Internal problem ID [9805]
Internal file name [OUTPUT/8748_Monday_June_06_2022_05_23_19_AM_90541664/index.tex]
Book: Differential Gleichungen, E. Kamke, 3rd ed. Chelsea Pub. NY, 1948
Section: Chapter 3, linear third order
Problem number: 1479.
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.
\[ \boxed {x y^{\prime \prime \prime }+\left (a +b \right ) y^{\prime \prime }-y^{\prime } x -a y=0} \] Unable to solve this ODE.
\[ \begin {array}{lll} & {} & \textrm {Let's solve}\hspace {3pt} \\ {} & {} & x \left (\frac {d}{d x}y^{\prime \prime }\right )+\left (a +b \right ) \left (\frac {d}{d x}y^{\prime }\right )-y^{\prime } x -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}=0\hspace {3pt}\textrm {is a regular singular point}\hspace {3pt} \\ {} & \circ & \textrm {Define functions}\hspace {3pt} \\ {} & {} & \left [P_{2}\left (x \right )=\frac {a +b}{x}, P_{3}\left (x \right )=-1, P_{4}\left (x \right )=-\frac {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}}}=a +b \\ {} & \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 {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^{\prime }\hspace {3pt}\textrm {to series expansion}\hspace {3pt} \\ {} & {} & x \cdot y^{\prime }=\moverset {\infty }{\munderset {k =0}{\sum }}a_{k} \left (k +r \right ) x^{k +r} \\ {} & \circ & \textrm {Convert}\hspace {3pt} \frac {d}{d x}y^{\prime }\hspace {3pt}\textrm {to series expansion}\hspace {3pt} \\ {} & {} & \frac {d}{d x}y^{\prime }=\moverset {\infty }{\munderset {k =0}{\sum }}a_{k} \left (k +r \right ) \left (k +r -1\right ) x^{k +r -2} \\ {} & \circ & \textrm {Shift index using}\hspace {3pt} k \mathrm {->}k +2 \\ {} & {} & \frac {d}{d x}y^{\prime }=\moverset {\infty }{\munderset {k =-2}{\sum }}a_{k +2} \left (k +2+r \right ) \left (k +1+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 (a +b -2+r \right ) x^{-2+r}+a_{1} \left (1+r \right ) r \left (a +b -1+r \right ) x^{-1+r}+\left (\moverset {\infty }{\munderset {k =0}{\sum }}\left (a_{k +2} \left (k +2+r \right ) \left (k +1+r \right ) \left (a +b +k +r \right )-a_{k} \left (k +r +a \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+r \right ) \left (a +b -2+r \right )=0 \\ \bullet & {} & \textrm {Values of r that satisfy the indicial equation}\hspace {3pt} \\ {} & {} & r \in \left \{0, 1, -a -b +2\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 (a +b +k +r \right )-a_{k} \left (k +r +a \right )=0 \\ \bullet & {} & \textrm {Recursion relation that defines series solution to ODE}\hspace {3pt} \\ {} & {} & a_{k +2}=\frac {a_{k} \left (k +r +a \right )}{\left (k +2+r \right ) \left (k +1+r \right ) \left (a +b +k +r \right )} \\ \bullet & {} & \textrm {Recursion relation for}\hspace {3pt} r =0 \\ {} & {} & a_{k +2}=\frac {a_{k} \left (k +a \right )}{\left (k +2\right ) \left (k +1\right ) \left (a +b +k \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} \left (k +a \right )}{\left (k +2\right ) \left (k +1\right ) \left (a +b +k \right )}, 0=0\right ] \\ \bullet & {} & \textrm {Recursion relation for}\hspace {3pt} r =1 \\ {} & {} & a_{k +2}=\frac {a_{k} \left (k +1+a \right )}{\left (k +3\right ) \left (k +2\right ) \left (a +b +k +1\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 {a_{k} \left (k +1+a \right )}{\left (k +3\right ) \left (k +2\right ) \left (a +b +k +1\right )}, 2 a_{1} \left (a +b \right )=0\right ] \\ \bullet & {} & \textrm {Recursion relation for}\hspace {3pt} r =-a -b +2 \\ {} & {} & a_{k +2}=\frac {a_{k} \left (k -b +2\right )}{\left (k +4-a -b \right ) \left (k +3-a -b \right ) \left (k +2\right )} \\ \bullet & {} & \textrm {Solution for}\hspace {3pt} r =-a -b +2 \\ {} & {} & \left [y=\moverset {\infty }{\munderset {k =0}{\sum }}a_{k} x^{k -a -b +2}, a_{k +2}=\frac {a_{k} \left (k -b +2\right )}{\left (k +4-a -b \right ) \left (k +3-a -b \right ) \left (k +2\right )}, a_{1} \left (3-a -b \right ) \left (-a -b +2\right )=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 -a -b +2}\right ), c_{k +2}=\frac {c_{k} \left (k +a \right )}{\left (k +2\right ) \left (k +1\right ) \left (a +b +k \right )}, 0=0, d_{k +2}=\frac {d_{k} \left (k +1+a \right )}{\left (k +3\right ) \left (k +2\right ) \left (a +b +k +1\right )}, 2 d_{1} \left (a +b \right )=0, e_{k +2}=\frac {e_{k} \left (k -b +2\right )}{\left (k +4-a -b \right ) \left (k +3-a -b \right ) \left (k +2\right )}, e_{1} \left (3-a -b \right ) \left (-a -b +2\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 <- pFq successful: received ODE is equivalent to the 1F2 ODE, case c = 0 `
✓ Solution by Maple
Time used: 0.015 (sec). Leaf size: 92
dsolve(x*diff(diff(diff(y(x),x),x),x)+(a+b)*diff(diff(y(x),x),x)-x*diff(y(x),x)-a*y(x)=0,y(x), singsol=all)
\[ y \left (x \right ) = c_{1} \operatorname {hypergeom}\left (\left [\frac {a}{2}\right ], \left [\frac {1}{2}, \frac {a}{2}+\frac {b}{2}\right ], \frac {x^{2}}{4}\right )+c_{2} x \operatorname {hypergeom}\left (\left [\frac {1}{2}+\frac {a}{2}\right ], \left [\frac {3}{2}, \frac {a}{2}+\frac {b}{2}+\frac {1}{2}\right ], \frac {x^{2}}{4}\right )+c_{3} x^{-a -b +2} \operatorname {hypergeom}\left (\left [1-\frac {b}{2}\right ], \left [2-\frac {b}{2}-\frac {a}{2}, -\frac {a}{2}-\frac {b}{2}+\frac {3}{2}\right ], \frac {x^{2}}{4}\right ) \]
✓ Solution by Mathematica
Time used: 0.076 (sec). Leaf size: 153
DSolve[-(a*y[x]) - x*y'[x] + (a + b)*y''[x] + x*Derivative[3][y][x] == 0,y[x],x,IncludeSingularSolutions -> True]
\[ y(x)\to \frac {1}{2} i c_2 x \, _1F_2\left (\frac {a}{2}+\frac {1}{2};\frac {3}{2},\frac {a}{2}+\frac {b}{2}+\frac {1}{2};\frac {x^2}{4}\right )+c_1 \, _1F_2\left (\frac {a}{2};\frac {1}{2},\frac {a}{2}+\frac {b}{2};\frac {x^2}{4}\right )+c_3 \left (\frac {i}{2}\right )^{-a-b+2} x^{-a-b+2} \, _1F_2\left (1-\frac {b}{2};-\frac {a}{2}-\frac {b}{2}+\frac {3}{2},-\frac {a}{2}-\frac {b}{2}+2;\frac {x^2}{4}\right ) \]