Internal problem ID [9824]
Internal file name [OUTPUT/8767_Monday_June_06_2022_05_25_27_AM_24055604/index.tex]
Book: Differential Gleichungen, E. Kamke, 3rd ed. Chelsea Pub. NY, 1948
Section: Chapter 3, linear third order
Problem number: 1498.
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.
\[ \boxed {x^{2} y^{\prime \prime \prime }-2 \left (n +1\right ) x y^{\prime \prime }+\left (a \,x^{2}+6 n \right ) y^{\prime }-2 y a x=0} \] 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 (n +1\right ) x \left (\frac {d}{d x}y^{\prime }\right )+\left (a \,x^{2}+6 n \right ) y^{\prime }-2 y a x =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 (n +1\right )}{x}, P_{3}\left (x \right )=\frac {a \,x^{2}+6 n}{x^{2}}, P_{4}\left (x \right )=-\frac {2 a}{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 n -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}}}=6 n \\ {} & \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 \cdot y\hspace {3pt}\textrm {to series expansion}\hspace {3pt} \\ {} & {} & x \cdot y=\moverset {\infty }{\munderset {k =0}{\sum }}a_{k} x^{k +r +1} \\ {} & \circ & \textrm {Shift index using}\hspace {3pt} k \mathrm {->}k -1 \\ {} & {} & x \cdot y=\moverset {\infty }{\munderset {k =1}{\sum }}a_{k -1} x^{k +r} \\ {} & \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 \cdot \left (\frac {d}{d x}y^{\prime }\right )\hspace {3pt}\textrm {to series expansion}\hspace {3pt} \\ {} & {} & x \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 -1} \\ {} & \circ & \textrm {Shift index using}\hspace {3pt} k \mathrm {->}k +1 \\ {} & {} & x \cdot \left (\frac {d}{d x}y^{\prime }\right )=\moverset {\infty }{\munderset {k =-1}{\sum }}a_{k +1} \left (k +r +1\right ) \left (k +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 +r +1\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 (-4+r \right ) \left (-1-2 n +r \right ) x^{-1+r}+a_{1} \left (1+r \right ) \left (-3+r \right ) \left (-2 n +r \right ) x^{r}+\left (\moverset {\infty }{\munderset {k =1}{\sum }}\left (a_{k +1} \left (k +r +1\right ) \left (k -3+r \right ) \left (k -2 n +r \right )+a_{k -1} a \left (k -3+r \right )\right ) x^{k +r}\right )=0 \\ \bullet & {} & a_{0}\textrm {cannot be 0 by assumption, giving the indicial equation}\hspace {3pt} \\ {} & {} & r \left (-4+r \right ) \left (-1-2 n +r \right )=0 \\ \bullet & {} & \textrm {Values of r that satisfy the indicial equation}\hspace {3pt} \\ {} & {} & r \in \left \{0, 4, 1+2 n \right \} \\ \bullet & {} & \textrm {Each term must be 0}\hspace {3pt} \\ {} & {} & a_{1} \left (1+r \right ) \left (-3+r \right ) \left (-2 n +r \right )=0 \\ \bullet & {} & \textrm {Each term in the series must be 0, giving the recursion relation}\hspace {3pt} \\ {} & {} & \left (k -3+r \right ) \left (a_{k +1} \left (k +r +1\right ) \left (k -2 n +r \right )+a_{k -1} a \right )=0 \\ \bullet & {} & \textrm {Shift index using}\hspace {3pt} k \mathrm {->}k +1 \\ {} & {} & \left (k +r -2\right ) \left (a_{k +2} \left (k +2+r \right ) \left (k +1-2 n +r \right )+a_{k} a \right )=0 \\ \bullet & {} & \textrm {Recursion relation that defines series solution to ODE}\hspace {3pt} \\ {} & {} & a_{k +2}=-\frac {a_{k} a}{\left (k +2+r \right ) \left (k +1-2 n +r \right )} \\ \bullet & {} & \textrm {Recursion relation for}\hspace {3pt} r =0 \\ {} & {} & a_{k +2}=-\frac {a_{k} a}{\left (k +2\right ) \left (k +1-2 n \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 {a_{k} a}{\left (k +2\right ) \left (k +1-2 n \right )}, 6 a_{1} n =0\right ] \\ \bullet & {} & \textrm {Recursion relation for}\hspace {3pt} r =4 \\ {} & {} & a_{k +2}=-\frac {a_{k} a}{\left (k +6\right ) \left (k +5-2 n \right )} \\ \bullet & {} & \textrm {Solution for}\hspace {3pt} r =4 \\ {} & {} & \left [y=\moverset {\infty }{\munderset {k =0}{\sum }}a_{k} x^{k +4}, a_{k +2}=-\frac {a_{k} a}{\left (k +6\right ) \left (k +5-2 n \right )}, 5 a_{1} \left (-2 n +4\right )=0\right ] \\ \bullet & {} & \textrm {Recursion relation for}\hspace {3pt} r =1+2 n \\ {} & {} & a_{k +2}=-\frac {a_{k} a}{\left (k +3+2 n \right ) \left (k +2\right )} \\ \bullet & {} & \textrm {Solution for}\hspace {3pt} r =1+2 n \\ {} & {} & \left [y=\moverset {\infty }{\munderset {k =0}{\sum }}a_{k} x^{k +1+2 n}, a_{k +2}=-\frac {a_{k} a}{\left (k +3+2 n \right ) \left (k +2\right )}, a_{1} \left (2 n +2\right ) \left (-2+2 n \right )=0\right ] \\ \bullet & {} & \textrm {Combine solutions and rename parameters}\hspace {3pt} \\ {} & {} & \left [y=\left (\moverset {\infty }{\munderset {k =0}{\sum }}b_{k} x^{k}\right )+\left (\moverset {\infty }{\munderset {k =0}{\sum }}c_{k} x^{k +4}\right )+\left (\moverset {\infty }{\munderset {k =0}{\sum }}d_{k} x^{k +1+2 n}\right ), b_{k +2}=-\frac {b_{k} a}{\left (k +2\right ) \left (k +1-2 n \right )}, 6 b_{1} n =0, c_{k +2}=-\frac {c_{k} a}{\left (k +6\right ) \left (k +5-2 n \right )}, 5 c_{1} \left (-2 n +4\right )=0, d_{k +2}=-\frac {d_{k} a}{\left (k +3+2 n \right ) \left (k +2\right )}, d_{1} \left (2 n +2\right ) \left (-2+2 n \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 -2*x/(x^2+2*(2*n-1)/a)*y(x)+diff(y(x),x), a*y(x)-2*n/x*diff(y(x),x)+diff(diff(y(x),x),x) trying differential order: 1; missing the dependent variable checking if the LODE is of Euler type checking if the LODE is of Euler type exponential solutions successful <- differential factorization 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: 53
dsolve(x^2*diff(diff(diff(y(x),x),x),x)-2*(n+1)*x*diff(diff(y(x),x),x)+(a*x^2+6*n)*diff(y(x),x)-2*y(x)*a*x=0,y(x), singsol=all)
\[ y \left (x \right ) = c_{1} x^{n +\frac {1}{2}} \operatorname {BesselJ}\left (-n -\frac {1}{2}, x \sqrt {a}\right )+c_{2} x^{n +\frac {1}{2}} \operatorname {BesselY}\left (-n -\frac {1}{2}, x \sqrt {a}\right )+c_{3} \left (a \,x^{2}+4 n -2\right ) \]
✓ Solution by Mathematica
Time used: 6.184 (sec). Leaf size: 353
DSolve[-2*a*x*y[x] + (6*n + a*x^2)*y'[x] - 2*(1 + n)*x*y''[x] + x^2*Derivative[3][y][x] == 0,y[x],x,IncludeSingularSolutions -> True]
\[ y(x)\to 2^{-n-\frac {3}{2}} \left (\pi c_3 4^n x^4 \sec (\pi n) \operatorname {Gamma}\left (\frac {3}{2}-n\right ) \left (\sqrt {a} x\right )^{-n-\frac {1}{2}} \operatorname {BesselJ}\left (n+\frac {1}{2},\sqrt {a} x\right ) \, _1\tilde {F}_2\left (\frac {3}{2}-n;\frac {1}{2}-n,\frac {5}{2}-n;-\frac {a x^2}{4}\right )+\frac {\operatorname {BesselY}\left (n+\frac {1}{2},\sqrt {a} x\right ) \left (2 \pi c_3 \left (4 n^2-1\right ) \left (\sqrt {a} x\right )^{n+\frac {1}{2}}+a 2^{n+\frac {1}{2}} \operatorname {Gamma}\left (n+\frac {3}{2}\right ) \left (2 a c_2 x^{n+\frac {1}{2}}-\pi \sqrt {a} c_3 x^3 \operatorname {BesselJ}\left (n-\frac {1}{2},\sqrt {a} x\right )-2 \pi c_3 x^2 \operatorname {BesselJ}\left (n-\frac {3}{2},\sqrt {a} x\right )\right )\right )+\operatorname {BesselJ}\left (n+\frac {1}{2},\sqrt {a} x\right ) \left (2 \pi c_3 \left (4 n^2-1\right ) \tan (\pi n) \left (\sqrt {a} x\right )^{n+\frac {1}{2}}+a 2^{n+\frac {1}{2}} \operatorname {Gamma}\left (n+\frac {3}{2}\right ) \left (2 a c_1 x^{n+\frac {1}{2}}-\pi \sqrt {a} c_3 x^3 \tan (\pi n) \operatorname {BesselJ}\left (n-\frac {1}{2},\sqrt {a} x\right )-2 \pi c_3 x^2 \tan (\pi n) \operatorname {BesselJ}\left (n-\frac {3}{2},\sqrt {a} x\right )\right )\right )}{a^2 \operatorname {Gamma}\left (n+\frac {3}{2}\right )}\right ) \]