Internal problem ID [9810]
Internal file name [OUTPUT/8753_Monday_June_06_2022_05_23_52_AM_78538692/index.tex]
Book: Differential Gleichungen, E. Kamke, 3rd ed. Chelsea Pub. NY, 1948
Section: Chapter 3, linear third order
Problem number: 1484.
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.
\[ \boxed {2 x y^{\prime \prime \prime }+3 \left (2 a x +k \right ) y^{\prime \prime }+6 \left (a k +b x \right ) y^{\prime }+\left (3 b k +2 c x \right ) y=0} \] 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 )+3 \left (2 a x +k \right ) \left (\frac {d}{d x}y^{\prime }\right )+6 \left (a k +b x \right ) y^{\prime }+\left (3 b k +2 c x \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 {3 \left (2 a x +k \right )}{2 x}, P_{3}\left (x \right )=\frac {3 \left (a k +b x \right )}{x}, P_{4}\left (x \right )=\frac {3 b k +2 c x}{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}}}=\frac {3 k}{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}}}=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 (6 a k +6 b x \right ) y^{\prime }+\left (6 a x +3 k \right ) \left (\frac {d}{d x}y^{\prime }\right )+2 x \left (\frac {d}{d x}y^{\prime \prime }\right )+\left (3 b k +2 c x \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\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 =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} \\ {} & {} & a_{0} r \left (-1+r \right ) \left (3 k -4+2 r \right ) x^{-2+r}+\left (a_{1} \left (1+r \right ) r \left (3 k -2+2 r \right )+6 a a_{0} r \left (-1+r +k \right )\right ) x^{-1+r}+\left (a_{2} \left (2+r \right ) \left (1+r \right ) \left (3 k +2 r \right )+6 a a_{1} \left (1+r \right ) \left (r +k \right )+3 a_{0} b \left (2 r +k \right )\right ) x^{r}+\left (\moverset {\infty }{\munderset {k =1}{\sum }}\left (a_{k +2} \left (k +2+r \right ) \left (k +1+r \right ) \left (3 k +2 k +2 r \right )+6 a a_{k +1} \left (k +1+r \right ) \left (k +r +k \right )+3 a_{k} b \left (2 k +2 r +k \right )+2 a_{k -1} c \right ) x^{k +r}\right )=0 \\ \bullet & {} & a_{0}\textrm {cannot be 0 by assumption, giving the indicial equation}\hspace {3pt} \\ {} & {} & r \left (-1+r \right ) \left (3 k -4+2 r \right )=0 \\ \bullet & {} & \textrm {Values of r that satisfy the indicial equation}\hspace {3pt} \\ {} & {} & r \in \left \{0, 1, -\frac {3 k}{2}+2\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 ) r \left (3 k -2+2 r \right )+6 a a_{0} r \left (-1+r +k \right )=0, a_{2} \left (2+r \right ) \left (1+r \right ) \left (3 k +2 r \right )+6 a a_{1} \left (1+r \right ) \left (r +k \right )+3 a_{0} b \left (2 r +k \right )=0\right ] \\ \bullet & {} & \textrm {Each term in the series must be 0, giving the recursion relation}\hspace {3pt} \\ {} & {} & a_{k +2} \left (k +2+r \right ) \left (k +1+r \right ) \left (3 k +2 k +2 r \right )+6 a a_{k +1} \left (k +1+r \right ) \left (k +r +k \right )+3 a_{k} b \left (2 k +2 r +k \right )+2 a_{k -1} c =0 \\ \bullet & {} & \textrm {Shift index using}\hspace {3pt} k \mathrm {->}k +1 \\ {} & {} & a_{k +3} \left (k +3+r \right ) \left (k +2+r \right ) \left (3 k +2 k +2+2 r \right )+6 a a_{k +2} \left (k +2+r \right ) \left (k +1+r +k \right )+3 a_{k +1} b \left (2 k +2+2 r +k \right )+2 a_{k} c =0 \\ \bullet & {} & \textrm {Recursion relation that defines series solution to ODE}\hspace {3pt} \\ {} & {} & a_{k +3}=-\frac {6 a k k a_{k +2}+6 a k r a_{k +2}+6 a \,k^{2} a_{k +2}+12 a k r a_{k +2}+6 a \,r^{2} a_{k +2}+12 a k a_{k +2}+18 a k a_{k +2}+18 a r a_{k +2}+3 b k a_{k +1}+6 b k a_{k +1}+6 b r a_{k +1}+12 a a_{k +2}+6 b a_{k +1}+2 a_{k} c}{\left (k +3+r \right ) \left (k +2+r \right ) \left (3 k +2 k +2+2 r \right )} \\ \bullet & {} & \textrm {Recursion relation for}\hspace {3pt} r =0 \\ {} & {} & a_{k +3}=-\frac {6 a k k a_{k +2}+6 a \,k^{2} a_{k +2}+12 a k a_{k +2}+18 a k a_{k +2}+3 b k a_{k +1}+6 b k a_{k +1}+12 a a_{k +2}+6 b a_{k +1}+2 a_{k} c}{\left (k +3\right ) \left (k +2\right ) \left (3 k +2 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 +3}=-\frac {6 a k k a_{k +2}+6 a \,k^{2} a_{k +2}+12 a k a_{k +2}+18 a k a_{k +2}+3 b k a_{k +1}+6 b k a_{k +1}+12 a a_{k +2}+6 b a_{k +1}+2 a_{k} c}{\left (k +3\right ) \left (k +2\right ) \left (3 k +2 k +2\right )}, 0=0, 6 a a_{1} k +3 a_{0} b k +6 a_{2} k =0\right ] \\ \bullet & {} & \textrm {Recursion relation for}\hspace {3pt} r =1 \\ {} & {} & a_{k +3}=-\frac {6 a k k a_{k +2}+6 a \,k^{2} a_{k +2}+18 a k a_{k +2}+30 a k a_{k +2}+3 b k a_{k +1}+6 b k a_{k +1}+36 a a_{k +2}+12 b a_{k +1}+2 a_{k} c}{\left (k +4\right ) \left (k +3\right ) \left (3 k +2 k +4\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 {6 a k k a_{k +2}+6 a \,k^{2} a_{k +2}+18 a k a_{k +2}+30 a k a_{k +2}+3 b k a_{k +1}+6 b k a_{k +1}+36 a a_{k +2}+12 b a_{k +1}+2 a_{k} c}{\left (k +4\right ) \left (k +3\right ) \left (3 k +2 k +4\right )}, 6 a a_{0} k +6 a_{1} k =0, 6 a_{2} \left (3 k +2\right )+12 a a_{1} \left (k +1\right )+3 a_{0} b \left (k +2\right )=0\right ] \\ \bullet & {} & \textrm {Recursion relation for}\hspace {3pt} r =-\frac {3 k}{2}+2 \\ {} & {} & a_{k +3}=-\frac {6 a k k a_{k +2}+6 a k \left (-\frac {3 k}{2}+2\right ) a_{k +2}+6 a \,k^{2} a_{k +2}+12 a k \left (-\frac {3 k}{2}+2\right ) a_{k +2}+6 a \left (-\frac {3 k}{2}+2\right )^{2} a_{k +2}+12 a k a_{k +2}+18 a k a_{k +2}+18 a \left (-\frac {3 k}{2}+2\right ) a_{k +2}+3 b k a_{k +1}+6 b k a_{k +1}+6 b \left (-\frac {3 k}{2}+2\right ) a_{k +1}+12 a a_{k +2}+6 b a_{k +1}+2 a_{k} c}{\left (k +5-\frac {3 k}{2}\right ) \left (k +4-\frac {3 k}{2}\right ) \left (2 k +6\right )} \\ \bullet & {} & \textrm {Solution for}\hspace {3pt} r =-\frac {3 k}{2}+2 \\ {} & {} & \left [y=\moverset {\infty }{\munderset {k =0}{\sum }}a_{k} x^{k -\frac {3 k}{2}+2}, a_{k +3}=-\frac {6 a k k a_{k +2}+6 a k \left (-\frac {3 k}{2}+2\right ) a_{k +2}+6 a \,k^{2} a_{k +2}+12 a k \left (-\frac {3 k}{2}+2\right ) a_{k +2}+6 a \left (-\frac {3 k}{2}+2\right )^{2} a_{k +2}+12 a k a_{k +2}+18 a k a_{k +2}+18 a \left (-\frac {3 k}{2}+2\right ) a_{k +2}+3 b k a_{k +1}+6 b k a_{k +1}+6 b \left (-\frac {3 k}{2}+2\right ) a_{k +1}+12 a a_{k +2}+6 b a_{k +1}+2 a_{k} c}{\left (k +5-\frac {3 k}{2}\right ) \left (k +4-\frac {3 k}{2}\right ) \left (2 k +6\right )}, 2 a_{1} \left (3-\frac {3 k}{2}\right ) \left (-\frac {3 k}{2}+2\right )+6 a a_{0} \left (-\frac {3 k}{2}+2\right ) \left (1-\frac {k}{2}\right )=0, 4 a_{2} \left (4-\frac {3 k}{2}\right ) \left (3-\frac {3 k}{2}\right )+6 a a_{1} \left (3-\frac {3 k}{2}\right ) \left (-\frac {k}{2}+2\right )+3 a_{0} b \left (-2 k +4\right )=0\right ] \\ \bullet & {} & \textrm {Combine solutions and rename parameters}\hspace {3pt} \\ {} & {} & \left [y=\left (\moverset {\infty }{\munderset {m =0}{\sum }}d_{m} x^{m}\right )+\left (\moverset {\infty }{\munderset {m =0}{\sum }}e_{m} x^{m +1}\right )+\left (\moverset {\infty }{\munderset {m =0}{\sum }}f_{m} x^{m -\frac {3 k}{2}+2}\right ), d_{m +3}=-\frac {6 a k m d_{m +2}+6 a \,m^{2} d_{m +2}+12 a k d_{m +2}+18 a m d_{m +2}+3 b k d_{m +1}+6 b m d_{m +1}+12 a d_{m +2}+6 b d_{m +1}+2 c d_{m}}{\left (m +3\right ) \left (m +2\right ) \left (3 k +2 m +2\right )}, 0=0, 6 a k d_{1}+3 b k d_{0}+6 k d_{2}=0, e_{m +3}=-\frac {6 a k m e_{m +2}+6 a \,m^{2} e_{m +2}+18 a k e_{m +2}+30 a m e_{m +2}+3 b k e_{m +1}+6 b m e_{m +1}+36 a e_{m +2}+12 b e_{m +1}+2 c e_{m}}{\left (m +4\right ) \left (m +3\right ) \left (3 k +2 m +4\right )}, 6 a k e_{0}+6 k e_{1}=0, 6 e_{2} \left (3 k +2\right )+12 a e_{1} \left (k +1\right )+3 e_{0} b \left (k +2\right )=0, f_{m +3}=-\frac {6 a k m f_{m +2}+6 a k \left (-\frac {3 k}{2}+2\right ) f_{m +2}+6 a \,m^{2} f_{m +2}+12 a m \left (-\frac {3 k}{2}+2\right ) f_{m +2}+6 a \left (-\frac {3 k}{2}+2\right )^{2} f_{m +2}+12 a k f_{m +2}+18 a m f_{m +2}+18 a \left (-\frac {3 k}{2}+2\right ) f_{m +2}+3 b k f_{m +1}+6 b m f_{m +1}+6 b \left (-\frac {3 k}{2}+2\right ) f_{m +1}+12 a f_{m +2}+6 b f_{m +1}+2 f_{m} c}{\left (m +5-\frac {3 k}{2}\right ) \left (m +4-\frac {3 k}{2}\right ) \left (2 m +6\right )}, 2 f_{1} \left (3-\frac {3 k}{2}\right ) \left (-\frac {3 k}{2}+2\right )+6 a f_{0} \left (-\frac {3 k}{2}+2\right ) \left (1-\frac {k}{2}\right )=0, 4 f_{2} \left (4-\frac {3 k}{2}\right ) \left (3-\frac {3 k}{2}\right )+6 a f_{1} \left (3-\frac {3 k}{2}\right ) \left (-\frac {k}{2}+2\right )+3 f_{0} b \left (-2 k +4\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 trying Louvillian solutions for 3rd order ODEs, imprimitive case -> pFq: Equivalence to the 3F2 or one of its 3 confluent cases under a power @ Moebius trying a solution in terms of MeijerG functions -> pFq: Equivalence to the 3F2 or one of its 3 confluent cases under a power @ Moebius trying a solution in terms of MeijerG functions checking if the LODE is of Euler type <- no solution through differential factorization was found trying reduction of order using simple exponentials trying differential order: 3; exact nonlinear --- Trying Lie symmetry methods, high order --- `, `-> Computing symmetries using: way = 3`[0, y]
✗ Solution by Maple
dsolve(2*x*diff(diff(diff(y(x),x),x),x)+3*(2*a*x+k)*diff(diff(y(x),x),x)+6*(a*k+b*x)*diff(y(x),x)+(3*b*k+2*c*x)*y(x)=0,y(x), singsol=all)
\[ \text {No solution found} \]
✗ Solution by Mathematica
Time used: 0.0 (sec). Leaf size: 0
DSolve[(3*b*k + 2*c*x)*y[x] + 6*(a*k + b*x)*y'[x] + 3*(k + 2*a*x)*y''[x] + 2*x*Derivative[3][y][x] == 0,y[x],x,IncludeSingularSolutions -> True]
Not solved