Internal problem ID [9828]
Internal file name [OUTPUT/8771_Monday_June_06_2022_05_25_55_AM_26674045/index.tex]
Book: Differential Gleichungen, E. Kamke, 3rd ed. Chelsea Pub. NY, 1948
Section: Chapter 3, linear third order
Problem number: 1502.
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.
\[ \boxed {x^{2} y^{\prime \prime \prime }-\left (x^{4}-6 x \right ) y^{\prime \prime }-\left (2 x^{3}-6\right ) y^{\prime }+2 y x^{2}=0} \] 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^{4}-6 x \right ) \left (\frac {d}{d x}y^{\prime }\right )-\left (2 x^{3}-6\right ) y^{\prime }+2 y x^{2}=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^{3}-6}{x}, P_{3}\left (x \right )=-\frac {2 \left (x^{3}-3\right )}{x^{2}}, P_{4}\left (x \right )=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}}}=6 \\ {} & \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 \\ {} & \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} \\ {} & {} & x^{2} \left (\frac {d}{d x}y^{\prime \prime }\right )-x \left (x^{3}-6\right ) \left (\frac {d}{d x}y^{\prime }\right )+\left (-2 x^{3}+6\right ) y^{\prime }+2 y x^{2}=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^{2}\cdot y\hspace {3pt}\textrm {to series expansion}\hspace {3pt} \\ {} & {} & x^{2}\cdot y=\moverset {\infty }{\munderset {k =0}{\sum }}a_{k} x^{k +r +2} \\ {} & \circ & \textrm {Shift index using}\hspace {3pt} k \mathrm {->}k -2 \\ {} & {} & x^{2}\cdot y=\moverset {\infty }{\munderset {k =2}{\sum }}a_{k -2} x^{k +r} \\ {} & \circ & \textrm {Convert}\hspace {3pt} x^{m}\cdot y^{\prime }\hspace {3pt}\textrm {to series expansion for}\hspace {3pt} m =0..3 \\ {} & {} & 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..4 \\ {} & {} & 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+r \right ) x^{-1+r}+a_{1} \left (1+r \right ) \left (3+r \right ) \left (2+r \right ) x^{r}+a_{2} \left (2+r \right ) \left (4+r \right ) \left (3+r \right ) x^{1+r}+\left (\moverset {\infty }{\munderset {k =2}{\sum }}\left (a_{k +1} \left (k +1+r \right ) \left (k +3+r \right ) \left (k +r +2\right )-a_{k -2} \left (k +r \right ) \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 (2+r \right ) \left (1+r \right )=0 \\ \bullet & {} & \textrm {Values of r that satisfy the indicial equation}\hspace {3pt} \\ {} & {} & r \in \left \{-2, -1, 0\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 ) \left (3+r \right ) \left (2+r \right )=0, a_{2} \left (2+r \right ) \left (4+r \right ) \left (3+r \right )=0\right ] \\ \bullet & {} & \textrm {Solve for the dependent coefficient(s)}\hspace {3pt} \\ {} & {} & \left \{a_{1}=0, a_{2}=0\right \} \\ \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 +3+r \right ) \left (k +r +2\right )-a_{k -2} \left (k +r \right ) \left (k -3+r \right )=0 \\ \bullet & {} & \textrm {Shift index using}\hspace {3pt} k \mathrm {->}k +2 \\ {} & {} & a_{k +3} \left (k +3+r \right ) \left (k +5+r \right ) \left (k +4+r \right )-a_{k} \left (k +r +2\right ) \left (k +r -1\right )=0 \\ \bullet & {} & \textrm {Recursion relation that defines series solution to ODE}\hspace {3pt} \\ {} & {} & a_{k +3}=\frac {a_{k} \left (k +r +2\right ) \left (k +r -1\right )}{\left (k +3+r \right ) \left (k +5+r \right ) \left (k +4+r \right )} \\ \bullet & {} & \textrm {Recursion relation for}\hspace {3pt} r =-2\hspace {3pt}\textrm {; series terminates at}\hspace {3pt} k =3 \\ {} & {} & a_{k +3}=\frac {a_{k} k \left (k -3\right )}{\left (k +1\right ) \left (k +3\right ) \left (k +2\right )} \\ \bullet & {} & \textrm {Solution for}\hspace {3pt} r =-2 \\ {} & {} & \left [y=\moverset {\infty }{\munderset {k =0}{\sum }}a_{k} x^{k -2}, a_{k +3}=\frac {a_{k} k \left (k -3\right )}{\left (k +1\right ) \left (k +3\right ) \left (k +2\right )}, a_{1}=0, a_{2}=0\right ] \\ \bullet & {} & \textrm {Recursion relation for}\hspace {3pt} r =-1 \\ {} & {} & a_{k +3}=\frac {a_{k} \left (k +1\right ) \left (k -2\right )}{\left (k +2\right ) \left (k +4\right ) \left (k +3\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 {a_{k} \left (k +1\right ) \left (k -2\right )}{\left (k +2\right ) \left (k +4\right ) \left (k +3\right )}, a_{1}=0, a_{2}=0\right ] \\ \bullet & {} & \textrm {Recursion relation for}\hspace {3pt} r =0 \\ {} & {} & a_{k +3}=\frac {a_{k} \left (k +2\right ) \left (k -1\right )}{\left (k +3\right ) \left (k +5\right ) \left (k +4\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 {a_{k} \left (k +2\right ) \left (k -1\right )}{\left (k +3\right ) \left (k +5\right ) \left (k +4\right )}, a_{1}=0, a_{2}=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}\right )+\left (\moverset {\infty }{\munderset {k =0}{\sum }}c_{k} x^{k}\right ), a_{k +3}=\frac {a_{k} k \left (k -3\right )}{\left (k +1\right ) \left (k +3\right ) \left (k +2\right )}, a_{1}=0, a_{2}=0, b_{k +3}=\frac {b_{k} \left (k +1\right ) \left (k -2\right )}{\left (k +2\right ) \left (k +4\right ) \left (k +3\right )}, b_{1}=0, b_{2}=0, c_{k +3}=\frac {c_{k} \left (k +2\right ) \left (k -1\right )}{\left (k +3\right ) \left (k +5\right ) \left (k +4\right )}, c_{1}=0, c_{2}=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 trying Louvillian solutions for 3rd order ODEs, imprimitive case Louvillian solutions for 3rd order ODEs, imprimitive case: input is reducible, switching to DFactorsols checking if the LODE is of Euler type expon. solutions partially successful. Result(s) =`, [1/x^2]`Calling dsolve with: 2*y(x)-x*(x^3-4)*diff(y(x),x)+x^2*diff(diff(y(x),x 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 -> elliptic -> Legendre <- Kummer successful <- special function solution successful`
✓ Solution by Maple
Time used: 0.156 (sec). Leaf size: 103
dsolve(x^2*diff(diff(diff(y(x),x),x),x)-(x^4-6*x)*diff(diff(y(x),x),x)-(2*x^3-6)*diff(y(x),x)+2*x^2*y(x)=0,y(x), singsol=all)
\[ y \left (x \right ) = \frac {c_{3} \left (\int {\mathrm e}^{\frac {x^{3}}{6}} \sqrt {x}\, \left (\operatorname {BesselK}\left (\frac {5}{6}, -\frac {x^{3}}{6}\right ) x^{3}-\operatorname {BesselK}\left (\frac {1}{6}, -\frac {x^{3}}{6}\right ) x^{3}+2 \operatorname {BesselK}\left (\frac {1}{6}, -\frac {x^{3}}{6}\right )\right )d x \right )+c_{2} \left (\int {\mathrm e}^{\frac {x^{3}}{6}} \sqrt {x}\, \left (\operatorname {BesselI}\left (\frac {1}{6}, -\frac {x^{3}}{6}\right ) x^{3}+\operatorname {BesselI}\left (-\frac {5}{6}, -\frac {x^{3}}{6}\right ) x^{3}-2 \operatorname {BesselI}\left (\frac {1}{6}, -\frac {x^{3}}{6}\right )\right )d x \right )+c_{1}}{x^{2}} \]
✓ Solution by Mathematica
Time used: 0.059 (sec). Leaf size: 98
DSolve[2*x^2*y[x] - (-6 + 2*x^3)*y'[x] - (-6*x + x^4)*y''[x] + x^2*Derivative[3][y][x] == 0,y[x],x,IncludeSingularSolutions -> True]
\[ y(x)\to \frac {c_2 \operatorname {Gamma}\left (\frac {1}{3}\right ) \, _2F_2\left (-\frac {2}{3},\frac {1}{3};\frac {2}{3},\frac {4}{3};\frac {x^3}{3}\right )}{3 x \operatorname {Gamma}\left (\frac {4}{3}\right )}+\frac {\sqrt [3]{-\frac {1}{3}} c_3 \operatorname {Gamma}\left (\frac {2}{3}\right ) \, _2F_2\left (-\frac {1}{3},\frac {2}{3};\frac {4}{3},\frac {5}{3};\frac {x^3}{3}\right )}{3 \operatorname {Gamma}\left (\frac {5}{3}\right )}+\frac {c_1}{x^2} \]