Internal problem ID [9877]
Internal file name [OUTPUT/8824_Monday_June_06_2022_05_34_25_AM_58750103/index.tex]
Book: Differential Gleichungen, E. Kamke, 3rd ed. Chelsea Pub. NY, 1948
Section: Chapter 4, linear fourth order
Problem number: 1555.
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.
\[ \boxed {x^{2} y^{\prime \prime \prime \prime }+6 x y^{\prime \prime \prime }+6 y^{\prime \prime }-\lambda ^{2} y=0} \] 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 )+6 \left (\frac {d}{d x}y^{\prime \prime }\right ) x +6 \frac {d}{d x}y^{\prime }-\lambda ^{2} 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}, P_{3}\left (x \right )=\frac {6}{x^{2}}, P_{4}\left (x \right )=0, P_{5}\left (x \right )=-\frac {\lambda ^{2}}{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}}}=6 \\ {} & \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}}}=6 \\ {} & \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 {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} \\ {} & {} & a_{0} r^{2} \left (1+r \right ) \left (-1+r \right ) x^{-2+r}+a_{1} \left (1+r \right )^{2} \left (2+r \right ) r \,x^{-1+r}+\left (\moverset {\infty }{\munderset {k =0}{\sum }}\left (a_{k +2} \left (k +2+r \right )^{2} \left (k +3+r \right ) \left (k +1+r \right )-\lambda ^{2} a_{k}\right ) x^{k +r}\right )=0 \\ \bullet & {} & a_{0}\textrm {cannot be 0 by assumption, giving the indicial equation}\hspace {3pt} \\ {} & {} & r^{2} \left (1+r \right ) \left (-1+r \right )=0 \\ \bullet & {} & \textrm {Values of r that satisfy the indicial equation}\hspace {3pt} \\ {} & {} & r \in \left \{-1, 0, 1\right \} \\ \bullet & {} & \textrm {Each term in the series must be 0, giving the recursion relation}\hspace {3pt} \\ {} & {} & a_{k +2} \left (k +2+r \right )^{2} \left (k +3+r \right ) \left (k +1+r \right )-\lambda ^{2} a_{k}=0 \\ \bullet & {} & \textrm {Recursion relation that defines series solution to ODE}\hspace {3pt} \\ {} & {} & a_{k +2}=\frac {\lambda ^{2} a_{k}}{\left (k +2+r \right )^{2} \left (k +3+r \right ) \left (k +1+r \right )} \\ \bullet & {} & \textrm {Recursion relation for}\hspace {3pt} r =-1 \\ {} & {} & a_{k +2}=\frac {\lambda ^{2} a_{k}}{\left (k +1\right )^{2} \left (k +2\right ) k} \\ \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 {\lambda ^{2} a_{k}}{\left (k +1\right )^{2} \left (k +2\right ) k}, 0=0\right ] \\ \bullet & {} & \textrm {Recursion relation for}\hspace {3pt} r =0 \\ {} & {} & a_{k +2}=\frac {\lambda ^{2} a_{k}}{\left (k +2\right )^{2} \left (k +3\right ) \left (k +1\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 {\lambda ^{2} a_{k}}{\left (k +2\right )^{2} \left (k +3\right ) \left (k +1\right )}, 0=0\right ] \\ \bullet & {} & \textrm {Recursion relation for}\hspace {3pt} r =1 \\ {} & {} & a_{k +2}=\frac {\lambda ^{2} a_{k}}{\left (k +3\right )^{2} \left (k +4\right ) \left (k +2\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 {\lambda ^{2} a_{k}}{\left (k +3\right )^{2} \left (k +4\right ) \left (k +2\right )}, 12 a_{1}=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}\right )+\left (\moverset {\infty }{\munderset {k =0}{\sum }}c_{k} x^{k +1}\right ), a_{k +2}=\frac {\lambda ^{2} a_{k}}{\left (k +1\right )^{2} \left (k +2\right ) k}, 0=0, b_{k +2}=\frac {\lambda ^{2} b_{k}}{\left (k +2\right )^{2} \left (k +3\right ) \left (k +1\right )}, 0=0, c_{k +2}=\frac {\lambda ^{2} c_{k}}{\left (k +3\right )^{2} \left (k +4\right ) \left (k +2\right )}, 12 c_{1}=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 LCLM of lambda/x*y(x)+2*diff(y(x),x)/x+diff(diff(y(x),x),x), -lambda/x*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: 0.125 (sec). Leaf size: 61
dsolve(x^2*diff(diff(diff(diff(y(x),x),x),x),x)+6*x*diff(diff(diff(y(x),x),x),x)+6*diff(diff(y(x),x),x)-lambda^2*y(x)=0,y(x), singsol=all)
\[ y \left (x \right ) = \frac {c_{1} \operatorname {BesselJ}\left (1, 2 \sqrt {\lambda }\, \sqrt {x}\right )+c_{2} \operatorname {BesselY}\left (1, 2 \sqrt {\lambda }\, \sqrt {x}\right )+c_{4} \operatorname {BesselY}\left (1, 2 \sqrt {-\lambda }\, \sqrt {x}\right )+c_{3} \operatorname {BesselJ}\left (1, 2 \sqrt {-\lambda }\, \sqrt {x}\right )}{\sqrt {x}} \]
✓ Solution by Mathematica
Time used: 0.05 (sec). Leaf size: 156
DSolve[-(\[Lambda]^2*y[x]) + 6*y''[x] + 6*x*Derivative[3][y][x] + x^2*Derivative[4][y][x] == 0,y[x],x,IncludeSingularSolutions -> True]
\[ y(x)\to c_4 G_{0,4}^{2,0}\left (\frac {x^2 \lambda ^2}{16}| \begin {array}{c} -\frac {1}{2},\frac {1}{2},0,0 \\ \end {array} \right )+c_2 G_{0,4}^{2,0}\left (\frac {x^2 \lambda ^2}{16}| \begin {array}{c} 0,0,-\frac {1}{2},\frac {1}{2} \\ \end {array} \right )+\frac {c_1 \left (\operatorname {BesselJ}\left (1,2 \sqrt {x} \sqrt {\lambda }\right )+\operatorname {BesselI}\left (1,2 \sqrt {x} \sqrt {\lambda }\right )\right )}{2 \sqrt {\lambda } \sqrt {x}}-\frac {i c_3 \left (\operatorname {BesselI}\left (1,2 \sqrt {x} \sqrt {\lambda }\right )-\operatorname {BesselJ}\left (1,2 \sqrt {x} \sqrt {\lambda }\right )\right )}{4 \sqrt {\lambda } \sqrt {x}} \]