Internal problem ID [9832]
Internal file name [OUTPUT/8777_Monday_June_06_2022_05_26_44_AM_34482524/index.tex]
Book: Differential Gleichungen, E. Kamke, 3rd ed. Chelsea Pub. NY, 1948
Section: Chapter 3, linear third order
Problem number: 1508.
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.
\[ \boxed {x^{3} y^{\prime \prime \prime }+\left (-\nu ^{2}+1\right ) x y^{\prime }+\left (a \,x^{3}+\nu ^{2}-1\right ) y=0} \] Unable to solve this ODE.
\[ \begin {array}{lll} & {} & \textrm {Let's solve}\hspace {3pt} \\ {} & {} & x^{3} \left (\frac {d}{d x}y^{\prime \prime }\right )+\left (-\nu ^{2}+1\right ) x y^{\prime }+\left (a \,x^{3}+\nu ^{2}-1\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 )=0, P_{3}\left (x \right )=-\frac {\nu ^{2}-1}{x^{2}}, P_{4}\left (x \right )=\frac {a \,x^{3}+\nu ^{2}-1}{x^{3}}\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}}}=0 \\ {} & \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}}}=-\nu ^{2}+1 \\ {} & \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}}}=\nu ^{2}-1 \\ {} & \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} \\ {} & {} & x^{3} \left (\frac {d}{d x}y^{\prime \prime }\right )-\left (\nu ^{2}-1\right ) x y^{\prime }+\left (a \,x^{3}+\nu ^{2}-1\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..3 \\ {} & {} & 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 \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} x^{3}\cdot \left (\frac {d}{d x}y^{\prime \prime }\right )\hspace {3pt}\textrm {to series expansion}\hspace {3pt} \\ {} & {} & x^{3}\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} \\ & {} & \textrm {Rewrite ODE with series expansions}\hspace {3pt} \\ {} & {} & a_{0} \left (r -1\right ) \left (r -1+\nu \right ) \left (r -1-\nu \right ) x^{r}+a_{1} r \left (r +\nu \right ) \left (r -\nu \right ) x^{1+r}+a_{2} \left (1+r \right ) \left (r +1+\nu \right ) \left (r +1-\nu \right ) x^{2+r}+\left (\moverset {\infty }{\munderset {k =3}{\sum }}\left (a_{k} \left (k +r -1\right ) \left (r -1+\nu +k \right ) \left (r -1-\nu +k \right )+a_{k -3} a \right ) x^{k +r}\right )=0 \\ \bullet & {} & a_{0}\textrm {cannot be 0 by assumption, giving the indicial equation}\hspace {3pt} \\ {} & {} & \left (r -1\right ) \left (r -1+\nu \right ) \left (r -1-\nu \right )=0 \\ \bullet & {} & \textrm {Values of r that satisfy the indicial equation}\hspace {3pt} \\ {} & {} & r \in \left \{1, 1-\nu , 1+\nu \right \} \\ \bullet & {} & \textrm {The coefficients of each power of}\hspace {3pt} x \hspace {3pt}\textrm {must be 0}\hspace {3pt} \\ {} & {} & \left [a_{1} r \left (r +\nu \right ) \left (r -\nu \right )=0, a_{2} \left (1+r \right ) \left (r +1+\nu \right ) \left (r +1-\nu \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} \left (k +r -1\right ) \left (r -1+\nu +k \right ) \left (r -1-\nu +k \right )+a_{k -3} a =0 \\ \bullet & {} & \textrm {Shift index using}\hspace {3pt} k \mathrm {->}k +3 \\ {} & {} & a_{k +3} \left (k +2+r \right ) \left (r +2+\nu +k \right ) \left (r +2-\nu +k \right )+a_{k} a =0 \\ \bullet & {} & \textrm {Recursion relation that defines series solution to ODE}\hspace {3pt} \\ {} & {} & a_{k +3}=-\frac {a_{k} a}{\left (k +2+r \right ) \left (r +2+\nu +k \right ) \left (r +2-\nu +k \right )} \\ \bullet & {} & \textrm {Recursion relation for}\hspace {3pt} r =1 \\ {} & {} & a_{k +3}=-\frac {a_{k} a}{\left (k +3\right ) \left (3+\nu +k \right ) \left (3-\nu +k \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 {a_{k} a}{\left (k +3\right ) \left (3+\nu +k \right ) \left (3-\nu +k \right )}, a_{1}=0, a_{2}=0\right ] \\ \bullet & {} & \textrm {Recursion relation for}\hspace {3pt} r =1-\nu \\ {} & {} & a_{k +3}=-\frac {a_{k} a}{\left (3-\nu +k \right ) \left (k +3\right ) \left (3-2 \nu +k \right )} \\ \bullet & {} & \textrm {Solution for}\hspace {3pt} r =1-\nu \\ {} & {} & \left [y=\moverset {\infty }{\munderset {k =0}{\sum }}a_{k} x^{k +1-\nu }, a_{k +3}=-\frac {a_{k} a}{\left (3-\nu +k \right ) \left (k +3\right ) \left (3-2 \nu +k \right )}, a_{1}=0, a_{2}=0\right ] \\ \bullet & {} & \textrm {Recursion relation for}\hspace {3pt} r =1+\nu \\ {} & {} & a_{k +3}=-\frac {a_{k} a}{\left (3+\nu +k \right ) \left (3+2 \nu +k \right ) \left (k +3\right )} \\ \bullet & {} & \textrm {Solution for}\hspace {3pt} r =1+\nu \\ {} & {} & \left [y=\moverset {\infty }{\munderset {k =0}{\sum }}a_{k} x^{k +1+\nu }, a_{k +3}=-\frac {a_{k} a}{\left (3+\nu +k \right ) \left (3+2 \nu +k \right ) \left (k +3\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 }}b_{k} x^{k +1}\right )+\left (\moverset {\infty }{\munderset {k =0}{\sum }}c_{k} x^{k +1-\nu }\right )+\left (\moverset {\infty }{\munderset {k =0}{\sum }}d_{k} x^{k +1+\nu }\right ), b_{k +3}=-\frac {b_{k} a}{\left (k +3\right ) \left (k +3+\nu \right ) \left (k +3-\nu \right )}, b_{1}=0, b_{2}=0, c_{k +3}=-\frac {c_{k} a}{\left (k +3-\nu \right ) \left (k +3\right ) \left (3-2 \nu +k \right )}, c_{1}=0, c_{2}=0, d_{k +3}=-\frac {d_{k} a}{\left (k +3+\nu \right ) \left (3+2 \nu +k \right ) \left (k +3\right )}, d_{1}=0, d_{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: 0.015 (sec). Leaf size: 78
dsolve(x^3*diff(diff(diff(y(x),x),x),x)+(-nu^2+1)*x*diff(y(x),x)+(a*x^3+nu^2-1)*y(x)=0,y(x), singsol=all)
\[ y \left (x \right ) = x \left (c_{1} \operatorname {hypergeom}\left (\left [\right ], \left [1-\frac {\nu }{3}, 1+\frac {\nu }{3}\right ], -\frac {x^{3} a}{27}\right )+c_{2} x^{-\nu } \operatorname {hypergeom}\left (\left [\right ], \left [1-\frac {2 \nu }{3}, 1-\frac {\nu }{3}\right ], -\frac {x^{3} a}{27}\right )+c_{3} x^{\nu } \operatorname {hypergeom}\left (\left [\right ], \left [1+\frac {2 \nu }{3}, 1+\frac {\nu }{3}\right ], -\frac {x^{3} a}{27}\right )\right ) \]
✓ Solution by Mathematica
Time used: 0.429 (sec). Leaf size: 143
DSolve[(-1 + nu^2 + a*x^3)*y[x] + (1 - nu^2)*x*y'[x] + x^3*Derivative[3][y][x] == 0,y[x],x,IncludeSingularSolutions -> True]
\[ y(x)\to 3^{-\nu -1} x a^{-\nu /3} \left (a^{\frac {\nu +1}{3}} \left (c_3 a^{\nu /3} x^{\nu } \, _0F_2\left (;\frac {\nu }{3}+1,\frac {2 \nu }{3}+1;-\frac {a x^3}{27}\right )+c_1 3^{\nu } \, _0F_2\left (;1-\frac {\nu }{3},\frac {\nu }{3}+1;-\frac {a x^3}{27}\right )\right )+\sqrt [3]{a} c_2 9^{\nu } x^{-\nu } \, _0F_2\left (;1-\frac {2 \nu }{3},1-\frac {\nu }{3};-\frac {a x^3}{27}\right )\right ) \]