Internal problem ID [9827]
Internal file name [OUTPUT/8770_Monday_June_06_2022_05_25_49_AM_61369090/index.tex]
Book: Differential Gleichungen, E. Kamke, 3rd ed. Chelsea Pub. NY, 1948
Section: Chapter 3, linear third order
Problem number: 1501.
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.
\[ \boxed {x^{2} y^{\prime \prime \prime }-2 \left (x^{2}-x \right ) y^{\prime \prime }+\left (x^{2}-2 x +\frac {1}{4}-\nu ^{2}\right ) y^{\prime }+\left (\nu ^{2}-\frac {1}{4}\right ) y=0} \] Unable to solve this ODE.
\[ \begin {array}{lll} & {} & \textrm {Let's solve}\hspace {3pt} \\ {} & {} & x^{2} \left (\frac {d}{d x}y^{\prime \prime }\right )-2 \left (x^{2}-x \right ) \left (\frac {d}{d x}y^{\prime }\right )+\left (x^{2}-2 x +\frac {1}{4}-\nu ^{2}\right ) y^{\prime }+\left (\nu ^{2}-\frac {1}{4}\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 {2 \left (x -1\right )}{x}, P_{3}\left (x \right )=-\frac {4 \nu ^{2}-4 x^{2}+8 x -1}{4 x^{2}}, P_{4}\left (x \right )=\frac {4 \nu ^{2}-1}{4 x^{2}}\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 \\ {} & \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}}}=\frac {1}{4}-\nu ^{2} \\ {} & \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} \\ {} & {} & 4 x^{2} \left (\frac {d}{d x}y^{\prime \prime }\right )-8 x \left (x -1\right ) \left (\frac {d}{d x}y^{\prime }\right )+\left (-4 \nu ^{2}+4 x^{2}-8 x +1\right ) y^{\prime }+\left (4 \nu ^{2}-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^{\prime }\hspace {3pt}\textrm {to series expansion for}\hspace {3pt} m =0..2 \\ {} & {} & 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 =1..2 \\ {} & {} & 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^{2}\cdot \left (\frac {d}{d x}y^{\prime \prime }\right )\hspace {3pt}\textrm {to series expansion}\hspace {3pt} \\ {} & {} & x^{2}\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 -1} \\ {} & \circ & \textrm {Shift index using}\hspace {3pt} k \mathrm {->}k +1 \\ {} & {} & x^{2}\cdot \left (\frac {d}{d x}y^{\prime \prime }\right )=\moverset {\infty }{\munderset {k =-1}{\sum }}a_{k +1} \left (k +1+r \right ) \left (k +r \right ) \left (k +r -1\right ) x^{k +r} \\ & {} & \textrm {Rewrite ODE with series expansions}\hspace {3pt} \\ {} & {} & a_{0} r \left (2 r -1+2 \nu \right ) \left (2 r -1-2 \nu \right ) x^{-1+r}+\left (a_{1} \left (1+r \right ) \left (2 r +1+2 \nu \right ) \left (2 r +1-2 \nu \right )-a_{0} \left (-4 \nu ^{2}+8 r^{2}+1\right )\right ) x^{r}+\left (\moverset {\infty }{\munderset {k =1}{\sum }}\left (a_{k +1} \left (k +1+r \right ) \left (2 r +1+2 \nu +2 k \right ) \left (2 r +1-2 \nu +2 k \right )-a_{k} \left (8 k^{2}+16 k r -4 \nu ^{2}+8 r^{2}+1\right )+4 a_{k -1} \left (k +r -1\right )\right ) x^{k +r}\right )=0 \\ \bullet & {} & a_{0}\textrm {cannot be 0 by assumption, giving the indicial equation}\hspace {3pt} \\ {} & {} & r \left (2 r -1+2 \nu \right ) \left (2 r -1-2 \nu \right )=0 \\ \bullet & {} & \textrm {Values of r that satisfy the indicial equation}\hspace {3pt} \\ {} & {} & r \in \left \{0, \frac {1}{2}-\nu , \frac {1}{2}+\nu \right \} \\ \bullet & {} & \textrm {Each term must be 0}\hspace {3pt} \\ {} & {} & a_{1} \left (1+r \right ) \left (2 r +1+2 \nu \right ) \left (2 r +1-2 \nu \right )-a_{0} \left (-4 \nu ^{2}+8 