Internal problem ID [9880]
Internal file name [OUTPUT/8827_Monday_June_06_2022_05_34_45_AM_37208121/index.tex]
Book: Differential Gleichungen, E. Kamke, 3rd ed. Chelsea Pub. NY, 1948
Section: Chapter 4, linear fourth order
Problem number: 1558.
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.
\[ \boxed {x^{2} y^{\prime \prime \prime \prime }+\left (2 n -2 \nu +4\right ) x y^{\prime \prime \prime }+\left (n -\nu +1\right ) \left (n -\nu +2\right ) y^{\prime \prime }-\frac {b^{4} y}{16}=0} \] Unable to solve this ODE.
\[ \begin {array}{lll} & {} & \textrm {Let's solve}\hspace {3pt} \\ {} & {} & x^{2} \left (\frac {d}{d x}\frac {d}{d x}y^{\prime \prime }\right )+\left (2 n -2 \nu +4\right ) x \left (\frac {d}{d x}y^{\prime \prime }\right )+\left (n -\nu +1\right ) \left (n -\nu +2\right ) \left (\frac {d}{d x}y^{\prime }\right )-\frac {b^{4} y}{16}=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 {2 \left (n -\nu +2\right )}{x}, P_{3}\left (x \right )=\frac {\left (n -\nu +1\right ) \left (n -\nu +2\right )}{x^{2}}, P_{4}\left (x \right )=0, P_{5}\left (x \right )=-\frac {b^{4}}{16 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}}}=2 n -2 \nu +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}}}=\left (n -\nu +1\right ) \left (n -\nu +2\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}}}=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} \\ {} & {} & 16 x^{2} \left (\frac {d}{d x}\frac {d}{d x}y^{\prime \prime }\right )+32 \left (n -\nu +2\right ) x \left (\frac {d}{d x}y^{\prime \prime }\right )+16 \left (n -\nu +1\right ) \left (n -\nu +2\right ) \left (\frac {d}{d x}y^{\prime }\right )-b^{4} 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} \frac {d}{d x}y^{\prime }\hspace {3pt}\textrm {to series expansion}\hspace {3pt} \\ {} & {} & \frac {d}{d x}y^{\prime }=\moverset {\infty }{\munderset {k =0}{\sum }}a_{k} \left (k +r \right ) \left (k +r -1\right ) x^{k +r -2} \\ {} & \circ & \textrm {Shift index using}\hspace {3pt} k \mathrm {->}k +2 \\ {} & {} & \frac {d}{d x}y^{\prime }=\moverset {\infty }{\munderset {k =-2}{\sum }}a_{k +2} \left (k +2+r \right ) \left (k +1+r \right ) x^{k +r} \\ {} & \circ & \textrm {Convert}\hspace {3pt} x \cdot \left (\frac {d}{d x}y^{\prime \prime }\right )\hspace {3pt}\textrm {to series expansion}\hspace {3pt} \\ {} & {} & x \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 -2} \\ {} & \circ & \textrm {Shift index using}\hspace {3pt} k \mathrm {->}k +2 \\ {} & {} & x \cdot \left (\frac {d}{d x}y^{\prime \prime }\right )=\moverset {\infty }{\munderset {k =-2}{\sum }}a_{k +2} \left (k +2+r \right ) \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}\frac {d}{d x}y^{\prime \prime }\right )\hspace {3pt}\textrm {to series expansion}\hspace {3pt} \\ {} & {} & x^{2}\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 -2} \\ {} & \circ & \textrm {Shift index using}\hspace {3pt} k \mathrm {->}k +2 \\ {} & {} & x^{2}\cdot \left (\frac {d}{d x}\frac {d}{d x}y^{\prime \prime }\right )=\moverset {\infty }{\munderset {k =-2}{\sum }}a_{k +2} \left (k +2+r \right ) \left (k +1+r \right ) \left (k +r \right ) \left (k +r -1\right ) x^{k +r} \\ & {} & \textrm {Rewrite ODE with series expansions}\hspace {3pt} \\ {} & {} & 16 a_{0} r \left (-1+r \right ) \left (n -\nu +r \right ) \left (r -1-\nu +n \right ) x^{-2+r}+16 a_{1} \left (1+r \right ) r \left (1+n -\nu +r \right ) \left (n -\nu +r \right ) x^{-1+r}+\left (\moverset {\infty }{\munderset {k =0}{\sum }}\left (16 a_{k +2} \left (k +2+r \right ) \left (k +1+r \right ) \left (k +2+n -\nu +r \right ) \left (r +1-\nu +n +k \right )-b^{4} a_{k}\right ) x^{k +r}\right )=0 \\ \bullet & {} & a_{0}\textrm {cannot be 0 by assumption, giving the indicial equation}\hspace {3pt} \\ {} & {} & 16 r \left (-1+r \right ) \left (n -\nu +r \right ) \left (r -1-\nu +n \right )=0 \\ \bullet & {} & \textrm {Values of r that satisfy the indicial equation}\hspace {3pt} \\ {} & {} & r \in \left \{0, 1, -n +\nu , 1+\nu -n \right \} \\ \bullet & {} & \textrm {Each term in the series must be 0, giving the recursion relation}\hspace {3pt} \\ {} & {} & 16 a_{k +2} \left (k +2+r \right ) \left (k +1+r \right ) \left (k +2+n -\nu +r \right ) \left (r +1-\nu +n +k \right )-b^{4} a_{k}=0 \\ \bullet & {} & \textrm {Recursion relation that defines series solution to ODE}\hspace {3pt} \\ {} & {} & a_{k +2}=\frac {b^{4} a_{k}}{16 \left (k +2+r \right ) \left (k +1+r \right ) \left (k +2+n -\nu +r \right ) \left (r +1-\nu +n +k \right )} \\ \bullet & {} & \textrm {Recursion relation for}\hspace {3pt} r =0 \\ {} & {} & a_{k +2}=\frac {b^{4} a_{k}}{16 \left (k +2\right ) \left (k +1\right ) \left (k +2+n -\nu \right ) \left (1-\nu +n +k \right )} \\ \bullet & {} & \textrm {Solution for}\hspace {3pt} r =0 \\ {} & {} & \left [y=\moverset {\infty }{\munderset {k =0}{\sum }}a_{k} x^{k}, a_{k +2}=\frac {b^{4} a_{k}}{16 \left (k +2\right ) \left (k +1\right ) \left (k +2+n -\nu \right ) \left (1-\nu +n +k \right )}, 0=0\right ] \\ \bullet & {} & \textrm {Recursion relation for}\hspace {3pt} r =1 \\ {} & {} & a_{k +2}=\frac {b^{4} a_{k}}{16 \left (k +3\right ) \left (k +2\right ) \left (k +3+n -\nu \right ) \left (k +2+n -\nu \right )} \\ \bullet & {} & \textrm {Solution for}\hspace {3pt} r =1 \\ {} & {} & \left [y=\moverset {\infty }{\munderset {k =0}{\sum }}a_{k} x^{k +1}, a_{k +2}=\frac {b^{4} a_{k}}{16 \left (k +3\right ) \left (k +2\right ) \left (k +3+n -\nu \right ) \left (k +2+n -\nu \right )}, 32 a_{1} \left (n -\nu +2\right ) \left (n -\nu +1\right )=0\right ] \\ \bullet & {} & \textrm {Recursion relation for}\hspace {3pt} r =-n +\nu \\ {} & {} & a_{k +2}=\frac {b^{4} a_{k}}{16 \left (k +2-n +\nu \right ) \left (k +1-n +\nu \right ) \left (k +2\right ) \left (k +1\right )} \\ \bullet & {} & \textrm {Solution for}\hspace {3pt} r =-n +\nu \\ {} & {} & \left [y=\moverset {\infty }{\munderset {k =0}{\sum }}a_{k} x^{k -n +\nu }, a_{k +2}=\frac {b^{4} a_{k}}{16 \left (k +2-n +\nu \right ) \left (k +1-n +\nu \right ) \left (k +2\right ) \left (k +1\right )}, 0=0\right ] \\ \bullet & {} & \textrm {Recursion relation for}\hspace {3pt} r =1+\nu -n \\ {} & {} & a_{k +2}=\frac {b^{4} a_{k}}{16 \left (k +3+\nu -n \right ) \left (k +2-n +\nu \right ) \left (k +3\right ) \left (k +2\right )} \\ \bullet & {} & \textrm {Solution for}\hspace {3pt} r =1+\nu -n \\ {} & {} & \left [y=\moverset {\infty }{\munderset {k =0}{\sum }}a_{k} x^{k +1-n +\nu }, a_{k +2}=\frac {b^{4} a_{k}}{16 \left (k +3+\nu -n \right ) \left (k +2-n +\nu \right ) \left (k +3\right ) \left (k +2\right )}, 32 a_{1} \left (2-n +\nu \right ) \left (1+\nu -n \right )=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 }}c_{k} x^{k +1}\right )+\left (\moverset {\infty }{\munderset {k =0}{\sum }}d_{k} x^{k -n +\nu }\right )+\left (\moverset {\infty }{\munderset {k =0}{\sum }}e_{k} x^{k +1-n +\nu }\right ), a_{k +2}=\frac {b^{4} a_{k}}{16 \left (k +2\right ) \left (k +1\right ) \left (k +2+n -\nu \right ) \left (1-\nu +n +k \right )}, 0=0, c_{k +2}=\frac {b^{4} c_{k}}{16 \left (k +3\right ) \left (k +2\right ) \left (k +3+n -\nu \right ) \left (k +2+n -\nu \right )}, 32 c_{1} \left (n -\nu +2\right ) \left (n -\nu +1\right )=0, d_{k +2}=\frac {b^{4} d_{k}}{16 \left (k +2-n +\nu \right ) \left (k +1-n +\nu \right ) \left (k +2\right ) \left (k +1\right )}, 0=0, e_{k +2}=\frac {b^{4} e_{k}}{16 \left (k +3+\nu -n \right ) \left (k +2-n +\nu \right ) \left (k +3\right ) \left (k +2\right )}, 32 e_{1} \left (2-n +\nu \right ) \left (1+\nu -n \right )=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((1/2)*(n-nu+1)/x, x))`Equation is the LCLM of 1/4*b^2/x*y(x)+(n-nu+1)/x*diff(y(x),x)+diff(diff(y( 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: 0.015 (sec). Leaf size: 67
dsolve(x^2*diff(diff(diff(diff(y(x),x),x),x),x)+(2*n-2*nu+4)*x*diff(diff(diff(y(x),x),x),x)+(n-nu+1)*(n-nu+2)*diff(diff(y(x),x),x)-1/16*b^4*y(x)=0,y(x), singsol=all)
\[ y \left (x \right ) = \left (\operatorname {BesselK}\left (n -\nu , b \sqrt {x}\right ) c_{3} +\operatorname {BesselI}\left (n -\nu , b \sqrt {x}\right ) c_{1} +\operatorname {BesselY}\left (n -\nu , b \sqrt {x}\right ) c_{4} +\operatorname {BesselJ}\left (n -\nu , b \sqrt {x}\right ) c_{2} \right ) x^{-\frac {n}{2}+\frac {\nu }{2}} \]
✓ Solution by Mathematica
Time used: 0.124 (sec). Leaf size: 222
DSolve[-1/16*(b^4*y[x]) + (1 + n - nu)*(2 + n - nu)*y''[x] + (4 + 2*n - 2*nu)*x*Derivative[3][y][x] + x^2*Derivative[4][y][x] == 0,y[x],x,IncludeSingularSolutions -> True]
\[ y(x)\to i^{-n} 2^{n-3 \nu -3} b^{\nu -n} x^{\frac {\nu -n}{2}} \left (i^n 4^{\nu } (4 c_1 \operatorname {Gamma}(n-\nu +1)-i c_2 \operatorname {Gamma}(n-\nu +2)) \operatorname {BesselJ}\left (n-\nu ,b \sqrt {x}\right )+i^n 4^{\nu } (4 c_1 \operatorname {Gamma}(n-\nu +1)+i c_2 \operatorname {Gamma}(n-\nu +2)) \operatorname {BesselI}\left (n-\nu ,b \sqrt {x}\right )+4^n i^{\nu } \left ((4 c_3 \operatorname {Gamma}(-n+\nu +1)-i c_4 \operatorname {Gamma}(-n+\nu +2)) \operatorname {BesselJ}\left (\nu -n,b \sqrt {x}\right )+(4 c_3 \operatorname {Gamma}(-n+\nu +1)+i c_4 \operatorname {Gamma}(-n+\nu +2)) \operatorname {BesselI}\left (\nu -n,b \sqrt {x}\right )\right )\right ) \]