Internal problem ID [9872]
Internal file name [OUTPUT/8819_Monday_June_06_2022_05_33_23_AM_96471917/index.tex]
Book: Differential Gleichungen, E. Kamke, 3rd ed. Chelsea Pub. NY, 1948
Section: Chapter 4, linear fourth order
Problem number: 1550.
ODE order: 4.
ODE degree: 1.
The type(s) of ODE detected by this program : "unknown"
Maple gives the following as the ode type
[[_high_order, _with_linear_symmetries]]
Unable to solve or complete the solution.
\[ \boxed {x y^{\prime \prime \prime \prime }-\left (6 x^{2}+1\right ) y^{\prime \prime \prime }+12 x^{3} y^{\prime \prime }-\left (9 x^{2}-7\right ) x^{2} y^{\prime }+2 \left (x^{2}-3\right ) x^{3} y=0} \] Unable to solve this ODE.
\[ \begin {array}{lll} & {} & \textrm {Let's solve}\hspace {3pt} \\ {} & {} & x \left (\frac {d}{d x}\frac {d}{d x}y^{\prime \prime }\right )-\left (6 x^{2}+1\right ) \left (\frac {d}{d x}y^{\prime \prime }\right )+12 x^{3} \left (\frac {d}{d x}y^{\prime }\right )-\left (9 x^{2}-7\right ) x^{2} y^{\prime }+2 \left (x^{2}-3\right ) x^{3} y=0 \\ \bullet & {} & \textrm {Highest derivative means the order of the ODE is}\hspace {3pt} 4 \\ {} & {} & \frac {d}{d x}\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 {6 x^{2}+1}{x}, P_{3}\left (x \right )=12 x^{2}, P_{4}\left (x \right )=-\left (9 x^{2}-7\right ) x , P_{5}\left (x \right )=2 \left (x^{2}-3\right ) x^{2}\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}}}=-1 \\ {} & \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^{4}\cdot P_{5}\left (x \right )\textrm {is analytic at}\hspace {3pt} x =0 \\ {} & {} & \left (x^{4}\cdot P_{5}\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 {Multiply by denominators}\hspace {3pt} \\ {} & {} & 2 \left (x^{2}-3\right ) x^{3} y-\left (9 x^{2}-7\right ) x^{2} y^{\prime }+12 x^{3} \left (\frac {d}{d x}y^{\prime }\right )+\left (-6 x^{2}-1\right ) \left (\frac {d}{d x}y^{\prime \prime }\right )+x \left (\frac {d}{d x}\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\hspace {3pt}\textrm {to series expansion for}\hspace {3pt} m =3..5 \\ {} & {} & 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^{m}\cdot y^{\prime }\hspace {3pt}\textrm {to series expansion for}\hspace {3pt} m =2..4 \\ {} & {} & 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^{3}\cdot \left (\frac {d}{d x}y^{\prime }\right )\hspace {3pt}\textrm {to series expansion}\hspace {3pt} \\ {} & {} & x^{3}\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^{3}\cdot \left (\frac {d}{d x}y^{\prime }\right )=\moverset {\infty }{\munderset {k =1}{\sum }}a_{k -1} \left (k +r -1\right ) \left (k -2+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 =0..