+k a_{k +1}+2 k a_{k +2}-4 v a_{k +2}-a_{k}+2 a_{k +1}}{\left (k +3\right ) \left (k +2\right ) \left (k +1-2 v \right )}, 0=0, -4 a_{2} v -a_{1} \left (-1-2 v \right )-a_{0}=0\right ] \\ \bullet & {} & \textrm {Recursion relation for}\hspace {3pt} r =1 \\ {} & {} & a_{k +3}=\frac {k^{2} a_{k +2}-2 k v a_{k +2}+k a_{k +1}+4 k a_{k +2}-6 v a_{k +2}-a_{k}+3 a_{k +1}+3 a_{k +2}}{\left (k +4\right ) \left (k +3\right ) \left (k +2-2 v \right )} \\ \bullet & {} & \textrm {Solution for}\hspace {3pt} r =1 \\ {} & {} & \left [y=\moverset {\infty }{\munderset {k =0}{\sum }}a_{k} x^{k +1}, a_{k +3}=\frac {k^{2} a_{k +2}-2 k v a_{k +2}+k a_{k +1}+4 k a_{k +2}-6 v a_{k +2}-a_{k}+3 a_{k +1}+3 a_{k +2}}{\left (k +4\right ) \left (k +3\right ) \left (k +2-2 v \right )}, -4 a_{1} v -a_{0} \left (-1-2 v \right )=0, 6 a_{2} \left (1-2 v \right )+4 a_{1} v -2 a_{0}=0\right ] \\ \bullet & {} & \textrm {Recursion relation for}\hspace {3pt} r =2+2 v \\ {} & {} & a_{k +3}=\frac {k^{2} a_{k +2}+2 k \left (2+2 v \right ) a_{k +2}-2 k v a_{k +2}+\left (2+2 v \right )^{2} a_{k +2}-2 \left (2+2 v \right ) v a_{k +2}+k a_{k +1}+2 k a_{k +2}+\left (2+2 v \right ) a_{k +1}+2 \left (2+2 v \right ) a_{k +2}-4 v a_{k +2}-a_{k}+2 a_{k +1}}{\left (k +5+2 v \right ) \left (k +4+2 v \right ) \left (k +3\right )} \\ \bullet & {} & \textrm {Solution for}\hspace {3pt} r =2+2 v \\ {} & {} & \left [y=\moverset {\infty }{\munderset {k =0}{\sum }}a_{k} x^{k +2+2 v}, a_{k +3}=\frac {k^{2} a_{k +2}+2 k \left (2+2 v \right ) a_{k +2}-2 k v a_{k +2}+\left (2+2 v \right )^{2} a_{k +2}-2 \left (2+2 v \right ) v a_{k +2}+k a_{k +1}+2 k a_{k +2}+\left (2+2 v \right ) a_{k +1}+2 \left (2+2 v \right ) a_{k +2}-4 v a_{k +2}-a_{k}+2 a_{k +1}}{\left (k +5+2 v \right ) \left (k +4+2 v \right ) \left (k +3\right )}, a_{1} \left (3+2 v \right ) \left (2+2 v \right )=0, 2 a_{2} \left (4+2 v \right ) \left (3+2 v \right )-a_{1} \left (3+2 v \right )-a_{0} \left (3+2 v \right )=0\right ] \\ \bullet & {} & \textrm {Combine solutions and rename parameters}\hspace {3pt} \\ {} & {} & \left [y=\left (\moverset {\infty }{\munderset {k =0}{\sum }}a_{k} x^{k}\right )+\left (\moverset {\infty }{\munderset {k =0}{\sum }}b_{k} x^{k +1}\right )+\left (\moverset {\infty }{\munderset {k =0}{\sum }}c_{k} x^{k +2+2 v}\right ), a_{k +3}=\frac {k^{2} a_{k +2}-2 k v a_{k +2}+k a_{k +1}+2 k a_{k +2}-4 v a_{k +2}-a_{k}+2 a_{k +1}}{\left (k +3\right ) \left (k +2\right ) \left (k +1-2 v \right )}, 0=0, -4 a_{2} v -a_{1} \left (-1-2 v \right )-a_{0}=0, b_{k +3}=\frac {k^{2} b_{k +2}-2 k v b_{k +2}+k b_{k +1}+4 k b_{k +2}-6 v b_{k +2}-b_{k}+3 b_{k +1}+3 b_{k +2}}{\left (k +4\right ) \left (k +3\right ) \left (k +2-2 v \right )}, -4 b_{1} v -b_{0} \left (-1-2 v \right )=0, 6 b_{2} \left (1-2 v \right )+4 b_{1} v -2 b_{0}=0, c_{k +3}=\frac {k^{2} c_{k +2}+2 k \left (2+2 v \right ) c_{k +2}-2 k v c_{k +2}+\left (2+2 v \right )^{2} c_{k +2}-2 \left (2+2 v \right ) v c_{k +2}+k c_{k +1}+2 k c_{k +2}+\left (2+2 v \right ) c_{k +1}+2 \left (2+2 v \right ) c_{k +2}-4 v c_{k +2}-c_{k}+2 c_{k +1}}{\left (k +5+2 v \right ) \left (k +4+2 v \right ) \left (k +3\right )}, c_{1} \left (3+2 v \right ) \left (2+2 v \right )=0, 2 c_{2} \left (4+2 v \right ) \left (3+2 v \right )-c_{1} \left (3+2 v \right )-c_{0} \left (3+2 v \right )=0\right ] \end {array} \]
Maple trace
`Methods for third order ODEs: --- Trying classification methods --- trying a quadrature checking if the LODE has constant coefficients checking if the LODE is of Euler type trying high order exact linear fully integrable trying to convert to a linear ODE with constant coefficients trying differential order: 3; missing the dependent variable Equation is the LCLM of -y(x)+diff(y(x),x), -y(x)-(2*v+1)/x*diff(y(x),x)+diff(diff(y(x),x),x) trying differential order: 1; missing the dependent variable checking if the LODE has constant coefficients <- constant coefficients successful 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 <- Bessel successful <- special function solution successful <- solving the LCLM ode successful `
✓ Solution by Maple
Time used: 0.031 (sec). Leaf size: 54
dsolve(x*diff(diff(diff(y(x),x),x),x)-(x+2*v)*diff(diff(y(x),x),x)-(x-2*v-1)*diff(y(x),x)+(x-1)*y(x)=0,y(x), singsol=all)
\[ y \left (x \right ) = \frac {x^{2+v} c_{2} \operatorname {BesselI}\left (-v +1, x\right )-2 \operatorname {BesselI}\left (-v , x\right ) x^{v +1} c_{2} v +x^{2+v} c_{3} \operatorname {BesselK}\left (v +1, x\right )+{\mathrm e}^{x} c_{1} x}{x} \]
✓ Solution by Mathematica
Time used: 0.127 (sec). Leaf size: 91
DSolve[(-1 + x)*y[x] - (-1 - 2*v + x)*y'[x] - (2*v + x)*y''[x] + x*Derivative[3][y][x] == 0,y[x],x,IncludeSingularSolutions -> True]
\[ y(x)\to \frac {1}{4} e^x \left (\frac {4 c_3 x^{2 v+2} \operatorname {Gamma}\left (v+\frac {3}{2}\right ) \, _1\tilde {F}_1\left (v+\frac {3}{2};2 v+3;-2 x\right )}{\operatorname {Gamma}\left (\frac {1}{2}-v\right )}+c_2 4^{-v} G_{2,3}^{2,1}\left (2 x\left | \begin {array}{c} 1,v+\frac {3}{2} \\ 1,2 (v+1),0 \\ \end {array} \right .\right )+4 c_1\right ) \]