Internal problem ID [9823]
Internal file name [OUTPUT/8766_Monday_June_06_2022_05_25_20_AM_89081094/index.tex]
Book: Differential Gleichungen, E. Kamke, 3rd ed. Chelsea Pub. NY, 1948
Section: Chapter 3, linear third order
Problem number: 1497.
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.
\[ \boxed {x^{2} y^{\prime \prime \prime }-3 \left (p +q \right ) x y^{\prime \prime }+3 p \left (3 q +1\right ) y^{\prime }-x^{2} y=0} \] 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 )-3 \left (p +q \right ) x \left (\frac {d}{d x}y^{\prime }\right )+3 p \left (3 q +1\right ) y^{\prime }-x^{2} 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 (p +q \right )}{x}, P_{3}\left (x \right )=\frac {3 p \left (3 q +1\right )}{x^{2}}, P_{4}\left (x \right )=-1\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}}}=-3 p -3 q \\ {} & \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}}}=3 p \left (3 q +1\right ) \\ {} & \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^{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} y^{\prime }\hspace {3pt}\textrm {to series expansion}\hspace {3pt} \\ {} & {} & y^{\prime }=\moverset {\infty }{\munderset {k =0}{\sum }}a_{k} \left (k +r \right ) x^{k +r -1} \\ {} & \circ & \textrm {Shift index using}\hspace {3pt} k \mathrm {->}k +1 \\ {} & {} & y^{\prime }=\moverset {\infty }{\munderset {k =-1}{\sum }}a_{k +1} \left (k +1+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 +1+r \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 +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 (-3 q +r -2\right ) \left (-1-3 p +r \right ) x^{-1+r}+a_{1} \left (1+r \right ) \left (-1-3 q +r \right ) \left (-3 p +r \right ) x^{r}+a_{2} \left (2+r \right ) \left (-3 q +r \right ) \left (1-3 p +r \right ) x^{1+r}+\left (\moverset {\infty }{\munderset {k =2}{\sum }}\left (a_{k +1} \left (k +1+r \right ) \left (k -1-3 q +r \right ) \left (k -3 p +r \right )-a_{k -2}\right ) x^{k +r}\right )=0 \\ \bullet & {} & a_{0}\textrm {cannot be 0 by assumption, giving the indicial equation}\hspace {3pt} \\ {} & {} & r \left (-3 q +r -2\right ) \left (-1-3 p +r \right )=0 \\ \bullet & {} & \textrm {Values of r that satisfy the indicial equation}\hspace {3pt} \\ {} & {} & r \in \left \{0, 1+3 p , 3 q +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 ) \left (-1-3 q +r \right ) \left (-3 p +r \right )=0, a_{2} \left (2+r \right ) \left (-3 q +r \right ) \left (1-3 p +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 -1-3 q +r \right ) \left (k -3 p +r \right )-a_{k -2}=0 \\ \bullet & {} & \textrm {Shift index using}\hspace {3pt} k \mathrm {->}k +2 \\ {} & {} & a_{k +3} \left (k +3+r \right ) \left (k +1-3 q +r \right ) \left (k +2-3 p +r \right )-a_{k}=0 \\ \bullet & {} & \textrm {Recursion relation that defines series solution to ODE}\hspace {3pt} \\ {} & {} & a_{k +3}=\frac {a_{k}}{\left (k +3+r \right ) \left (k +1-3 q +r \right ) \left (k +2-3 p +r \right )} \\ \bullet & {} & \textrm {Recursion relation for}\hspace {3pt} r =0 \\ {} & {} & a_{k +3}=\frac {a_{k}}{\left (k +3\right ) \left (k +1-3 q \right ) \left (k +2-3 p \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 +3\right ) \left (k +1-3 q \right ) \left (k +2-3 p \right )}, a_{1}=0, a_{2}=0\right ] \\ \bullet & {} & \textrm {Recursion relation for}\hspace {3pt} r =1+3 p \\ {} & {} & a_{k +3}=\frac {a_{k}}{\left (k +4+3 p \right ) \left (k +2-3 q +3 p \right ) \left (k +3\right )} \\ \bullet & {} & \textrm {Solution for}\hspace {3pt} r =1+3 p \\ {} & {} & \left [y=\moverset {\infty }{\munderset {k =0}{\sum }}a_{k} x^{k +1+3 p}, a_{k +3}=\frac {a_{k}}{\left (k +4+3 p \right ) \left (k +2-3 q +3 p \right ) \left (k +3\right )}, a_{1}=0, a_{2}=0\right ] \\ \bullet & {} & \textrm {Recursion relation for}\hspace {3pt} r =3 q +2 \\ {} & {} & a_{k +3}=\frac {a_{k}}{\left (k +5+3 q \right ) \left (k +3\right ) \left (k +4-3 p +3 q \right )} \\ \bullet & {} & \textrm {Solution for}\hspace {3pt} r =3 q +2 \\ {} & {} & \left [y=\moverset {\infty }{\munderset {k =0}{\sum }}a_{k} x^{k +3 q +2}, a_{k +3}=\frac {a_{k}}{\left (k +5+3 q \right ) \left (k +3\right ) \left (k +4-3 p +3 q \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}\right )+\left (\moverset {\infty }{\munderset {k =0}{\sum }}b_{k} x^{k +1+3 p}\right )+\left (\moverset {\infty }{\munderset {k =0}{\sum }}c_{k} x^{k +3 q +2}\right ), a_{k +3}=\frac {a_{k}}{\left (k +3\right ) \left (k +1-3 q \right ) \left (k +2-3 p \right )}, a_{1}=0, a_{2}=0, b_{k +3}=\frac {b_{k}}{\left (k +4+3 p \right ) \left (k +2-3 q +3 p \right ) \left (k +3\right )}, b_{1}=0, b_{2}=0, c_{k +3}=\frac {c_{k}}{\left (k +5+3 q \right ) \left (k +3\right ) \left (k +4-3 p +3 q \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 -> pFq: Equivalence to the 3F2 or one of its 3 confluent cases under a power @ Moebius <- pFq successful: received ODE is equivalent to the 0F2 ODE, case c = 0 `
✓ Solution by Maple
Time used: 0.016 (sec). Leaf size: 77
dsolve(x^2*diff(diff(diff(y(x),x),x),x)-3*(p+q)*x*diff(diff(y(x),x),x)+3*p*(3*q+1)*diff(y(x),x)-x^2*y(x)=0,y(x), singsol=all)
\[ y \left (x \right ) = c_{1} \operatorname {hypergeom}\left (\left [\right ], \left [-p +\frac {2}{3}, -q +\frac {1}{3}\right ], \frac {x^{3}}{27}\right )+c_{2} x^{2+3 q} \operatorname {hypergeom}\left (\left [\right ], \left [q +\frac {5}{3}, \frac {4}{3}+q -p \right ], \frac {x^{3}}{27}\right )+c_{3} x^{3 p +1} \operatorname {hypergeom}\left (\left [\right ], \left [p +\frac {4}{3}, \frac {2}{3}-q +p \right ], \frac {x^{3}}{27}\right ) \]
✓ Solution by Mathematica
Time used: 0.257 (sec). Leaf size: 127
DSolve[-(x^2*y[x]) + 3*p*(1 + 3*q)*y'[x] - 3*(p + q)*x*y''[x] + x^2*Derivative[3][y][x] == 0,y[x],x,IncludeSingularSolutions -> True]
\[ y(x)\to c_1 \, _0F_2\left (;\frac {2}{3}-p,\frac {1}{3}-q;\frac {x^3}{27}\right )+c_2 (-1)^{p+\frac {1}{3}} 3^{-3 p-1} x^{3 p+1} \, _0F_2\left (;p+\frac {4}{3},p-q+\frac {2}{3};\frac {x^3}{27}\right )+c_3 (-1)^{q+\frac {2}{3}} 3^{-3 q-2} x^{3 q+2} \, _0F_2\left (;q+\frac {5}{3},-p+q+\frac {4}{3};\frac {x^3}{27}\right ) \]