Internal problem ID [9826]
Internal file name [OUTPUT/8769_Monday_June_06_2022_05_25_41_AM_41779492/index.tex]
Book: Differential Gleichungen, E. Kamke, 3rd ed. Chelsea Pub. NY, 1948
Section: Chapter 3, linear third order
Problem number: 1500.
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.
\[ \boxed {x^{2} y^{\prime \prime \prime }-\left (x +\nu \right ) x y^{\prime \prime }+\nu \left (2 x +1\right ) y^{\prime }-\nu \left (x +1\right ) y=0} \] 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 )-\left (x +\nu \right ) x \left (\frac {d}{d x}y^{\prime }\right )+\nu \left (2 x +1\right ) y^{\prime }-\nu \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 +\nu }{x}, P_{3}\left (x \right )=\frac {\nu \left (2 x +1\right )}{x^{2}}, P_{4}\left (x \right )=-\frac {\nu \left (x +1\right )}{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}}}=-\nu \\ {} & \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}}}=\nu \\ {} & \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 {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 =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 \right ) \left (-1-\nu +r \right ) x^{-1+r}+\left (a_{1} \left (1+r \right ) \left (-1+r \right ) \left (-\nu +r \right )-a_{0} \left (-2 \nu r +r^{2}+\nu -r \right )\right ) x^{r}+\left (\moverset {\infty }{\munderset {k =1}{\sum }}\left (a_{k +1} \left (k +1+r \right ) \left (k +r -1\right ) \left (k -\nu +r \right )-a_{k} \left (k^{2}-2 \nu k +2 k r -2 \nu r +r^{2}-k +\nu -r \right )-\nu 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 (-2+r \right ) \left (-1-\nu +r \right )=0 \\ \bullet & {} & \textrm {Values of r that satisfy the indicial equation}\hspace {3pt} \\ {} & {} & r \in \left \{0, 2, 1+\nu \right \} \\ \bullet & {} & \textrm {Each term must be 0}\hspace {3pt} \\ {} & {} & a_{1} \left (1+r \right ) \left (-1+r \right ) \left (-\nu +r \right )-a_{0} \left (-2 \nu r +r^{2}+\nu -r \right )=0 \\ \bullet & {} & \textrm {Each term in the series must be 0, giving the recursion relation}\hspace {3pt} \\ {} & {} & a_{k +1} \left (k +1+r \right ) \left (k +r -1\right ) \left (k -\nu +r \right )-k^{2} a_{k}-2 \left (r -\nu -\frac {1}{2}\right ) a_{k} k -r^{2} a_{k}+a_{k} \left (2 \nu +1\right ) r -\nu \left (a_{k}+a_{k -1}\right )=0 \\ \bullet & {} & \textrm {Shift index using}\hspace {3pt} k \mathrm {->}k +1 \\ {} & {} & a_{k +2} \left (k +2+r \right ) \left (k +r \right ) \left (k +1-\nu +r \right )-\left (k +1\right )^{2} a_{k +1}-2 \left (r -\nu -\frac {1}{2}\right ) a_{k +1} \left (k +1\right )-r^{2} a_{k +1}+a_{k +1} \left (2 \nu +1\right ) r -\nu \left (a_{k +1}+a_{k}\right )=0 \\ \bullet & {} & \textrm {Recursion relation that defines series solution to ODE}\hspace {3pt} \\ {} & {} & a_{k +2}=\frac {k^{2} a_{k +1}-2 k \nu a_{k +1}+2 k r a_{k +1}-2 \nu r a_{k +1}+r^{2} a_{k +1}+k a_{k +1}+\nu a_{k}-\nu a_{k +1}+r a_{k +1}}{\left (k +2+r \right ) \left (k +r \right ) \left (k +1-\nu +r \right )} \\ \bullet & {} & \textrm {Recursion relation for}\hspace {3pt} r =0 \\ {} & {} & a_{k +2}=\frac {k^{2} a_{k +1}-2 k \nu a_{k +1}+k a_{k +1}+\nu a_{k}-\nu a_{k +1}}{\left (k +2\right ) k \left (k +1-\nu \right )} \\ \bullet & {} & \textrm {Series not valid for}\hspace {3pt} r =0\hspace {3pt}\textrm {, division by}\hspace {3pt} 0\hspace {3pt}\textrm {in the recursion relation at}\hspace {3pt} k =0 \\ {} & {} & a_{k +2}=\frac {k^{2} a_{k +1}-2 k \nu a_{k +1}+k a_{k +1}+\nu a_{k}-\nu a_{k +1}}{\left (k +2\right ) k \left (k +1-\nu \right )} \\ \bullet & {} & \textrm {Recursion relation for}\hspace {3pt} r =2 \\ {} & {} & a_{k +2}=\frac {k^{2} a_{k +1}-2 k \nu a_{k +1}+5 k a_{k +1}+\nu a_{k}-5 \nu a_{k +1}+6 a_{k +1}}{\left (k +4\right ) \left (k +2\right ) \left (k +3-\nu \right )} \\ \bullet & {} & \textrm {Solution for}\hspace {3pt} r =2 \\ {} & {} & \left [y=\moverset {\infty }{\munderset {k =0}{\sum }}a_{k} x^{k +2}, a_{k +2}=\frac {k^{2} a_{k +1}-2 k \nu a_{k +1}+5 k a_{k +1}+\nu a_{k}-5 \nu a_{k +1}+6 a_{k +1}}{\left (k +4\right ) \left (k +2\right ) \left (k +3-\nu \right )}, 3 a_{1} \left (-\nu +2\right )-a_{0} \left (-3 \nu +2\right )=0\right ] \\ \bullet & {} & \textrm {Recursion relation for}\hspace {3pt} r =1+\nu \\ {} & {} & a_{k +2}=\frac {k^{2} a_{k +1}-2 k \nu a_{k +1}+2 k \left (1+\nu \right ) a_{k +1}-2 \nu \left (1+\nu \right ) a_{k +1}+\left (1+\nu \right )^{2} a_{k +1}+k a_{k +1}+\nu a_{k}-\nu a_{k +1}+\left (1+\nu \right ) a_{k +1}}{\left (k +3+\nu \right ) \left (k +1+\nu \right ) \left (k +2\right )} \\ \bullet & {} & \textrm {Solution for}\hspace {3pt} r =1+\nu \\ {} & {} & \left [y=\moverset {\infty }{\munderset {k =0}{\sum }}a_{k} x^{k +1+\nu }, a_{k +2}=\frac {k^{2} a_{k +1}-2 k \nu a_{k +1}+2 k \left (1+\nu \right ) a_{k +1}-2 \nu \left (1+\nu \right ) a_{k +1}+\left (1+\nu \right )^{2} a_{k +1}+k a_{k +1}+\nu a_{k}-\nu a_{k +1}+\left (1+\nu \right ) a_{k +1}}{\left (k +3+\nu \right ) \left (k +1+\nu \right ) \left (k +2\right )}, a_{1} \left (2+\nu \right ) \nu -a_{0} \left (-2 \nu \left (1+\nu \right )+\left (1+\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 +2}\right )+\left (\moverset {\infty }{\munderset {k =0}{\sum }}b_{k} x^{k +1+\nu }\right ), a_{k +2}=\frac {k^{2} a_{k +1}-2 k \nu a_{k +1}+5 k a_{k +1}+\nu a_{k}-5 \nu a_{k +1}+6 a_{k +1}}{\left (k +4\right ) \left (k +2\right ) \left (k +3-\nu \right )}, 3 a_{1} \left (-\nu +2\right )-a_{0} \left (-3 \nu +2\right )=0, b_{k +2}=\frac {k^{2} b_{k +1}-2 k \nu b_{k +1}+2 k \left (1+\nu \right ) b_{k +1}-2 \nu \left (1+\nu \right ) b_{k +1}+\left (1+\nu \right )^{2} b_{k +1}+k b_{k +1}+\nu b_{k}-\nu b_{k +1}+\left (1+\nu \right ) b_{k +1}}{\left (k +3+\nu \right ) \left (k +1+\nu \right ) \left (k +2\right )}, b_{1} \left (2+\nu \right ) \nu -b_{0} \left (-2 \nu \left (1+\nu \right )+\left (1+\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/x*nu*y(x)-1/x*nu*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: 114
dsolve(x^2*diff(diff(diff(y(x),x),x),x)-(x+nu)*x*diff(diff(y(x),x),x)+nu*(2*x+1)*diff(y(x),x)-nu*(x+1)*y(x)=0,y(x), singsol=all)
\[ y \left (x \right ) = \frac {-c_{3} x^{\frac {\nu }{2}+1} \operatorname {BesselY}\left (-\nu +1, 2 \sqrt {\nu }\, \sqrt {x}\right )-c_{2} x^{\frac {\nu }{2}+1} \operatorname {BesselJ}\left (-\nu +1, 2 \sqrt {\nu }\, \sqrt {x}\right )-x^{\frac {\nu }{2}+\frac {1}{2}} \sqrt {\nu }\, \operatorname {BesselY}\left (-\nu , 2 \sqrt {\nu }\, \sqrt {x}\right ) c_{3} -x^{\frac {\nu }{2}+\frac {1}{2}} \sqrt {\nu }\, \operatorname {BesselJ}\left (-\nu , 2 \sqrt {\nu }\, \sqrt {x}\right ) c_{2} +c_{1} {\mathrm e}^{x} \sqrt {x}}{\sqrt {x}} \]
✗ Solution by Mathematica
Time used: 0.0 (sec). Leaf size: 0
DSolve[-(nu*(1 + x)*y[x]) + nu*(1 + 2*x)*y'[x] - x*(v + x)*y''[x] + x^2*Derivative[3][y][x] == 0,y[x],x,IncludeSingularSolutions -> True]
Not solved