r^{2}+1\right )=0 \\ \bullet & {} & \textrm {Each term in the series must be 0, giving the recursion relation}\hspace {3pt} \\ {} & {} & 4 \left (r +\frac {1}{2}+\nu +k \right ) \left (r +\frac {1}{2}-\nu +k \right ) \left (k +1+r \right ) a_{k +1}-8 k^{2} a_{k}+4 \left (-4 r a_{k}+a_{k -1}\right ) k -8 r^{2} a_{k}+4 a_{k} \nu ^{2}+4 a_{k -1} r -a_{k}-4 a_{k -1}=0 \\ \bullet & {} & \textrm {Shift index using}\hspace {3pt} k \mathrm {->}k +1 \\ {} & {} & 4 \left (r +\frac {3}{2}+\nu +k \right ) \left (r +\frac {3}{2}-\nu +k \right ) \left (k +2+r \right ) a_{k +2}-8 \left (k +1\right )^{2} a_{k +1}+4 \left (-4 r a_{k +1}+a_{k}\right ) \left (k +1\right )-8 r^{2} a_{k +1}+4 \nu ^{2} a_{k +1}+4 r a_{k}-a_{k +1}-4 a_{k}=0 \\ \bullet & {} & \textrm {Recursion relation that defines series solution to ODE}\hspace {3pt} \\ {} & {} & a_{k +2}=\frac {8 k^{2} a_{k +1}+16 k r a_{k +1}-4 \nu ^{2} a_{k +1}+8 r^{2} a_{k +1}-4 k a_{k}+16 k a_{k +1}-4 r a_{k}+16 r a_{k +1}+9 a_{k +1}}{\left (2 r +3+2 \nu +2 k \right ) \left (2 r +3-2 \nu +2 k \right ) \left (k +2+r \right )} \\ \bullet & {} & \textrm {Recursion relation for}\hspace {3pt} r =0 \\ {} & {} & a_{k +2}=\frac {8 k^{2} a_{k +1}-4 \nu ^{2} a_{k +1}-4 k a_{k}+16 k a_{k +1}+9 a_{k +1}}{\left (3+2 \nu +2 k \right ) \left (3-2 \nu +2 k \right ) \left (k +2\right )} \\ \bullet & {} & \textrm {Solution for}\hspace {3pt} r =0 \\ {} & {} & \left [y=\moverset {\infty }{\munderset {k =0}{\sum }}a_{k} x^{k}, a_{k +2}=\frac {8 k^{2} a_{k +1}-4 \nu ^{2} a_{k +1}-4 k a_{k}+16 k a_{k +1}+9 a_{k +1}}{\left (3+2 \nu +2 k \right ) \left (3-2 \nu +2 k \right ) \left (k +2\right )}, a_{1} \left (2 \nu +1\right ) \left (1-2 \nu \right )-a_{0} \left (-4 \nu ^{2}+1\right )=0\right ] \\ \bullet & {} & \textrm {Recursion relation for}\hspace {3pt} r =\frac {1}{2}-\nu \\ {} & {} & a_{k +2}=\frac {8 k^{2} a_{k +1}+16 k \left (\frac {1}{2}-\nu \right ) a_{k +1}-4 \nu ^{2} a_{k +1}+8 \left (\frac {1}{2}-\nu \right )^{2} a_{k +1}-4 k a_{k}+16 k a_{k +1}-4 \left (\frac {1}{2}-\nu \right ) a_{k}+16 \left (\frac {1}{2}-\nu \right ) a_{k +1}+9 a_{k +1}}{\left (4+2 k \right ) \left (4-4 \nu +2 k \right ) \left (k +\frac {5}{2}-\nu \right )} \\ \bullet & {} & \textrm {Solution for}\hspace {3pt} r =\frac {1}{2}-\nu \\ {} & {} & \left [y=\moverset {\infty }{\munderset {k =0}{\sum }}a_{k} x^{k +\frac {1}{2}-\nu }, a_{k +2}=\frac {8 k^{2} a_{k +1}+16 k \left (\frac {1}{2}-\nu \right ) a_{k +1}-4 \nu ^{2} a_{k +1}+8 \left (\frac {1}{2}-\nu \right )^{2} a_{k +1}-4 k a_{k}+16 k a_{k +1}-4 \left (\frac {1}{2}-\nu \right ) a_{k}+16 \left (\frac {1}{2}-\nu \right ) a_{k +1}+9 a_{k +1}}{\left (4+2 k \right ) \left (4-4 \nu +2 k \right ) \left (k +\frac {5}{2}-\nu \right )}, 2 a_{1} \left (\frac {3}{2}-\nu \right ) \left (2-4 \nu \right )-a_{0} \left (-4 \nu ^{2}+8 \left (\frac {1}{2}-\nu \right )^{2}+1\right )=0\right ] \\ \bullet & {} & \textrm {Recursion relation for}\hspace {3pt} r =\frac {1}{2}+\nu \\ {} & {} & a_{k +2}=\frac {8 k^{2} a_{k +1}+16 k \left (\frac {1}{2}+\nu \right ) a_{k +1}-4 \nu ^{2} a_{k +1}+8 \left (\frac {1}{2}+\nu \right )^{2} a_{k +1}-4 k a_{k}+16 k a_{k +1}-4 \left (\frac {1}{2}+\nu \right ) a_{k}+16 \left (\frac {1}{2}+\nu \right ) a_{k +1}+9 a_{k +1}}{\left (4+4 \nu +2 k \right ) \left (4+2 k \right ) \left (k +\frac {5}{2}+\nu \right )} \\ \bullet & {} & \textrm {Solution for}\hspace {3pt} r =\frac {1}{2}+\nu \\ {} & {} & \left [y=\moverset {\infty }{\munderset {k =0}{\sum }}a_{k} x^{k +\frac {1}{2}+\nu }, a_{k +2}=\frac {8 k^{2} a_{k +1}+16 k \left (\frac {1}{2}+\nu \right ) a_{k +1}-4 \nu ^{2} a_{k +1}+8 \left (\frac {1}{2}+\nu \right )^{2} a_{k +1}-4 k a_{k}+16 k a_{k +1}-4 \left (\frac {1}{2}+\nu \right ) a_{k}+16 \left (\frac {1}{2}+\nu \right ) a_{k +1}+9 a_{k +1}}{\left (4+4 \nu +2 k \right ) \left (4+2 k \right ) \left (k +\frac {5}{2}+\nu \right )}, 2 a_{1} \left (\frac {3}{2}+\nu \right ) \left (2+4 \nu \right )-a_{0} \left (-4 \nu ^{2}+8 \left (\frac {1}{2}+\nu \right )^{2}+1\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 +\frac {1}{2}-\nu }\right )+\left (\moverset {\infty }{\munderset {k =0}{\sum }}c_{k} x^{k +\frac {1}{2}+\nu }\right ), a_{k +2}=\frac {8 k^{2} a_{k +1}-4 \nu ^{2} a_{k +1}-4 k a_{k}+16 k a_{k +1}+9 a_{k +1}}{\left (3+2 \nu +2 k \right ) \left (3-2 \nu +2 k \right ) \left (k +2\right )}, a_{1} \left (2 \nu +1\right ) \left (1-2 \nu \right )-a_{0} \left (-4 \nu ^{2}+1\right )=0, b_{k +2}=\frac {8 k^{2} b_{k +1}+16 k \left (\frac {1}{2}-\nu \right ) b_{k +1}-4 \nu ^{2} b_{k +1}+8 \left (\frac {1}{2}-\nu \right )^{2} b_{k +1}-4 k b_{k}+16 k b_{k +1}-4 \left (\frac {1}{2}-\nu \right ) b_{k}+16 \left (\frac {1}{2}-\nu \right ) b_{k +1}+9 b_{k +1}}{\left (4+2 k \right ) \left (4-4 \nu +2 k \right ) \left (k +\frac {5}{2}-\nu \right )}, 2 b_{1} \left (\frac {3}{2}-\nu \right ) \left (2-4 \nu \right )-b_{0} \left (-4 \nu ^{2}+8 \left (\frac {1}{2}-\nu \right )^{2}+1\right )=0, c_{k +2}=\frac {8 k^{2} c_{k +1}+16 k \left (\frac {1}{2}+\nu \right ) c_{k +1}-4 \nu ^{2} c_{k +1}+8 \left (\frac {1}{2}+\nu \right )^{2} c_{k +1}-4 k c_{k}+16 k c_{k +1}-4 \left (\frac {1}{2}+\nu \right ) c_{k}+16 \left (\frac {1}{2}+\nu \right ) c_{k +1}+9 c_{k +1}}{\left (4+4 \nu +2 k \right ) \left (4+2 k \right ) \left (k +\frac {5}{2}+\nu \right )}, 2 c_{1} \left (\frac {3}{2}+\nu \right ) \left (2+4 \nu \right )-c_{0} \left (-4 \nu ^{2}+8 \left (\frac {1}{2}+\nu \right )^{2}+1\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), -1/4*(4*nu^2-1)/x^2*y(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: 0.016 (sec). Leaf size: 37
dsolve(x^2*diff(diff(diff(y(x),x),x),x)-2*(x^2-x)*diff(diff(y(x),x),x)+(x^2-2*x+1/4-nu^2)*diff(y(x),x)+(nu^2-1/4)*y(x)=0,y(x), singsol=all)
\[ y \left (x \right ) = c_{1} {\mathrm e}^{x}+c_{2} {\mathrm e}^{\frac {x}{2}} \sqrt {x}\, \operatorname {BesselI}\left (\nu , \frac {x}{2}\right )+c_{3} {\mathrm e}^{\frac {x}{2}} \sqrt {x}\, \operatorname {BesselK}\left (\nu , \frac {x}{2}\right ) \]
✓ Solution by Mathematica
Time used: 0.117 (sec). Leaf size: 80
DSolve[(-1/4 + nu^2)*y[x] + (1/4 - nu^2 - 2*x + x^2)*y'[x] - 2*(-x + x^2)*y''[x] + x^2*Derivative[3][y][x] == 0,y[x],x,IncludeSingularSolutions -> True]
\[ y(x)\to e^x \left (\frac {c_3 x^{\nu +\frac {1}{2}} \operatorname {Gamma}\left (\nu +\frac {1}{2}\right ) \, _1\tilde {F}_1\left (\nu +\frac {1}{2};2 \nu +1;-x\right )}{\operatorname {Gamma}\left (\frac {3}{2}-\nu \right )}+c_2 G_{2,3}^{2,1}\left (x\left | \begin {array}{c} 1,0 \\ \frac {1}{2}-\nu ,\nu +\frac {1}{2},0 \\ \end {array} \right .\right )+c_1\right ) \]