Internal problem ID [9883]
Internal file name [OUTPUT/8830_Monday_June_06_2022_05_35_05_AM_18793291/index.tex]
Book: Differential Gleichungen, E. Kamke, 3rd ed. Chelsea Pub. NY, 1948
Section: Chapter 4, linear fourth order
Problem number: 1561.
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.
\[ \boxed {x^{4} y^{\prime \prime \prime \prime }-2 n \left (n +1\right ) x^{2} y^{\prime \prime }+4 n \left (n +1\right ) x y^{\prime }+\left (a \,x^{4}+n \left (n +1\right ) \left (n +3\right ) \left (n -2\right )\right ) y=0} \] Unable to solve this ODE.
\[ \begin {array}{lll} & {} & \textrm {Let's solve}\hspace {3pt} \\ {} & {} & x^{4} \left (\frac {d}{d x}\frac {d}{d x}y^{\prime \prime }\right )-2 n \left (n +1\right ) x^{2} \left (\frac {d}{d x}y^{\prime }\right )+4 n \left (n +1\right ) x y^{\prime }+\left (a \,x^{4}+n \left (n +1\right ) \left (n +3\right ) \left (n -2\right )\right ) 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 )=0, P_{3}\left (x \right )=-\frac {2 n \left (n +1\right )}{x^{2}}, P_{4}\left (x \right )=\frac {4 n \left (n +1\right )}{x^{3}}, P_{5}\left (x \right )=\frac {a \,x^{4}+n^{4}+2 n^{3}-5 n^{2}-6 n}{x^{4}}\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}}}=-2 n \left (n +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}}}=4 n \left (n +1\right ) \\ {} & \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}}}=n^{4}+2 n^{3}-5 n^{2}-6 n \\ {} & \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^{4} \left (\frac {d}{d x}\frac {d}{d x}y^{\prime \prime }\right )-2 n \left (n +1\right ) x^{2} \left (\frac {d}{d x}y^{\prime }\right )+4 n \left (n +1\right ) x y^{\prime }+\left (a \,x^{4}+n^{4}+2 n^{3}-5 n^{2}-6 n \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..4 \\ {} & {} & 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^{2}\cdot \left (\frac {d}{d x}y^{\prime }\right )\hspace {3pt}\textrm {to series expansion}\hspace {3pt} \\ {} & {} & x^{2}\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} \\ {} & \circ & \textrm {Convert}\hspace {3pt} x^{4}\cdot \left (\frac {d}{d x}\frac {d}{d x}y^{\prime \prime }\right )\hspace {3pt}\textrm {to series expansion}\hspace {3pt} \\ {} & {} & x^{4}\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 +r -2\right ) \left (k +r -3\right ) x^{k +r} \\ & {} & \textrm {Rewrite ODE with series expansions}\hspace {3pt} \\ {} & {} & a_{0} \left (r -1-n \right ) \left (r -2+n \right ) \left (r -3-n \right ) \left (r +n \right ) x^{r}+a_{1} \left (r -n \right ) \left (r -1+n \right ) \left (r -2-n \right ) \left (r +n +1\right ) x^{1+r}+a_{2} \left (r +1-n \right ) \left (r +n \right ) \left (r -1-n \right ) \left (r +n +2\right ) x^{2+r}+a_{3} \left (r +2-n \right ) \left (r +n +1\right ) \left (r -n \right ) \left (r +n +3\right ) x^{3+r}+\left (\moverset {\infty }{\munderset {k =4}{\sum }}\left (a_{k} \left (r -1-n +k \right ) \left (r -2+n +k \right ) \left (r -3-n +k \right ) \left (r +n +k \right )+a_{k -4} 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-n \right ) \left (r -2+n \right ) \left (r -3-n \right ) \left (r +n \right )=0 \\ \bullet & {} & \textrm {Values of r that satisfy the indicial equation}\hspace {3pt} \\ {} & {} & r \in \left \{-n , -n +2, n +1, n +3\right \} \\ \bullet & {} & \textrm {The coefficients of each power of}\hspace {3pt} x \hspace {3pt}\textrm {must be 0}\hspace {3pt} \\ {} & {} & \left [a_{1} \left (r -n \right ) \left (r -1+n \right ) \left (r -2-n \right ) \left (r +n +1\right )=0, a_{2} \left (r +1-n \right ) \left (r +n \right ) \left (r -1-n \right ) \left (r +n +2\right )=0, a_{3} \left (r +2-n \right ) \left (r +n +1\right ) \left (r -n \right ) \left (r +n +3\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} \left (r -1-n +k \right ) \left (r -2+n +k \right ) \left (r -3-n +k \right ) \left (r +n +k \right )+a_{k -4} a =0 \\ \bullet & {} & \textrm {Shift index using}\hspace {3pt} k \mathrm {->}k +4 \\ {} & {} & a_{k +4} \left (r +3-n +k \right ) \left (r +2+n +k \right ) \left (r +1-n +k \right ) \left (r +n +k +4\right )+a_{k} a =0 \\ \bullet & {} & \textrm {Recursion relation that defines series solution to ODE}\hspace {3pt} \\ {} & {} & a_{k +4}=-\frac {a_{k} a}{\left (r +3-n +k \right ) \left (r +2+n +k \right ) \left (r +1-n +k \right ) \left (r +n +k +4\right )} \\ \bullet & {} & \textrm {Recursion relation for}\hspace {3pt} r =-n \\ {} & {} & a_{k +4}=-\frac {a_{k} a}{\left (-2 n +3+k \right ) \left (2+k \right ) \left (-2 n +1+k \right ) \left (k +4\right )} \\ \bullet & {} & \textrm {Solution for}\hspace {3pt} r =-n \\ {} & {} & \left [y=\moverset {\infty }{\munderset {k =0}{\sum }}a_{k} x^{k -n}, a_{k +4}=-\frac {a_{k} a}{\left (-2 n +3+k \right ) \left (2+k \right ) \left (-2 n +1+k \right ) \left (k +4\right )}, a_{1}=0, a_{2}=0, a_{3}=0\right ] \\ \bullet & {} & \textrm {Recursion relation for}\hspace {3pt} r =-n +2 \\ {} & {} & a_{k +4}=-\frac {a_{k} a}{\left (-2 n +5+k \right ) \left (k +4\right ) \left (-2 n +3+k \right ) \left (6+k \right )} \\ \bullet & {} & \textrm {Solution for}\hspace {3pt} r =-n +2 \\ {} & {} & \left [y=\moverset {\infty }{\munderset {k =0}{\sum }}a_{k} x^{k -n +2}, a_{k +4}=-\frac {a_{k} a}{\left (-2 n +5+k \right ) \left (k +4\right ) \left (-2 n +3+k \right ) \left (6+k \right )}, a_{1}=0, a_{2}=0, a_{3}=0\right ] \\ \bullet & {} & \textrm {Recursion relation for}\hspace {3pt} r =n +1 \\ {} & {} & a_{k +4}=-\frac {a_{k} a}{\left (k +4\right ) \left (2 n +3+k \right ) \left (2+k \right ) \left (2 n +5+k \right )} \\ \bullet & {} & \textrm {Solution for}\hspace {3pt} r =n +1 \\ {} & {} & \left [y=\moverset {\infty }{\munderset {k =0}{\sum }}a_{k} x^{k +n +1}, a_{k +4}=-\frac {a_{k} a}{\left (k +4\right ) \left (2 n +3+k \right ) \left (2+k \right ) \left (2 n +5+k \right )}, a_{1}=0, a_{2}=0, a_{3}=0\right ] \\ \bullet & {} & \textrm {Recursion relation for}\hspace {3pt} r =n +3 \\ {} & {} & a_{k +4}=-\frac {a_{k} a}{\left (6+k \right ) \left (2 n +5+k \right ) \left (k +4\right ) \left (2 n +7+k \right )} \\ \bullet & {} & \textrm {Solution for}\hspace {3pt} r =n +3 \\ {} & {} & \left [y=\moverset {\infty }{\munderset {k =0}{\sum }}a_{k} x^{k +n +3}, a_{k +4}=-\frac {a_{k} a}{\left (6+k \right ) \left (2 n +5+k \right ) \left (k +4\right ) \left (2 n +7+k \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 -n}\right )+\left (\moverset {\infty }{\munderset {k =0}{\sum }}c_{k} x^{k -n +2}\right )+\left (\moverset {\infty }{\munderset {k =0}{\sum }}d_{k} x^{k +n +1}\right )+\left (\moverset {\infty }{\munderset {k =0}{\sum }}e_{k} x^{k +n +3}\right ), b_{k +4}=-\frac {b_{k} a}{\left (-2 n +3+k \right ) \left (k +2\right ) \left (-2 n +1+k \right ) \left (k +4\right )}, b_{1}=0, b_{2}=0, b_{3}=0, c_{k +4}=-\frac {c_{k} a}{\left (-2 n +5+k \right ) \left (k +4\right ) \left (-2 n +3+k \right ) \left (6+k \right )}, c_{1}=0, c_{2}=0, c_{3}=0, d_{k +4}=-\frac {d_{k} a}{\left (k +4\right ) \left (2 n +3+k \right ) \left (k +2\right ) \left (2 n +5+k \right )}, d_{1}=0, d_{2}=0, d_{3}=0, e_{k +4}=-\frac {e_{k} a}{\left (6+k \right ) \left (2 n +5+k \right ) \left (k +4\right ) \left (2 n +7+k \right )}, e_{1}=0, e_{2}=0, e_{3}=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 Equation is the LCLM of ((-a)^(1/2)*x^2-n^2-n)/x^2*y(x)+diff(diff(y(x),x),x), -((-a)^(1/2)*x^2+n^2+n)/x^2*y(x)+diff(diff(y(x),x),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 <- Bessel successful <- special function solution successful 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 <- Bessel successful <- special function solution successful <- solving the LCLM ode successful `
✓ Solution by Maple
Time used: 0.031 (sec). Leaf size: 69
dsolve(x^4*diff(diff(diff(diff(y(x),x),x),x),x)-2*n*(n+1)*x^2*diff(diff(y(x),x),x)+4*n*(n+1)*x*diff(y(x),x)+(a*x^4+n*(n+1)*(n+3)*(n-2))*y(x)=0,y(x), singsol=all)
\[ y \left (x \right ) = \sqrt {x}\, \left (\operatorname {BesselY}\left (n +\frac {1}{2}, \sqrt {-\sqrt {-a}}\, x \right ) c_{4} +\operatorname {BesselJ}\left (n +\frac {1}{2}, \sqrt {-\sqrt {-a}}\, x \right ) c_{3} +\operatorname {BesselJ}\left (n +\frac {1}{2}, \left (-a \right )^{\frac {1}{4}} x \right ) c_{1} +\operatorname {BesselY}\left (n +\frac {1}{2}, \left (-a \right )^{\frac {1}{4}} x \right ) c_{2} \right ) \]
✓ Solution by Mathematica
Time used: 3.44 (sec). Leaf size: 310
DSolve[((-2 + n)*n*(1 + n)*(3 + n) + a*x^4)*y[x] + 4*n*(1 + n)*x*y'[x] - 2*n*(1 + n)*x^2*y''[x] + x^4*Derivative[4][y][x] == 0,y[x],x,IncludeSingularSolutions -> True]
\[ y(x)\to \sqrt [8]{a} 2^{-n-\frac {7}{2}} \sqrt {x} \left (2^{2 n+1} \text {ber}_{-n-\frac {1}{2}}\left (\sqrt [4]{a} x\right ) \left (4 c_2 \cos \left (\frac {3}{8} \pi (2 n+1)\right ) \operatorname {Gamma}\left (\frac {1}{2}-n\right )-c_1 \cos \left (\frac {3}{8} \pi (2 n-3)\right ) \operatorname {Gamma}\left (\frac {3}{2}-n\right )\right )+\text {ber}_{n+\frac {1}{2}}\left (\sqrt [4]{a} x\right ) \left (4 c_3 \cos \left (\frac {3}{8} \pi (2 n+1)\right ) \operatorname {Gamma}\left (n+\frac {3}{2}\right )-c_4 \cos \left (\frac {3}{8} \pi (2 n+5)\right ) \operatorname {Gamma}\left (n+\frac {5}{2}\right )\right )+c_1 2^{2 n+1} \sin \left (\frac {3}{8} \pi (2 n-3)\right ) \operatorname {Gamma}\left (\frac {3}{2}-n\right ) \text {bei}_{-n-\frac {1}{2}}\left (\sqrt [4]{a} x\right )-c_2 2^{2 n+3} \sin \left (\frac {3}{8} \pi (2 n+1)\right ) \operatorname {Gamma}\left (\frac {1}{2}-n\right ) \text {bei}_{-n-\frac {1}{2}}\left (\sqrt [4]{a} x\right )+4 c_3 \sin \left (\frac {3}{8} \pi (2 n+1)\right ) \operatorname {Gamma}\left (n+\frac {3}{2}\right ) \text {bei}_{n+\frac {1}{2}}\left (\sqrt [4]{a} x\right )-c_4 \sin \left (\frac {3}{8} \pi (2 n+5)\right ) \operatorname {Gamma}\left (n+\frac {5}{2}\right ) \text {bei}_{n+\frac {1}{2}}\left (\sqrt [4]{a} x\right )\right ) \]