Internal problem ID [9881]
Internal file name [OUTPUT/8828_Monday_June_06_2022_05_34_52_AM_83336273/index.tex]
Book: Differential Gleichungen, E. Kamke, 3rd ed. Chelsea Pub. NY, 1948
Section: Chapter 4, linear fourth order
Problem number: 1559.
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.
\[ \boxed {x^{3} y^{\prime \prime \prime \prime }+2 x^{2} y^{\prime \prime \prime }-x y^{\prime \prime }+y^{\prime }-a^{4} x^{3} y=0} \] 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 )+2 x^{2} \left (\frac {d}{d x}y^{\prime \prime }\right )-x \left (\frac {d}{d x}y^{\prime }\right )+y^{\prime }-a^{4} x^{3} 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 {2}{x}, P_{3}\left (x \right )=-\frac {1}{x^{2}}, P_{4}\left (x \right )=\frac {1}{x^{3}}, P_{5}\left (x \right )=-a^{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}}}=2 \\ {} & \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}}}=-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}}}=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 {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^{2} \left (-2+r \right )^{2} x^{-1+r}+a_{1} \left (1+r \right )^{2} \left (-1+r \right )^{2} x^{r}+a_{2} \left (2+r \right )^{2} r^{2} x^{1+r}+a_{3} \left (3+r \right )^{2} \left (1+r \right )^{2} x^{2+r}+\left (\moverset {\infty }{\munderset {k =3}{\sum }}\left (a_{k +1} \left (k +1+r \right )^{2} \left (k +r -1\right )^{2}-a^{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^{2} \left (-2+r \right )^{2}=0 \\ \bullet & {} & \textrm {Values of r that satisfy the indicial equation}\hspace {3pt} \\ {} & {} & r \in \left \{0, 2\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 )^{2} \left (-1+r \right )^{2}=0, a_{2} \left (2+r \right )^{2} r^{2}=0, a_{3} \left (3+r \right )^{2} \left (1+r \right )^{2}=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 )^{2} \left (k +r -1\right )^{2}-a^{4} a_{k -3}=0 \\ \bullet & {} & \textrm {Shift index using}\hspace {3pt} k \mathrm {->}k +3 \\ {} & {} & a_{k +4} \left (k +4+r \right )^{2} \left (k +2+r \right )^{2}-a^{4} a_{k}=0 \\ \bullet & {} & \textrm {Recursion relation that defines series solution to ODE}\hspace {3pt} \\ {} & {} & a_{k +4}=\frac {a^{4} a_{k}}{\left (k +4+r \right )^{2} \left (k +2+r \right )^{2}} \\ \bullet & {} & \textrm {Recursion relation for}\hspace {3pt} r =0 \\ {} & {} & a_{k +4}=\frac {a^{4} a_{k}}{\left (k +4\right )^{2} \left (k +2\right )^{2}} \\ \bullet & {} & \textrm {Solution for}\hspace {3pt} r =0 \\ {} & {} & \left [y=\moverset {\infty }{\munderset {k =0}{\sum }}a_{k} x^{k}, a_{k +4}=\frac {a^{4} a_{k}}{\left (k +4\right )^{2} \left (k +2\right )^{2}}, a_{1}=0, a_{2}=0, a_{3}=0\right ] \\ \bullet & {} & \textrm {Recursion relation for}\hspace {3pt} r =2 \\ {} & {} & a_{k +4}=\frac {a^{4} a_{k}}{\left (k +6\right )^{2} \left (k +4\right )^{2}} \\ \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 {a^{4} a_{k}}{\left (k +6\right )^{2} \left (k +4\right )^{2}}, 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}\right )+\left (\moverset {\infty }{\munderset {k =0}{\sum }}c_{k} x^{k +2}\right ), b_{k +4}=\frac {a^{4} b_{k}}{\left (k +4\right )^{2} \left (k +2\right )^{2}}, b_{1}=0, b_{2}=0, b_{3}=0, c_{k +4}=\frac {a^{4} c_{k}}{\left (k +6\right )^{2} \left (k +4\right )^{2}}, c_{1}=0, c_{2}=0, c_{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((1/2)/x, x))`Equation is the LCLM of y(x)*a^2+diff(y(x),x)/x+diff(diff(y(x),x),x), -y(x)*a^2+diff 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: 0.016 (sec). Leaf size: 33
dsolve(x^3*diff(diff(diff(diff(y(x),x),x),x),x)+2*x^2*diff(diff(diff(y(x),x),x),x)-x*diff(diff(y(x),x),x)+diff(y(x),x)-a^4*x^3*y(x)=0,y(x), singsol=all)
\[ y \left (x \right ) = c_{1} \operatorname {BesselI}\left (0, a x \right )+c_{2} \operatorname {BesselJ}\left (0, a x \right )+c_{3} \operatorname {BesselK}\left (0, a x \right )+c_{4} \operatorname {BesselY}\left (0, a x \right ) \]
✓ Solution by Mathematica
Time used: 0.221 (sec). Leaf size: 100
DSolve[-(a^4*x^3*y[x]) + y'[x] - x*y''[x] + 2*x^2*Derivative[3][y][x] + x^3*Derivative[4][y][x] == 0,y[x],x,IncludeSingularSolutions -> True]
\[ y(x)\to c_4 G_{0,4}^{2,0}\left (\frac {a^4 x^4}{256}| \begin {array}{c} 0,0,\frac {1}{2},\frac {1}{2} \\ \end {array} \right )+c_2 G_{0,4}^{2,0}\left (\frac {a^4 x^4}{256}| \begin {array}{c} \frac {1}{2},\frac {1}{2},0,0 \\ \end {array} \right )+\frac {1}{8} i c_1 (\operatorname {BesselI}(0,a x)-\operatorname {BesselJ}(0,a x))+\frac {1}{2} c_3 (\operatorname {BesselJ}(0,a x)+\operatorname {BesselI}(0,a x)) \]