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 -2+r \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} \\ {} & \circ & \textrm {Convert}\hspace {3pt} x \cdot \left (\frac {d}{d x}\frac {d}{d x}y^{\prime \prime }\right )\hspace {3pt}\textrm {to series expansion}\hspace {3pt} \\ {} & {} & x \cdot \left (\frac {d}{d x}\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 -2+r \right ) \left (k +r -3\right ) x^{k +r -3} \\ {} & \circ & \textrm {Shift index using}\hspace {3pt} k \mathrm {->}k +3 \\ {} & {} & x \cdot \left (\frac {d}{d x}\frac {d}{d x}y^{\prime \prime }\right )=\moverset {\infty }{\munderset {k =-3}{\sum }}a_{k +3} \left (k +3+r \right ) \left (k +2+r \right ) \left (k +r +1\right ) \left (k +r \right ) x^{k +r} \\ & {} & \textrm {Rewrite ODE with series expansions}\hspace {3pt} \\ {} & {} & a_{0} r \left (-1+r \right ) \left (-2+r \right ) \left (-4+r \right ) x^{-3+r}+a_{1} \left (1+r \right ) r \left (-1+r \right ) \left (-3+r \right ) x^{-2+r}+\left (a_{2} \left (2+r \right ) \left (1+r \right ) r \left (-2+r \right )-6 a_{0} r \left (-1+r \right ) \left (-2+r \right )\right ) x^{-1+r}+\left (a_{3} \left (3+r \right ) \left (2+r \right ) \left (1+r \right ) \left (-1+r \right )-6 a_{1} \left (1+r \right ) r \left (-1+r \right )\right ) x^{r}+\left (a_{4} \left (4+r \right ) \left (3+r \right ) \left (2+r \right ) r -6 a_{2} \left (2+r \right ) \left (1+r \right ) r +a_{0} r \left (-5+12 r \right )\right ) x^{1+r}+\left (a_{5} \left (5+r \right ) \left (4+r \right ) \left (3+r \right ) \left (1+r \right )-6 a_{3} \left (3+r \right ) \left (2+r \right ) \left (1+r \right )+a_{1} \left (1+r \right ) \left (7+12 r \right )\right ) x^{2+r}+\left (a_{6} \left (6+r \right ) \left (5+r \right ) \left (4+r \right ) \left (2+r \right )-6 a_{4} \left (4+r \right ) \left (3+r \right ) \left (2+r \right )+a_{2} \left (2+r \right ) \left (19+12 r \right )-3 a_{0} \left (2+3 r \right )\right ) x^{3+r}+\left (a_{7} \left (7+r \right ) \left (6+r \right ) \left (5+r \right ) \left (3+r \right )-6 a_{5} \left (5+r \right ) \left (4+r \right ) \left (3+r \right )+a_{3} \left (3+r \right ) \left (31+12 r \right )-3 a_{1} \left (5+3 r \right )\right ) x^{4+r}+\left (\moverset {\infty }{\munderset {k =5}{\sum }}\left (a_{k +3} \left (k +3+r \right ) \left (k +2+r \right ) \left (k +r +1\right ) \left (k +r -1\right )-6 a_{k +1} \left (k +r +1\right ) \left (k +r \right ) \left (k +r -1\right )+a_{k -1} \left (k +r -1\right ) \left (12 k -17+12 r \right )-3 a_{k -3} \left (3 k -7+3 r \right )+2 a_{k -5}\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 (-2+r \right ) \left (-4+r \right )=0 \\ \bullet & {} & \textrm {Values of r that satisfy the indicial equation}\hspace {3pt} \\ {} & {} & r \in \left \{0, 1, 2, 4\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 ) r \left (-1+r \right ) \left (-3+r \right )=0, a_{2} \left (2+r \right ) \left (1+r \right ) r \left (-2+r \right )-6 a_{0} r \left (-1+r \right ) \left (-2+r \right )=0, a_{3} \left (3+r \right ) \left (2+r \right ) \left (1+r \right ) \left (-1+r \right )-6 a_{1} \left (1+r \right ) r \left (-1+r \right )=0, a_{4} \left (4+r \right ) \left (3+r \right ) \left (2+r \right ) r -6 a_{2} \left (2+r \right ) \left (1+r \right ) r +a_{0} r \left (-5+12 r \right )=0, a_{5} \left (5+r \right ) \left (4+r \right ) \left (3+r \right ) \left (1+r \right )-6 a_{3} \left (3+r \right ) \left (2+r \right ) \left (1+r \right )+a_{1} \left (1+r \right ) \left (7+12 r \right )=0, a_{6} \left (6+r \right ) \left (5+r \right ) \left (4+r \right ) \left (2+r \right )-6 a_{4} \left (4+r \right ) \left (3+r \right ) \left (2+r \right )+a_{2} \left (2+r \right ) \left (19+12 r \right )-3 a_{0} \left (2+3 r \right )=0, a_{7} \left (7+r \right ) \left (6+r \right ) \left (5+r \right ) \left (3+r \right )-6 a_{5} \left (5+r \right ) \left (4+r \right ) \left (3+r \right )+a_{3} \left (3+r \right ) \left (31+12 r \right )-3 a_{1} \left (5+3 r \right )=0\right ] \\ \bullet & {} & \textrm {Solve for the dependent coefficient(s)}\hspace {3pt} \\ {} & {} & \left \{a_{1}=0, a_{2}=\frac {6 a_{0} \left (-1+r \right )}{r^{2}+3 r +2}, a_{3}=0, a_{4}=\frac {a_{0} \left (24 r -31\right )}{r^{3}+9 r^{2}+26 r +24}, a_{5}=0, a_{6}=\frac {3 a_{0} \left (27 r^{2}-23 r -22\right )}{r^{5}+18 r^{4}+121 r^{3}+372 r^{2}+508 r +240}, a_{7}=0\right \} \\ \bullet & {} & \textrm {Each term in the series must be 0, giving the recursion relation}\hspace {3pt} \\ {} & {} & a_{k +3} \left (k +3+r \right ) \left (k +2+r \right ) \left (k +r +1\right ) \left (k +r -1\right )-6 a_{k +1} \left (k +r +1\right ) \left (k +r \right ) \left (k +r -1\right )+a_{k -1} \left (k +r -1\right ) \left (12 k -17+12 r \right )-3 a_{k -3} \left (3 k -7+3 r \right )+2 a_{k -5}=0 \\ \bullet & {} & \textrm {Shift index using}\hspace {3pt} k \mathrm {->}k +5 \\ {} & {} & a_{k +8} \left (k +8+r \right ) \left (k +7+r \right ) \left (k +6+r \right ) \left (k +4+r \right )-6 a_{k +6} \left (k +6+r \right ) \left (k +5+r \right ) \left (k +4+r \right )+a_{k +4} \left (k +4+r \right ) \left (12 k +43+12 r \right )-3 a_{k +2} \left (3 k +8+3 r \right )+2 a_{k}=0 \\ \bullet & {} & \textrm {Recursion relation that defines series solution to ODE}\hspace {3pt} \\ {} & {} & a_{k +8}=\frac {6 k^{3} a_{k +6}+18 k^{2} r a_{k +6}+18 k \,r^{2} a_{k +6}+6 r^{3} a_{k +6}-12 k^{2} a_{k +4}+90 k^{2} a_{k +6}-24 k r a_{k +4}+180 k r a_{k +6}-12 r^{2} a_{k +4}+90 r^{2} a_{k +6}+9 k a_{k +2}-91 k a_{k +4}+444 k a_{k +6}+9 r a_{k +2}-91 r a_{k +4}+444 r a_{k +6}-2 a_{k}+24 a_{k +2}-172 a_{k +4}+720 a_{k +6}}{\left (k +8+r \right ) \left (k +7+r \right ) \left (k +6+r \right ) \left (k +4+r \right )} \\ \bullet & {} & \textrm {Recursion relation for}\hspace {3pt} r =0 \\ {} & {} & a_{k +8}=\frac {6 k^{3} a_{k +6}-12 k^{2} a_{k +4}+90 k^{2} a_{k +6}+9 k a_{k +2}-91 k a_{k +4}+444 k a_{k +6}-2 a_{k}+24 a_{k +2}-172 a_{k +4}+720 a_{k +6}}{\left (k +8\right ) \left (k +7\right ) \left (k +6\right ) \left (k +4\right )} \\ \bullet & {} & \textrm {Solution for}\hspace {3pt} r =0 \\ {} & {} & \left [y=\moverset {\infty }{\munderset {k =0}{\sum }}a_{k} x^{k}, a_{k +8}=\frac {6 k^{3} a_{k +6}-12 k^{2} a_{k +4}+90 k^{2} a_{k +6}+9 k a_{k +2}-91 k a_{k +4}+444 k a_{k +6}-2 a_{k}+24 a_{k +2}-172 a_{k +4}+720 a_{k +6}}{\left (k +8\right ) \left (k +7\right ) \left (k +6\right ) \left (k +4\right )}, a_{1}=0, a_{2}=-3 a_{0}, a_{3}=0, a_{4}=-\frac {31 a_{0}}{24}, a_{5}=0, a_{6}=-\frac {11 a_{0}}{40}, a_{7}=0\right ] \\ \bullet & {} & \textrm {Recursion relation for}\hspace {3pt} r =1 \\ {} & {} & a_{k +8}=\frac {6 k^{3} a_{k +6}-12 k^{2} a_{k +4}+108 k^{2} a_{k +6}+9 k a_{k +2}-115 k a_{k +4}+642 k a_{k +6}-2 a_{k}+33 a_{k +2}-275 a_{k +4}+1260 a_{k +6}}{\left (k +9\right ) \left (k +8\right ) \left (k +7\right ) \left (k +5\right )} \\ \bullet & {} & \textrm {Solution for}\hspace {3pt} r =1 \\ {} & {} & \left [y=\moverset {\infty }{\munderset {k =0}{\sum }}a_{k} x^{k +1}, a_{k +8}=\frac {6 k^{3} a_{k +6}-12 k^{2} a_{k +4}+108 k^{2} a_{k +6}+9 k a_{k +2}-115 k a_{k +4}+642 k a_{k +6}-2 a_{k}+33 a_{k +2}-275 a_{k +4}+1260 a_{k +6}}{\left (k +9\right ) \left (k +8\right ) \left (k +7\right ) \left (k +5\right )}, a_{1}=0, a_{2}=0, a_{3}=0, a_{4}=-\frac {7 a_{0}}{60}, a_{5}=0, a_{6}=-\frac {3 a_{0}}{70}, a_{7}=0\right ] \\ \bullet & {} & \textrm {Recursion relation for}\hspace {3pt} r =2 \\ {} & {} & a_{k +8}=\frac {6 k^{3} a_{k +6}-12 k^{2} a_{k +4}+126 k^{2} a_{k +6}+9 k a_{k +2}-139 k a_{k +4}+876 k a_{k +6}-2 a_{k}+42 a_{k +2}-402 a_{k +4}+2016 a_{k +6}}{\left (k +10\right ) \left (k +9\right ) \left (k +8\right ) \left (k +6\right )} \\ \bullet & {} & \textrm {Solution for}\hspace {3pt} r =2 \\ {} & {} & \left [y=\moverset {\infty }{\munderset {k =0}{\sum }}a_{k} x^{k +2}, a_{k +8}=\frac {6 k^{3} a_{k +6}-12 k^{2} a_{k +4}+126 k^{2} a_{k +6}+9 k a_{k +2}-139 k a_{k +4}+876 k a_{k +6}-2 a_{k}+42 a_{k +2}-402 a_{k +4}+2016 a_{k +6}}{\left (k +10\right ) \left (k +9\right ) \left (k +8\right ) \left (k +6\right )}, a_{1}=0, a_{2}=\frac {a_{0}}{2}, a_{3}=0, a_{4}=\frac {17 a_{0}}{120}, a_{5}=0, a_{6}=\frac {5 a_{0}}{168}, a_{7}=0\right ] \\ \bullet & {} & \textrm {Recursion relation for}\hspace {3pt} r =4 \\ {} & {} & a_{k +8}=\frac {6 k^{3} a_{k +6}-12 k^{2} a_{k +4}+162 k^{2} a_{k +6}+9 k a_{k +2}-187 k a_{k +4}+1452 k a_{k +6}-2 a_{k}+60 a_{k +2}-728 a_{k +4}+4320 a_{k +6}}{\left (k +12\right ) \left (k +11\right ) \left (k +10\right ) \left (k +8\right )} \\ \bullet & {} & \textrm {Solution for}\hspace {3pt} r =4 \\ {} & {} & \left [y=\moverset {\infty }{\munderset {k =0}{\sum }}a_{k} x^{k +4}, a_{k +8}=\frac {6 k^{3} a_{k +6}-12 k^{2} a_{k +4}+162 k^{2} a_{k +6}+9 k a_{k +2}-187 k a_{k +4}+1452 k a_{k +6}-2 a_{k}+60 a_{k +2}-728 a_{k +4}+4320 a_{k +6}}{\left (k +12\right ) \left (k +11\right ) \left (k +10\right ) \left (k +8\right )}, a_{1}=0, a_{2}=\frac {3 a_{0}}{5}, a_{3}=0, a_{4}=\frac {65 a_{0}}{336}, a_{5}=0, a_{6}=\frac {53 a_{0}}{1200}, a_{7}=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}\right )+\left (\moverset {\infty }{\munderset {k =0}{\sum }}c_{k} x^{k +2}\right )+\left (\moverset {\infty }{\munderset {k =0}{\sum }}d_{k} x^{k +4}\right ), a_{k +8}=\frac {6 k^{3} a_{k +6}-12 k^{2} a_{k +4}+90 k^{2} a_{k +6}+9 k a_{k +2}-91 k a_{k +4}+444 k a_{k +6}-2 a_{k}+24 a_{k +2}-172 a_{k +4}+720 a_{k +6}}{\left (k +8\right ) \left (k +7\right ) \left (k +6\right ) \left (k +4\right )}, a_{1}=0, a_{2}=-3 a_{0}, a_{3}=0, a_{4}=-\frac {31 a_{0}}{24}, a_{5}=0, a_{6}=-\frac {11 a_{0}}{40}, a_{7}=0, b_{k +8}=\frac {6 k^{3} b_{k +6}-12 k^{2} b_{k +4}+108 k^{2} b_{k +6}+9 k b_{k +2}-115 k b_{k +4}+642 k b_{k +6}-2 b_{k}+33 b_{k +2}-275 b_{k +4}+1260 b_{k +6}}{\left (k +9\right ) \left (k +8\right ) \left (k +7\right ) \left (k +5\right )}, b_{1}=0, b_{2}=0, b_{3}=0, b_{4}=-\frac {7 b_{0}}{60}, b_{5}=0, b_{6}=-\frac {3 b_{0}}{70}, b_{7}=0, c_{k +8}=\frac {6 k^{3} c_{k +6}-12 k^{2} c_{k +4}+126 k^{2} c_{k +6}+9 k c_{k +2}-139 k c_{k +4}+876 k c_{k +6}-2 c_{k}+42 c_{k +2}-402 c_{k +4}+2016 c_{k +6}}{\left (k +10\right ) \left (k +9\right ) \left (k +8\right ) \left (k +6\right )}, c_{1}=0, c_{2}=\frac {c_{0}}{2}, c_{3}=0, c_{4}=\frac {17 c_{0}}{120}, c_{5}=0, c_{6}=\frac {5 c_{0}}{168}, c_{7}=0, d_{k +8}=\frac {6 k^{3} d_{k +6}-12 k^{2} d_{k +4}+162 k^{2} d_{k +6}+9 k d_{k +2}-187 k d_{k +4}+1452 k d_{k +6}-2 d_{k}+60 d_{k +2}-728 d_{k +4}+4320 d_{k +6}}{\left (k +12\right ) \left (k +11\right ) \left (k +10\right ) \left (k +8\right )}, d_{1}=0, d_{2}=\frac {3 d_{0}}{5}, d_{3}=0, d_{4}=\frac {65 d_{0}}{336}, d_{5}=0, d_{6}=\frac {53 d_{0}}{1200}, d_{7}=0\right ] \end {array} \]
Maple trace
