Internal problem ID [9831]
Internal file name [OUTPUT/8774_Monday_June_06_2022_05_26_15_AM_69577401/index.tex]
Book: Differential Gleichungen, E. Kamke, 3rd ed. Chelsea Pub. NY, 1948
Section: Chapter 3, linear third order
Problem number: 1505.
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.
\[ \boxed {2 x \left (x -1\right ) y^{\prime \prime \prime }+3 \left (2 x -1\right ) y^{\prime \prime }+\left (2 a x +b \right ) y^{\prime }+a y=0} \] Unable to solve this ODE.
\[ \begin {array}{lll} & {} & \textrm {Let's solve}\hspace {3pt} \\ {} & {} & 2 x \left (x -1\right ) \left (\frac {d}{d x}y^{\prime \prime }\right )+3 \left (2 x -1\right ) \left (\frac {d}{d x}y^{\prime }\right )+\left (2 a x +b \right ) y^{\prime }+a 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}\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 x -1\right )}{2 x \left (x -1\right )}, P_{3}\left (x \right )=\frac {2 a x +b}{2 x \left (x -1\right )}, P_{4}\left (x \right )=\frac {a}{2 x \left (x -1\right )}\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}{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}\hspace {3pt}\textrm {is a regular singular point}\hspace {3pt} \\ {} & {} & x_{0}=0 \\ \bullet & {} & \textrm {Multiply by denominators}\hspace {3pt} \\ {} & {} & a y+\left (2 a x +b \right ) y^{\prime }+\left (6 x -3\right ) \left (\frac {d}{d x}y^{\prime }\right )+2 x \left (x -1\right ) \left (\frac {d}{d x}y^{\prime \prime }\right )=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^{m}\cdot \left (\frac {d}{d x}y^{\prime \prime }\right )\hspace {3pt}\textrm {to series expansion for}\hspace {3pt} m =1..2 \\ {} & {} & x^{m}\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 -3+m} \\ {} & \circ & \textrm {Shift index using}\hspace {3pt} k \mathrm {->}k +3-m \\ {} & {} & x^{m}\cdot \left (\frac {d}{d x}y^{\prime \prime }\right )=\moverset {\infty }{\munderset {k =-3+m}{\sum }}a_{k +3-m} \left (k +3-m +r \right ) \left (k +2-m +r \right ) \left (k +1-m +r \right ) x^{k +r} \\ & {} & \textrm {Rewrite ODE with series expansions}\hspace {3pt} \\ {} & {} & -a_{0} r \left (-1+2 r \right ) \left (-1+r \right ) x^{-2+r}+\left (-a_{1} \left (1+r \right ) \left (1+2 r \right ) r +a_{0} r \left (2 r^{2}+b -2\right )\right ) x^{-1+r}+\left (\moverset {\infty }{\munderset {k =0}{\sum }}\left (-a_{k +2} \left (k +2+r \right ) \left (2 k +3+2 r \right ) \left (k +1+r \right )+a_{k +1} \left (k +1+r \right ) \left (2 \left (k +1\right )^{2}+4 \left (k +1\right ) r +2 r^{2}+b -2\right )+a a_{k} \left (2 k +2 r +1\right )\right ) x^{k +r}\right )=0 \\ \bullet & {} & a_{0}\textrm {cannot be 0 by assumption, giving the indicial equation}\hspace {3pt} \\ {} & {} & -r \left (-1+2 r \right ) \left (-1+r \right )=0 \\ \bullet & {} & \textrm {Values of r that satisfy the indicial equation}\hspace {3pt} \\ {} & {} & r \in \left \{0, 1, \frac {1}{2}\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 +\frac {3}{2}+r \right ) \left (k +1+r \right )+a_{k +1} \left (k +1+r \right ) \left (2 k^{2}+4 k r +2 r^{2}+b +4 k +4 r \right )+a a_{k} \left (2 k +2 r +1\right )=0 \\ \bullet & {} & \textrm {Recursion relation that defines series solution to ODE}\hspace {3pt} \\ {} & {} & a_{k +2}=\frac {2 k^{3} a_{k +1}+6 k^{2} r a_{k +1}+6 k \,r^{2} a_{k +1}+2 r^{3} a_{k +1}+2 a a_{k} k +2 a a_{k} r +b k a_{k +1}+b r a_{k +1}+6 k^{2} a_{k +1}+12 k r a_{k +1}+6 r^{2} a_{k +1}+a a_{k}+b a_{k +1}+4 k a_{k +1}+4 r a_{k +1}}{\left (k +2+r \right ) \left (2 k +3+2 r \right ) \left (k +1+r \right )} \\ \bullet & {} & \textrm {Recursion relation for}\hspace {3pt} r =0 \\ {} & {} & a_{k +2}=\frac {2 k^{3} a_{k +1}+2 a a_{k} k +b k a_{k +1}+6 k^{2} a_{k +1}+a a_{k}+b a_{k +1}+4 k a_{k +1}}{\left (k +2\right ) \left (2 k +3\right ) \left (k +1\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 {2 k^{3} a_{k +1}+2 a a_{k} k +b k a_{k +1}+6 k^{2} a_{k +1}+a a_{k}+b a_{k +1}+4 k a_{k +1}}{\left (k +2\right ) \left (2 k +3\right ) \left (k +1\right )}, 0=0\right ] \\ \bullet & {} & \textrm {Recursion relation for}\hspace {3pt} r =1 \\ {} & {} & a_{k +2}=\frac {2 k^{3} a_{k +1}+2 a a_{k} k +b k a_{k +1}+12 k^{2} a_{k +1}+3 a a_{k}+2 b a_{k +1}+22 k a_{k +1}+12 a_{k +1}}{\left (k +3\right ) \left (2 k +5\right ) \left (k +2\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 {2 k^{3} a_{k +1}+2 a a_{k} k +b k a_{k +1}+12 k^{2} a_{k +1}+3 a a_{k}+2 b a_{k +1}+22 k a_{k +1}+12 a_{k +1}}{\left (k +3\right ) \left (2 k +5\right ) \left (k +2\right )}, a_{0} b -6 a_{1}=0\right ] \\ \bullet & {} & \textrm {Recursion relation for}\hspace {3pt} r =\frac {1}{2} \\ {} & {} & a_{k +2}=\frac {2 k^{3} a_{k +1}+2 a a_{k} k +b k a_{k +1}+9 k^{2} a_{k +1}+2 a a_{k}+\frac {3}{2} b a_{k +1}+\frac {23}{2} k a_{k +1}+\frac {15}{4} a_{k +1}}{\left (k +\frac {5}{2}\right ) \left (2 k +4\right ) \left (k +\frac {3}{2}\right )} \\ \bullet & {} & \textrm {Solution for}\hspace {3pt} r =\frac {1}{2} \\ {} & {} & \left [y=\moverset {\infty }{\munderset {k =0}{\sum }}a_{k} x^{k +\frac {1}{2}}, a_{k +2}=\frac {2 k^{3} a_{k +1}+2 a a_{k} k +b k a_{k +1}+9 k^{2} a_{k +1}+2 a a_{k}+\frac {3}{2} b a_{k +1}+\frac {23}{2} k a_{k +1}+\frac {15}{4} a_{k +1}}{\left (k +\frac {5}{2}\right ) \left (2 k +4\right ) \left (k +\frac {3}{2}\right )}, -\frac {3 a_{1}}{2}+\frac {a_{0} \left (b -\frac {3}{2}\right )}{2}=0\right ] \\ \bullet & {} & \textrm {Combine solutions and rename parameters}\hspace {3pt} \\ {} & {} & \left [y=\left (\moverset {\infty }{\munderset {k =0}{\sum }}c_{k} x^{k}\right )+\left (\moverset {\infty }{\munderset {k =0}{\sum }}d_{k} x^{k +1}\right )+\left (\moverset {\infty }{\munderset {k =0}{\sum }}e_{k} x^{k +\frac {1}{2}}\right ), c_{k +2}=\frac {2 k^{3} c_{k +1}+2 a k c_{k}+b k c_{k +1}+6 k^{2} c_{k +1}+a c_{k}+b c_{k +1}+4 k c_{k +1}}{\left (k +2\right ) \left (2 k +3\right ) \left (k +1\right )}, 0=0, d_{k +2}=\frac {2 k^{3} d_{k +1}+2 a k d_{k}+b k d_{k +1}+12 k^{2} d_{k +1}+3 a d_{k}+2 b d_{k +1}+22 k d_{k +1}+12 d_{k +1}}{\left (k +3\right ) \left (2 k +5\right ) \left (k +2\right )}, b d_{0}-6 d_{1}=0, e_{k +2}=\frac {2 k^{3} e_{k +1}+2 a e_{k} k +b k e_{k +1}+9 k^{2} e_{k +1}+2 a e_{k}+\frac {3}{2} b e_{k +1}+\frac {23}{2} k e_{k +1}+\frac {15}{4} e_{k +1}}{\left (k +\frac {5}{2}\right ) \left (4+2 k \right ) \left (k +\frac {3}{2}\right )}, -\frac {3 e_{1}}{2}+\frac {e_{0} \left (b -\frac {3}{2}\right )}{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 Equation is the 2nd symmetric power of diff(diff(y(x),x),x)+1/2*(-1+2*x)/x/(x-1)*diff(y(x),x)+1/8*(2*a*x+b-2)/x/(x-1)*y(x) = 0 -> Attempting now to solve this lower order ODE 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 -> hyper3: Equivalence to 1F1 under a power @ Moebius -> hypergeometric -> heuristic approach -> hyper3: Equivalence to 2F1, 1F1 or 0F1 under a power @ Moebius -> Mathieu -> Equivalence to the rational form of Mathieu ODE under a power @ Moebius Equivalence transformation and function parameters: {x = t}, {kappa = -2*b, mu = 4*a} <- Equivalence to the rational form of Mathieu ODE successful <- Mathieu successful <- special function solution successful`
✓ Solution by Maple
Time used: 0.453 (sec). Leaf size: 79
dsolve(2*x*(x-1)*diff(diff(diff(y(x),x),x),x)+3*(2*x-1)*diff(diff(y(x),x),x)+(2*a*x+b)*diff(y(x),x)+a*y(x)=0,y(x), singsol=all)
\[ y \left (x \right ) = c_{1} \operatorname {MathieuC}\left (1-\frac {a}{2}-\frac {b}{2}, \frac {a}{4}, \arccos \left (\sqrt {x}\right )\right )^{2}+c_{2} \operatorname {MathieuS}\left (1-\frac {a}{2}-\frac {b}{2}, \frac {a}{4}, \arccos \left (\sqrt {x}\right )\right )^{2}+c_{3} \operatorname {MathieuC}\left (1-\frac {a}{2}-\frac {b}{2}, \frac {a}{4}, \arccos \left (\sqrt {x}\right )\right ) \operatorname {MathieuS}\left (1-\frac {a}{2}-\frac {b}{2}, \frac {a}{4}, \arccos \left (\sqrt {x}\right )\right ) \]
✓ Solution by Mathematica
Time used: 60.212 (sec). Leaf size: 115
DSolve[a*y[x] + (b + 2*a*x)*y'[x] + 3*(-1 + 2*x)*y''[x] + 2*(-1 + x)*x*Derivative[3][y][x] == 0,y[x],x,IncludeSingularSolutions -> True]
\[ y(x)\to c_3 \text {MathieuC}\left [-\frac {a}{2}-\frac {b}{2}+1,\frac {a}{4},\arccos \left (\sqrt {x}\right )\right ] \text {MathieuS}\left [-\frac {a}{2}-\frac {b}{2}+1,\frac {a}{4},\arccos \left (\sqrt {x}\right )\right ]+c_1 \text {MathieuC}\left [-\frac {a}{2}-\frac {b}{2}+1,\frac {a}{4},\arccos \left (\sqrt {x}\right )\right ]^2+c_2 \text {MathieuS}\left [-\frac {a}{2}-\frac {b}{2}+1,\frac {a}{4},\arccos \left (\sqrt {x}\right )\right ]^2 \]