Internal problem ID [9809]
Internal file name [OUTPUT/8752_Monday_June_06_2022_05_23_45_AM_5319887/index.tex]
Book: Differential Gleichungen, E. Kamke, 3rd ed. Chelsea Pub. NY, 1948
Section: Chapter 3, linear third order
Problem number: 1483.
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.
\[ \boxed {2 x y^{\prime \prime \prime }-4 \left (x +\nu -1\right ) y^{\prime \prime }+\left (2 x +6 \nu -5\right ) y^{\prime }+\left (1-2 \nu \right ) y=0} \] Unable to solve this ODE.
\[ \begin {array}{lll} & {} & \textrm {Let's solve}\hspace {3pt} \\ {} & {} & 2 x \left (\frac {d}{d x}y^{\prime \prime }\right )-4 \left (x +\nu -1\right ) \left (\frac {d}{d x}y^{\prime }\right )+\left (2 x +6 \nu -5\right ) y^{\prime }+\left (1-2 \nu \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 +\nu -1\right )}{x}, P_{3}\left (x \right )=\frac {2 x +6 \nu -5}{2 x}, P_{4}\left (x \right )=-\frac {2 \nu -1}{2 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}}}=2-2 \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}}}=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 x +6 \nu -5\right ) y^{\prime }+\left (-4 \nu -4 x +4\right ) \left (\frac {d}{d x}y^{\prime }\right )+2 x \left (\frac {d}{d x}y^{\prime \prime }\right )+\left (1-2 \nu \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..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} \\ {} & {} & 2 a_{0} r \left (-1+r \right ) \left (-2 \nu +r \right ) x^{-2+r}+\left (2 a_{1} \left (1+r \right ) r \left (1-2 \nu +r \right )-a_{0} r \left (1+4 r -6 \nu \right )\right ) x^{-1+r}+\left (\moverset {\infty }{\munderset {k =0}{\sum }}\left (2 a_{k +2} \left (k +2+r \right ) \left (k +1+r \right ) \left (k +2-2 \nu +r \right )-a_{k +1} \left (k +1+r \right ) \left (4 k +5+4 r -6 \nu \right )+a_{k} \left (2 k +2 r -2 \nu +1\right )\right ) x^{k +r}\right )=0 \\ \bullet & {} & a_{0}\textrm {cannot be 0 by assumption, giving the indicial equation}\hspace {3pt} \\ {} & {} & 2 r \left (-1+r \right ) \left (-2 \nu +r \right )=0 \\ \bullet & {} & \textrm {Values of r that satisfy the indicial equation}\hspace {3pt} \\ {} & {} & r \in \left \{0, 1, 2 \nu \right \} \\ \bullet & {} & \textrm {Each term in the series must be 0, giving the recursion relation}\hspace {3pt} \\ {} & {} & 2 a_{k +2} \left (k +2+r \right ) \left (k +1+r \right ) \left (k +2-2 \nu +r \right )-a_{k +1} \left (k +1+r \right ) \left (4 k +5+4 r -6 \nu \right )+a_{k} \left (2 k +2 r -2 \nu +1\right )=0 \\ \bullet & {} & \textrm {Recursion relation that defines series solution to ODE}\hspace {3pt} \\ {} & {} & a_{k +2}=\frac {4 k^{2} a_{k +1}-6 k \nu a_{k +1}+8 k r a_{k +1}-6 \nu r a_{k +1}+4 r^{2} a_{k +1}-2 a_{k} k +9 k a_{k +1}+2 a_{k} \nu -6 \nu a_{k +1}-2 a_{k} r +9 r a_{k +1}-a_{k}+5 a_{k +1}}{2 \left (k +2+r \right ) \left (k +1+r \right ) \left (k +2-2 \nu +r \right )} \\ \bullet & {} & \textrm {Recursion relation for}\hspace {3pt} r =0 \\ {} & {} & a_{k +2}=\frac {4 k^{2} a_{k +1}-6 k \nu a_{k +1}-2 a_{k} k +9 k a_{k +1}+2 a_{k} \nu -6 \nu a_{k +1}-a_{k}+5 a_{k +1}}{2 \left (k +2\right ) \left (k +1\right ) \left (k +2-2 \nu \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 {4 k^{2} a_{k +1}-6 k \nu a_{k +1}-2 a_{k} k +9 k a_{k +1}+2 a_{k} \nu -6 \nu a_{k +1}-a_{k}+5 a_{k +1}}{2 \left (k +2\right ) \left (k +1\right ) \left (k +2-2 \nu \right )}, 0=0\right ] \\ \bullet & {} & \textrm {Recursion relation for}\hspace {3pt} r =1 \\ {} & {} & a_{k +2}=\frac {4 k^{2} a_{k +1}-6 k \nu a_{k +1}-2 a_{k} k +17 k a_{k +1}+2 a_{k} \nu -12 \nu a_{k +1}-3 a_{k}+18 a_{k +1}}{2 \left (k +3\right ) \left (k +2\right ) \left (k +3-2 \nu \right )} \\ \bullet & {} & \textrm {Solution for}\hspace {3pt} r =1 \\ {} & {} & \left [y=\moverset {\infty }{\munderset {k =0}{\sum }}a_{k} x^{k +1}, a_{k +2}=\frac {4 k^{2} a_{k +1}-6 k \nu a_{k +1}-2 a_{k} k +17 k a_{k +1}+2 a_{k} \nu -12 \nu a_{k +1}-3 a_{k}+18 a_{k +1}}{2 \left (k +3\right ) \left (k +2\right ) \left (k +3-2 \nu \right )}, 4 a_{1} \left (2-2 \nu \right )-a_{0} \left (5-6 \nu \right )=0\right ] \\ \bullet & {} & \textrm {Recursion relation for}\hspace {3pt} r =2 \nu \\ {} & {} & a_{k +2}=\frac {4 k^{2} a_{k +1}+10 k \nu a_{k +1}+4 \nu ^{2} a_{k +1}-2 a_{k} k +9 k a_{k +1}-2 a_{k} \nu +12 \nu a_{k +1}-a_{k}+5 a_{k +1}}{2 \left (k +2+2 \nu \right ) \left (k +1+2 \nu \right ) \left (k +2\right )} \\ \bullet & {} & \textrm {Solution for}\hspace {3pt} r =2 \nu \\ {} & {} & \left [y=\moverset {\infty }{\munderset {k =0}{\sum }}a_{k} x^{k +2 \nu }, a_{k +2}=\frac {4 k^{2} a_{k +1}+10 k \nu a_{k +1}+4 \nu ^{2} a_{k +1}-2 a_{k} k +9 k a_{k +1}-2 a_{k} \nu +12 \nu a_{k +1}-a_{k}+5 a_{k +1}}{2 \left (k +2+2 \nu \right ) \left (k +1+2 \nu \right ) \left (k +2\right )}, 4 a_{1} \left (2 \nu +1\right ) \nu -2 a_{0} \nu \left (2 \nu +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 +1}\right )+\left (\moverset {\infty }{\munderset {k =0}{\sum }}c_{k} x^{k +2 \nu }\right ), a_{k +2}=\frac {4 k^{2} a_{k +1}-6 k \nu a_{k +1}-2 k a_{k}+9 k a_{k +1}+2 \nu a_{k}-6 \nu a_{k +1}-a_{k}+5 a_{k +1}}{2 \left (k +2\right ) \left (k +1\right ) \left (k +2-2 \nu \right )}, 0=0, b_{k +2}=\frac {4 k^{2} b_{k +1}-6 k \nu b_{k +1}-2 k b_{k}+17 k b_{k +1}+2 \nu b_{k}-12 \nu b_{k +1}-3 b_{k}+18 b_{k +1}}{2 \left (k +3\right ) \left (k +2\right ) \left (k +3-2 \nu \right )}, 4 b_{1} \left (2-2 \nu \right )-b_{0} \left (5-6 \nu \right )=0, c_{k +2}=\frac {4 k^{2} c_{k +1}+10 k \nu c_{k +1}+4 \nu ^{2} c_{k +1}-2 k c_{k}+9 k c_{k +1}-2 \nu c_{k}+12 \nu c_{k +1}-c_{k}+5 c_{k +1}}{2 \left (k +2+2 \nu \right ) \left (k +1+2 \nu \right ) \left (k +2\right )}, 4 c_{1} \left (2 \nu +1\right ) \nu -2 c_{0} \nu \left (2 \nu +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), 1/2*(-1+2*nu)/x*y(x)-(x+2*nu-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: 0.015 (sec). Leaf size: 37
dsolve(2*x*diff(diff(diff(y(x),x),x),x)-4*(x+nu-1)*diff(diff(y(x),x),x)+(2*x+6*nu-5)*diff(y(x),x)+(1-2*nu)*y(x)=0,y(x), singsol=all)
\[ y \left (x \right ) = c_{1} {\mathrm e}^{x}+c_{2} {\mathrm e}^{\frac {x}{2}} x^{\nu } \operatorname {BesselI}\left (\nu , \frac {x}{2}\right )+c_{3} {\mathrm e}^{\frac {x}{2}} x^{\nu } \operatorname {BesselK}\left (\nu , \frac {x}{2}\right ) \]
✓ Solution by Mathematica
Time used: 0.085 (sec). Leaf size: 105
DSolve[(1 - 2*nu)*y[x] + (-5 + 6*nu + 2*x)*y'[x] - 4*(-1 + nu + x)*y''[x] + 2*x*Derivative[3][y][x] == 0,y[x],x,IncludeSingularSolutions -> True]
\[ y(x)\to e^x \left (\frac {2 c_3 \operatorname {Gamma}\left (\frac {5}{2}-3 \nu \right ) \left (\operatorname {Gamma}(2-2 \nu ) \, _1\tilde {F}_1\left (\frac {3}{2}-3 \nu ;1-2 \nu ;-x\right )+2 \nu -1\right )}{3 (2 \nu -1) \operatorname {Gamma}(2-2 \nu ) \operatorname {Gamma}\left (\frac {3}{2}-\nu \right )}+c_2 G_{2,3}^{2,1}\left (x\left | \begin {array}{c} 1,3 \nu -\frac {1}{2} \\ 1,2 \nu ,0 \\ \end {array} \right .\right )+c_1\right ) \]