`Methods for high 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: 4; missing the dependent variable trying a solution in terms of MeijerG functions trying a solution in terms of MeijerG functions checking if the LODE is of Euler type expon. solutions partially successful. Result(s) =`, [exp(x^2), exp((1/2)*x^2)]` Calling dsolve with: (x^4+3*x^2-2)/x*y(x)-3*x^ 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 -> Whittaker -> hyper3: Equivalence to 1F1 under a power @ Moebius <- hyper3 successful: received ODE is equivalent to the 1F1 ODE <- Whittaker successful <- special function solution successful <- differential factorization successful`
✓ Solution by Maple
Time used: 0.344 (sec). Leaf size: 157
dsolve(x*diff(diff(diff(diff(y(x),x),x),x),x)-(6*x^2+1)*diff(diff(diff(y(x),x),x),x)+12*x^3*diff(diff(y(x),x),x)-(9*x^2-7)*x^2*diff(y(x),x)+2*(x^2-3)*x^3*y(x)=0,y(x), singsol=all)
\[ y \left (x \right ) = -{\mathrm e}^{x^{2}} \left (\int \frac {\operatorname {WhittakerM}\left (\frac {9 \sqrt {5}}{20}, \frac {3}{4}, \frac {\sqrt {5}\, x^{2}}{2}\right ) {\mathrm e}^{-\frac {x^{2}}{4}}}{x^{\frac {3}{2}}}d x \right ) c_{3} -{\mathrm e}^{x^{2}} \left (\int \frac {\operatorname {WhittakerW}\left (\frac {9 \sqrt {5}}{20}, \frac {3}{4}, \frac {\sqrt {5}\, x^{2}}{2}\right ) {\mathrm e}^{-\frac {x^{2}}{4}}}{x^{\frac {3}{2}}}d x \right ) c_{4} +\left (\int \frac {\operatorname {WhittakerM}\left (\frac {9 \sqrt {5}}{20}, \frac {3}{4}, \frac {\sqrt {5}\, x^{2}}{2}\right ) {\mathrm e}^{\frac {x^{2}}{4}}}{x^{\frac {3}{2}}}d x \right ) {\mathrm e}^{\frac {x^{2}}{2}} c_{3} +{\mathrm e}^{\frac {x^{2}}{2}} \left (\int \frac {\operatorname {WhittakerW}\left (\frac {9 \sqrt {5}}{20}, \frac {3}{4}, \frac {\sqrt {5}\, x^{2}}{2}\right ) {\mathrm e}^{\frac {x^{2}}{4}}}{x^{\frac {3}{2}}}d x \right ) c_{4} +c_{1} {\mathrm e}^{x^{2}}+c_{2} {\mathrm e}^{\frac {x^{2}}{2}} \]
✓ Solution by Mathematica
Time used: 1.968 (sec). Leaf size: 216
DSolve[2*x^3*(-3 + x^2)*y[x] - x^2*(-7 + 9*x^2)*y'[x] + 12*x^3*y''[x] - (1 + 6*x^2)*Derivative[3][y][x] + x*Derivative[4][y][x] == 0,y[x],x,IncludeSingularSolutions -> True]
\[ y(x)\to e^{\frac {x^2}{2}} \left (c_3 \int _1^x\frac {e^{\frac {K[1]^2}{2}} \left (\int \frac {e^{\frac {1}{4} \left (-1+\sqrt {5}\right ) K[1]^2} \operatorname {HypergeometricU}\left (-\frac {1}{4}+\frac {9}{4 \sqrt {5}},-\frac {1}{2},-\frac {1}{2} \sqrt {5} K[1]^2\right ) \left (K[1]^2\right )^{3/4}}{K[1]^{7/2}} \, dK[1]\right ) K[1]}{\sqrt [4]{2}}dK[1]+c_4 \int _1^x\frac {e^{\frac {K[2]^2}{2}} \left (\int \frac {e^{\frac {1}{4} \left (-1+\sqrt {5}\right ) K[2]^2} \left (K[2]^2\right )^{3/4} L_{\frac {1}{4}-\frac {9}{4 \sqrt {5}}}^{-\frac {3}{2}}\left (-\frac {1}{2} \sqrt {5} K[2]^2\right )}{K[2]^{7/2}} \, dK[2]\right ) K[2]}{\sqrt [4]{2}}dK[2]+c_2 e^{\frac {x^2}{2}}+c_1\right ) \]