5.28 problem 1563

Internal problem ID [9885]
Internal file name [OUTPUT/8832_Monday_June_06_2022_05_35_23_AM_21682553/index.tex]

Book: Differential Gleichungen, E. Kamke, 3rd ed. Chelsea Pub. NY, 1948
Section: Chapter 4, linear fourth order
Problem number: 1563.
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.

\[ \boxed {x^{4} y^{\prime \prime \prime \prime }+4 x^{3} y^{\prime \prime \prime }-\left (4 n^{2}-1\right ) x^{2} y^{\prime \prime }-\left (4 n^{2}-1\right ) x y^{\prime }+\left (-4 x^{4}+4 n^{2}-1\right ) y=0} \] Unable to solve this ODE.

5.28.1 Maple step by step solution

\[ \begin {array}{lll} & {} & \textrm {Let's solve}\hspace {3pt} \\ {} & {} & x^{4} \left (\frac {d}{d x}\frac {d}{d x}y^{\prime \prime }\right )+4 x^{3} \left (\frac {d}{d x}y^{\prime \prime }\right )-\left (4 n^{2}-1\right ) x^{2} \left (\frac {d}{d x}y^{\prime }\right )-\left (4 n^{2}-1\right ) x y^{\prime }+\left (-4 x^{4}+4 n^{2}-1\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 )=\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 )=\frac {-4 x^{4}+4 n^{2}-1}{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}}}=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}}}=4 n^{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 {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^{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} \\ {} & \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 (1+r \right ) \left (-1+r \right ) \left (r -1+2 n \right ) \left (r -1-2 n \right ) x^{r}+a_{1} \left (2+r \right ) r \left (r +2 n \right ) \left (r -2 n \right ) x^{1+r}+a_{2} \left (3+r \right ) \left (1+r \right ) \left (r +1+2 n \right ) \left (r +1-2 n \right ) x^{2+r}+a_{3} \left (4+r \right ) \left (2+r \right ) \left (r +2+2 n \right ) \left (r +2-2 n \right ) x^{3+r}+\left (\moverset {\infty }{\munderset {k =4}{\sum }}\left (a_{k} \left (k +r +1\right ) \left (k +r -1\right ) \left (r -1+2 n +k \right ) \left (r -1-2 n +k \right )-4 a_{k -4}\right ) x^{k +r}\right )=0 \\ \bullet & {} & a_{0}\textrm {cannot be 0 by assumption, giving the indicial equation}\hspace {3pt} \\ {} & {} & \left (1+r \right ) \left (-1+r \right ) \left (r -1+2 n \right ) \left (r -1-2 n \right )=0 \\ \bullet & {} & \textrm {Values of r that satisfy the indicial equation}\hspace {3pt} \\ {} & {} & r \in \left \{-1, 1, -2 n +1, 2 n +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 (2+r \right ) r \left (r +2 n \right ) \left (r -2 n \right )=0, a_{2} \left (3+r \right ) \left (1+r \right ) \left (r +1+2 n \right ) \left (r +1-2 n \right )=0, a_{3} \left (4+r \right ) \left (2+r \right ) \left (r +2+2 n \right ) \left (r +2-2 n \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 (k +r +1\right ) \left (k +r -1\right ) \left (r -1+2 n +k \right ) \left (r -1-2 n +k \right )-4 a_{k -4}=0 \\ \bullet & {} & \textrm {Shift index using}\hspace {3pt} k \mathrm {->}k +4 \\ {} & {} & a_{k +4} \left (k +5+r \right ) \left (k +3+r \right ) \left (r +3+2 n +k \right ) \left (r +3-2 n +k \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 +5+r \right ) \left (k +3+r \right ) \left (r +3+2 n +k \right ) \left (r +3-2 n +k \right )} \\ \bullet & {} & \textrm {Recursion relation for}\hspace {3pt} r =-1 \\ {} & {} & a_{k +4}=\frac {4 a_{k}}{\left (k +4\right ) \left (k +2\right ) \left (2+2 n +k \right ) \left (2-2 n +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 +4}=\frac {4 a_{k}}{\left (k +4\right ) \left (k +2\right ) \left (2+2 n +k \right ) \left (2-2 n +k \right )}, a_{1}=0, a_{2}=0, a_{3}=0\right ] \\ \bullet & {} & \textrm {Recursion relation for}\hspace {3pt} r =1 \\ {} & {} & a_{k +4}=\frac {4 a_{k}}{\left (k +6\right ) \left (k +4\right ) \left (4+2 n +k \right ) \left (4-2 n +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 +4}=\frac {4 a_{k}}{\left (k +6\right ) \left (k +4\right ) \left (4+2 n +k \right ) \left (4-2 n +k \right )}, a_{1}=0, a_{2}=0, a_{3}=0\right ] \\ \bullet & {} & \textrm {Recursion relation for}\hspace {3pt} r =-2 n +1 \\ {} & {} & a_{k +4}=\frac {4 a_{k}}{\left (k +6-2 n \right ) \left (4-2 n +k \right ) \left (k +4\right ) \left (-4 n +4+k \right )} \\ \bullet & {} & \textrm {Solution for}\hspace {3pt} r =-2 n +1 \\ {} & {} & \left [y=\moverset {\infty }{\munderset {k =0}{\sum }}a_{k} x^{k -2 n +1}, a_{k +4}=\frac {4 a_{k}}{\left (k +6-2 n \right ) \left (4-2 n +k \right ) \left (k +4\right ) \left (-4 n +4+k \right )}, a_{1}=0, a_{2}=0, a_{3}=0\right ] \\ \bullet & {} & \textrm {Recursion relation for}\hspace {3pt} r =2 n +1 \\ {} & {} & a_{k +4}=\frac {4 a_{k}}{\left (k +6+2 n \right ) \left (4+2 n +k \right ) \left (4 n +4+k \right ) \left (k +4\right )} \\ \bullet & {} & \textrm {Solution for}\hspace {3pt} r =2 n +1 \\ {} & {} & \left [y=\moverset {\infty }{\munderset {k =0}{\sum }}a_{k} x^{k +2 n +1}, a_{k +4}=\frac {4 a_{k}}{\left (k +6+2 n \right ) \left (4+2 n +k \right ) \left (4 n +4+k \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 -1}\right )+\left (\moverset {\infty }{\munderset {k =0}{\sum }}b_{k} x^{k +1}\right )+\left (\moverset {\infty }{\munderset {k =0}{\sum }}c_{k} x^{-2 n +1+k}\right )+\left (\moverset {\infty }{\munderset {k =0}{\sum }}d_{k} x^{k +2 n +1}\right ), a_{k +4}=\frac {4 a_{k}}{\left (k +4\right ) \left (k +2\right ) \left (k +2+2 n \right ) \left (k +2-2 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 +4+2 n \right ) \left (k +4-2 n \right )}, b_{1}=0, b_{2}=0, b_{3}=0, c_{k +4}=\frac {4 c_{k}}{\left (k +6-2 n \right ) \left (k +4-2 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 +6+2 n \right ) \left (k +4+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 
trying a solution in terms of MeijerG functions 
<- MeijerG function solution successful`
 

