Internal problem ID [9804]
Internal file name [OUTPUT/8747_Monday_June_06_2022_05_23_13_AM_24448173/index.tex]
Book: Differential Gleichungen, E. Kamke, 3rd ed. Chelsea Pub. NY, 1948
Section: Chapter 3, linear third order
Problem number: 1478.
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.
\[ \boxed {x y^{\prime \prime \prime }+3 y^{\prime \prime }-a \,x^{2} y=0} \] Unable to solve this ODE.
\[ \begin {array}{lll} & {} & \textrm {Let's solve}\hspace {3pt} \\ {} & {} & x \left (\frac {d}{d x}y^{\prime \prime }\right )+3 \frac {d}{d x}y^{\prime }-a \,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}{x}, P_{3}\left (x \right )=0, P_{4}\left (x \right )=-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}}}=3 \\ {} & \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^{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} \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 +r +2\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 +r +2\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 (-1+r \right ) x^{-2+r}+a_{1} \left (1+r \right ) \left (2+r \right ) r \,x^{-1+r}+a_{2} \left (2+r \right ) \left (3+r \right ) \left (1+r \right ) x^{r}+a_{3} \left (3+r \right ) \left (4+r \right ) \left (2+r \right ) x^{1+r}+\left (\moverset {\infty }{\munderset {k =2}{\sum }}\left (a_{k +2} \left (k +r +2\right ) \left (k +3+r \right ) \left (k +1+r \right )-a 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 (1+r \right ) \left (-1+r \right )=0 \\ \bullet & {} & \textrm {Values of r that satisfy the indicial equation}\hspace {3pt} \\ {} & {} & r \in \left \{-1, 0, 1\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 (2+r \right ) r =0, a_{2} \left (2+r \right ) \left (3+r \right ) \left (1+r \right )=0, a_{3} \left (3+r \right ) \left (4+r \right ) \left (2+r \right )=0\right ] \\ \bullet & {} & \textrm {Solve for the dependent coefficient(s)}\hspace {3pt} \\ {} & {} & \left \{a_{1}=0, a_{2}=0, a_{3}=0\right \} \\ \bullet & {} & \textrm {Each term in the series must be 0, giving the recursion relation}\hspace {3pt} \\ {} & {} & a_{k +2} \left (k +r +2\right ) \left (k +3+r \right ) \left (k +1+r \right )-a a_{k -2}=0 \\ \bullet & {} & \textrm {Shift index using}\hspace {3pt} k \mathrm {->}k +2 \\ {} & {} & a_{k +4} \left (k +4+r \right ) \left (k +5+r \right ) \left (k +3+r \right )-a a_{k}=0 \\ \bullet & {} & \textrm {Recursion relation that defines series solution to ODE}\hspace {3pt} \\ {} & {} & a_{k +4}=\frac {a a_{k}}{\left (k +4+r \right ) \left (k +5+r \right ) \left (k +3+r \right )} \\ \bullet & {} & \textrm {Recursion relation for}\hspace {3pt} r =-1 \\ {} & {} & a_{k +4}=\frac {a a_{k}}{\left (k +3\right ) \left (k +4\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 +4}=\frac {a a_{k}}{\left (k +3\right ) \left (k +4\right ) \left (k +2\right )}, a_{1}=0, a_{2}=0, a_{3}=0\right ] \\ \bullet & {} & \textrm {Recursion relation for}\hspace {3pt} r =0 \\ {} & {} & a_{k +4}=\frac {a a_{k}}{\left (k +4\right ) \left (k +5\right ) \left (k +3\right )} \\ \bullet & {} & \textrm {Solution for}\hspace {3pt} r =0 \\ {} & {} & \left [y=\moverset {\infty }{\munderset {k =0}{\sum }}a_{k} x^{k}, a_{k +4}=\frac {a a_{k}}{\left (k +4\right ) \left (k +5\right ) \left (k +3\right )}, a_{1}=0, a_{2}=0, a_{3}=0\right ] \\ \bullet & {} & \textrm {Recursion relation for}\hspace {3pt} r =1 \\ {} & {} & a_{k +4}=\frac {a a_{k}}{\left (k +5\right ) \left (k +6\right ) \left (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 +4}=\frac {a a_{k}}{\left (k +5\right ) \left (k +6\right ) \left (k +4\right )}, a_{1}=0, a_{2}=0, a_{3}=0\right ] \\ \bullet & {} & \textrm {Combine solutions and rename parameters}\hspace {3pt} \\ {} & {} & \left [y=\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 )+\left (\moverset {\infty }{\munderset {k =0}{\sum }}d_{k} x^{k +1}\right ), b_{k +4}=\frac {a b_{k}}{\left (k +3\right ) \left (k +4\right ) \left (k +2\right )}, b_{1}=0, b_{2}=0, b_{3}=0, c_{k +4}=\frac {a c_{k}}{\left (k +4\right ) \left (k +5\right ) \left (k +3\right )}, c_{1}=0, c_{2}=0, c_{3}=0, d_{k +4}=\frac {a d_{k}}{\left (k +5\right ) \left (k +6\right ) \left (k +4\right )}, d_{1}=0, d_{2}=0, d_{3}=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: 0.015 (sec). Leaf size: 48
dsolve(x*diff(diff(diff(y(x),x),x),x)+3*diff(diff(y(x),x),x)-a*x^2*y(x)=0,y(x), singsol=all)
\[ y \left (x \right ) = c_{1} \operatorname {hypergeom}\left (\left [\right ], \left [\frac {3}{4}, \frac {5}{4}\right ], \frac {x^{4} a}{64}\right )+\frac {c_{2} \operatorname {hypergeom}\left (\left [\right ], \left [\frac {1}{2}, \frac {3}{4}\right ], \frac {x^{4} a}{64}\right )}{x}+c_{3} x \operatorname {hypergeom}\left (\left [\right ], \left [\frac {5}{4}, \frac {3}{2}\right ], \frac {x^{4} a}{64}\right ) \]
✓ Solution by Mathematica
Time used: 0.017 (sec). Leaf size: 90
DSolve[-(a*x^2*y[x]) + 3*y''[x] + x*Derivative[3][y][x] == 0,y[x],x,IncludeSingularSolutions -> True]
\[ y(x)\to \frac {(2-2 i) c_1 \, _0F_2\left (;\frac {1}{2},\frac {3}{4};\frac {a x^4}{64}\right )}{\sqrt [4]{a} x}+c_2 \, _0F_2\left (;\frac {3}{4},\frac {5}{4};\frac {a x^4}{64}\right )+\left (\frac {1}{4}+\frac {i}{4}\right ) \sqrt [4]{a} c_3 x \, _0F_2\left (;\frac {5}{4},\frac {3}{2};\frac {a x^4}{64}\right ) \]