Internal problem ID [9884]
Internal file name [OUTPUT/8831_Monday_June_06_2022_05_35_16_AM_51245361/index.tex]
Book: Differential Gleichungen, E. Kamke, 3rd ed. Chelsea Pub. NY, 1948
Section: Chapter 4, linear fourth order
Problem number: 1562.
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.
\[ \boxed {x^{4} y^{\prime \prime \prime \prime }+4 x^{3} y^{\prime \prime \prime }-\left (4 n^{2}-1\right ) x^{2} y^{\prime \prime }+\left (4 n^{2}-1\right ) x y^{\prime }-4 y x^{4}=0} \] Unable to solve this ODE.
\[ \begin {array}{lll} & {} & \textrm {Let's solve}\hspace {3pt} \\ {} & {} & x^{3} \left (\frac {d}{d x}\frac {d}{d x}y^{\prime \prime }\right )-4 \left (\frac {d}{d x}y^{\prime }\right ) n^{2} x -4 y x^{3}+4 x^{2} \left (\frac {d}{d x}y^{\prime \prime }\right )+4 y^{\prime } n^{2}+x \left (\frac {d}{d x}y^{\prime }\right )-y^{\prime }=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 {4}{x}, P_{3}\left (x \right )=-\frac {4 n^{2}-1}{x^{2}}, P_{4}\left (x \right )=\frac {4 n^{2}-1}{x^{3}}, P_{5}\left (x \right )=-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}}}=4 \\ {} & \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}}}=-4 n^{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}}}=4 n^{2}-1 \\ {} & \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} \\ {} & {} & x^{3} \left (\frac {d}{d x}\frac {d}{d x}y^{\prime \prime }\right )+4 x^{2} \left (\frac {d}{d x}y^{\prime \prime }\right )-x \left (4 n^{2}-1\right ) \left (\frac {d}{d x}y^{\prime }\right )+\left (4 n^{2}-1\right ) y^{\prime }-4 y x^{3}=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^{3}\cdot y\hspace {3pt}\textrm {to series expansion}\hspace {3pt} \\ {} & {} & x^{3}\cdot y=\moverset {\infty }{\munderset {k =0}{\sum }}a_{k} x^{k +r +3} \\ {} & \circ & \textrm {Shift index using}\hspace {3pt} k \mathrm {->}k -3 \\ {} & {} & x^{3}\cdot y=\moverset {\infty }{\munderset {k =3}{\sum }}a_{k -3} 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} \\ {} & \circ & \textrm {Convert}\hspace {3pt} x^{3}\cdot \left (\frac {d}{d x}\frac {d}{d x}y^{\prime \prime }\right )\hspace {3pt}\textrm {to series expansion}\hspace {3pt} \\ {} & {} & x^{3}\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 -1} \\ {} & \circ & \textrm {Shift index using}\hspace {3pt} k \mathrm {->}k +1 \\ {} & {} & x^{3}\cdot \left (\frac {d}{d x}\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 ) \left (k +r -2\right ) x^{k +r} \\ & {} & \textrm {Rewrite ODE with series expansions}\hspace {3pt} \\ {} & {} & a_{0} r \left (-2+r \right ) \left (2 n +r \right ) \left (-2 n +r \right ) x^{-1+r}+a_{1} \left (1+r \right ) \left (-1+r \right ) \left (1+2 n +r \right ) \left (1-2 n +r \right ) x^{r}+a_{2} \left (2+r \right ) r \left (2+2 n +r \right ) \left (2-2 n +r \right ) x^{1+r}+a_{3} \left (3+r \right ) \left (1+r \right ) \left (3+2 n +r \right ) \left (3-2 n +r \right ) x^{2+r}+\left (\moverset {\infty }{\munderset {k =3}{\sum }}\left (a_{k +1} \left (k +1+r \right ) \left (k +r -1\right ) \left (k +1+2 n +r \right ) \left (k +1-2 n +r \right )-4 a_{k -3}\right ) x^{k +r}\right )=0 \\ \bullet & {} & a_{0}\textrm {cannot be 0 by assumption, giving the indicial equation}\hspace {3pt} \\ {} & {} & r \left (-2+r \right ) \left (2 n +r \right ) \left (-2 n +r \right )=0 \\ \bullet & {} & \textrm {Values of r that satisfy the indicial equation}\hspace {3pt} \\ {} & {} & r \in \left \{0, 2, -2 n , 2 n \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+r \right ) \left (1+2 n +r \right ) \left (1-2 n +r \right )=0, a_{2} \left (2+r \right ) r \left (2+2 n +r \right ) \left (2-2 n +r \right )=0, a_{3} \left (3+r \right ) \left (1+r \right ) \left (3+2 n +r \right ) \left (3-2 n +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 +1} \left (k +1+r \right ) \left (k +r -1\right ) \left (k +1+2 n +r \right ) \left (k +1-2 n +r \right )-4 a_{k -3}=0 \\ \bullet & {} & \textrm {Shift index using}\hspace {3pt} k \mathrm {->}k +3 \\ {} & {} & a_{k +4} \left (k +4+r \right ) \left (k +2+r \right ) \left (k +4+2 n +r \right ) \left (k +4-2 n +r \right )-4 a_{k}=0 \\ \bullet & {} & \textrm {Recursion relation that defines series solution to ODE}\hspace {3pt} \\ {} & {} & a_{k +4}=\frac {4 a_{k}}{\left (k +4+r \right ) \left (k +2+r \right ) \left (k +4+2 n +r \right ) \left (k +4-2 n +r \right )} \\ \bullet & {} & \textrm {Recursion relation for}\hspace {3pt} r =0 \\ {} & {} & a_{k +4}=\frac {4 a_{k}}{\left (k +4\right ) \left (k +2\right ) \left (k +4+2 n \right ) \left (k +4-2 n \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 {4 a_{k}}{\left (k +4\right ) \left (k +2\right ) \left (k +4+2 n \right ) \left (k +4-2 n \right )}, a_{1}=0, a_{2}=0, a_{3}=0\right ] \\ \bullet & {} & \textrm {Recursion relation for}\hspace {3pt} r =2 \\ {} & {} & a_{k +4}=\frac {4 a_{k}}{\left (k +6\right ) \left (k +4\right ) \left (k +6+2 n \right ) \left (k +6-2 n \right )} \\ \bullet & {} & \textrm {Solution for}\hspace {3pt} r =2 \\ {} & {} & \left [y=\moverset {\infty }{\munderset {k =0}{\sum }}a_{k} x^{k +2}, a_{k +4}=\frac {4 a_{k}}{\left (k +6\right ) \left (k +4\right ) \left (k +6+2 n \right ) \left (k +6-2 n \right )}, a_{1}=0, a_{2}=0, a_{3}=0\right ] \\ \bullet & {} & \textrm {Recursion relation for}\hspace {3pt} r =-2 n \\ {} & {} & a_{k +4}=\frac {4 a_{k}}{\left (k +4-2 n \right ) \left (k +2-2 n \right ) \left (k +4\right ) \left (k +4-4 n \right )} \\ \bullet & {} & \textrm {Solution for}\hspace {3pt} r =-2 n \\ {} & {} & \left [y=\moverset {\infty }{\munderset {k =0}{\sum }}a_{k} x^{k -2 n}, a_{k +4}=\frac {4 a_{k}}{\left (k +4-2 n \right ) \left (k +2-2 n \right ) \left (k +4\right ) \left (k +4-4 n \right )}, a_{1}=0, a_{2}=0, a_{3}=0\right ] \\ \bullet & {} & \textrm {Recursion relation for}\hspace {3pt} r =2 n \\ {} & {} & a_{k +4}=\frac {4 a_{k}}{\left (k +4+2 n \right ) \left (k +2+2 n \right ) \left (k +4+4 n \right ) \left (k +4\right )} \\ \bullet & {} & \textrm {Solution for}\hspace {3pt} r =2 n \\ {} & {} & \left [y=\moverset {\infty }{\munderset {k =0}{\sum }}a_{k} x^{k +2 n}, a_{k +4}=\frac {4 a_{k}}{\left (k +4+2 n \right ) \left (k +2+2 n \right ) \left (k +4+4 n \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 }}a_{k} x^{k}\right )+\left (\moverset {\infty }{\munderset {k =0}{\sum }}b_{k} x^{k +2}\right )+\left (\moverset {\infty }{\munderset {k =0}{\sum }}c_{k} x^{k -2 n}\right )+\left (\moverset {\infty }{\munderset {k =0}{\sum }}d_{k} x^{k +2 n}\right ), a_{k +4}=\frac {4 a_{k}}{\left (k +4\right ) \left (k +2\right ) \left (k +4+2 n \right ) \left (k +4-2 n \right )}, a_{1}=0, a_{2}=0, a_{3}=0, b_{k +4}=\frac {4 b_{k}}{\left (6+k \right ) \left (k +4\right ) \left (k +6+2 n \right ) \left (k +6-2 n \right )}, b_{1}=0, b_{2}=0, b_{3}=0, c_{k +4}=\frac {4 c_{k}}{\left (k +4-2 n \right ) \left (k +2-2 n \right ) \left (k +4\right ) \left (k +4-4 n \right )}, c_{1}=0, c_{2}=0, c_{3}=0, d_{k +4}=\frac {4 d_{k}}{\left (k +4+2 n \right ) \left (k +2+2 n \right ) \left (k +4+4 n \right ) \left (k +4\right )}, d_{1}=0, d_{2}=0, d_{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 Multiplying solutions by`, exp(Int(1/x, x))`Equation is the symmetric product of`, diff(diff(y(x), x), x)-(1/4)*((4*I)*x^2+4*n^2-1)* 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`
✓ Solution by Maple
Time used: 0.031 (sec). Leaf size: 77
dsolve(x^4*diff(diff(diff(diff(y(x),x),x),x),x)+4*x^3*diff(diff(diff(y(x),x),x),x)-(4*n^2-1)*x^2*diff(diff(y(x),x),x)+(4*n^2-1)*x*diff(y(x),x)-4*y(x)*x^4=0,y(x), singsol=all)
\[ y \left (x \right ) = \left (\operatorname {BesselY}\left (n , \left (\frac {1}{2}-\frac {i}{2}\right ) \sqrt {2}\, x \right ) c_{3} +\operatorname {BesselJ}\left (n , \left (\frac {1}{2}-\frac {i}{2}\right ) \sqrt {2}\, x \right ) c_{1} \right ) \operatorname {BesselJ}\left (n , \left (\frac {1}{2}+\frac {i}{2}\right ) \sqrt {2}\, x \right )+\operatorname {BesselY}\left (n , \left (\frac {1}{2}+\frac {i}{2}\right ) \sqrt {2}\, x \right ) \left (\operatorname {BesselY}\left (n , \left (\frac {1}{2}-\frac {i}{2}\right ) \sqrt {2}\, x \right ) c_{4} +c_{2} \operatorname {BesselJ}\left (n , \left (\frac {1}{2}-\frac {i}{2}\right ) \sqrt {2}\, x \right )\right ) \]
✓ Solution by Mathematica
Time used: 0.916 (sec). Leaf size: 140
DSolve[-4*x^4*y[x] + (-1 + 4*n^2)*x*y'[x] - (-1 + 4*n^2)*x^2*y''[x] + 4*x^3*Derivative[3][y][x] + x^4*Derivative[4][y][x] == 0,y[x],x,IncludeSingularSolutions -> True]
\[ y(x)\to c_1 \, _0F_3\left (;\frac {1}{2},1-\frac {n}{2},\frac {n}{2}+1;\frac {x^4}{64}\right )+\frac {1}{8} i c_2 x^2 \, _0F_3\left (;\frac {3}{2},\frac {3}{2}-\frac {n}{2},\frac {n}{2}+\frac {3}{2};\frac {x^4}{64}\right )+c_3 \left (\frac {i}{2}\right )^{-n} \operatorname {Gamma}(1-n)^2 \left (\text {ber}_{-n}(x){}^2+\text {bei}_{-n}(x){}^2\right )+c_4 \left (\frac {i}{2}\right )^n \operatorname {Gamma}(n+1)^2 \left (\text {ber}_n(x){}^2+\text {bei}_n(x){}^2\right ) \]