✓ Solution by Maple

Time used: 0.047 (sec). Leaf size: 87

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*x^4+4*n^2-1)*y(x)=0,y(x), singsol=all)
 

\[ y \left (x \right ) = \frac {c_{4} \operatorname {hypergeom}\left (\left [\right ], \left [\frac {1}{2}, \frac {n}{2}+\frac {1}{2}, -\frac {n}{2}+\frac {1}{2}\right ], \frac {x^{4}}{64}\right )+\left (c_{3} \operatorname {hypergeom}\left (\left [\right ], \left [\frac {3}{2}, -\frac {n}{2}+1, \frac {n}{2}+1\right ], \frac {x^{4}}{64}\right )+c_{2} \operatorname {KelvinBei}\left (-n , x\right )^{2}+\operatorname {KelvinBer}\left (-n , x\right )^{2} c_{2} +c_{1} \left (\operatorname {KelvinBer}\left (n , x\right )^{2}+\operatorname {KelvinBei}\left (n , x\right )^{2}\right )\right ) x^{2}}{x} \]

✓ Solution by Mathematica

Time used: 1.725 (sec). Leaf size: 187

DSolve[(-1 + 4*n^2 - 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]
 

\[ y(x)\to \frac {\sqrt [4]{-1} \left (x^2 \left (c_2 \, _0F_3\left (;\frac {3}{2},1-\frac {n}{2},\frac {n}{2}+1;\frac {x^4}{64}\right )+c_3 \left (\frac {i}{8}\right )^{-n} x^{-2 n} \, _0F_3\left (;1-n,1-\frac {n}{2},\frac {3}{2}-\frac {n}{2};\frac {x^4}{64}\right )+c_4 \left (\frac {i}{8}\right )^n x^{2 n} \, _0F_3\left (;\frac {n}{2}+1,\frac {n}{2}+\frac {3}{2},n+1;\frac {x^4}{64}\right )\right )-8 i c_1 \, _0F_3\left (;\frac {1}{2},\frac {1}{2}-\frac {n}{2},\frac {n}{2}+\frac {1}{2};\frac {x^4}{64}\right )\right )}{2 \sqrt {2} x} \]