Internal problem ID [9842]
Internal file name [OUTPUT/8787_Monday_June_06_2022_05_29_08_AM_46542294/index.tex]
Book: Differential Gleichungen, E. Kamke, 3rd ed. Chelsea Pub. NY, 1948
Section: Chapter 3, linear third order
Problem number: 1518.
ODE order: 3.
ODE degree: 1.
The type(s) of ODE detected by this program : "unknown"
Maple gives the following as the ode type
[[_3rd_order, _with_linear_symmetries]]
Unable to solve or complete the solution.
\[ \boxed {\left (x^{2}+1\right ) x y^{\prime \prime \prime }+3 \left (2 x^{2}+1\right ) y^{\prime \prime }-12 y=0} \] Unable to solve this ODE.
\[ \begin {array}{lll} & {} & \textrm {Let's solve}\hspace {3pt} \\ {} & {} & \left (x^{2}+1\right ) x \left (\frac {d}{d x}y^{\prime \prime }\right )+3 \left (2 x^{2}+1\right ) \left (\frac {d}{d x}y^{\prime }\right )-12 y=0 \\ \bullet & {} & \textrm {Highest derivative means the order of the ODE is}\hspace {3pt} 3 \\ {} & {} & \frac {d}{d x}y^{\prime \prime } \\ \square & {} & \textrm {Check to see if}\hspace {3pt} x_{0}\hspace {3pt}\textrm {is a regular singular point}\hspace {3pt} \\ {} & \circ & \textrm {Define functions}\hspace {3pt} \\ {} & {} & \left [P_{2}\left (x \right )=\frac {3 \left (2 x^{2}+1\right )}{\left (x^{2}+1\right ) x}, P_{3}\left (x \right )=0, P_{4}\left (x \right )=-\frac {12}{\left (x^{2}+1\right ) x}\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}}}=3 \\ {} & \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}}}=0 \\ {} & \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 =0\textrm {is a regular singular point}\hspace {3pt} \\ & {} & \textrm {Check to see if}\hspace {3pt} x_{0}\hspace {3pt}\textrm {is a regular singular point}\hspace {3pt} \\ {} & {} & x_{0}=0 \\ \bullet & {} & \textrm {Multiply by denominators}\hspace {3pt} \\ {} & {} & -12 y+\left (6 x^{2}+3\right ) \left (\frac {d}{d x}y^{\prime }\right )+\left (x^{2}+1\right ) x \left (\frac {d}{d x}y^{\prime \prime }\right )=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 \left (\frac {d}{d x}y^{\prime }\right )\hspace {3pt}\textrm {to series expansion for}\hspace {3pt} m =0..2 \\ {} & {} & x^{m}\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 -2+m} \\ {} & \circ & \textrm {Shift index using}\hspace {3pt} k \mathrm {->}k +2-m \\ {} & {} & x^{m}\cdot \left (\frac {d}{d x}y^{\prime }\right )=\moverset {\infty }{\munderset {k =-2+m}{\sum }}a_{k +2-m} \left (k +2-m +r \right ) \left (k +1-m +r \right ) x^{k +r} \\ {} & \circ & \textrm {Convert}\hspace {3pt} x^{m}\cdot \left (\frac {d}{d x}y^{\prime \prime }\right )\hspace {3pt}\textrm {to series expansion for}\hspace {3pt} m =1..3 \\ {} & {} & x^{m}\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 -3+m} \\ {} & \circ & \textrm {Shift index using}\hspace {3pt} k \mathrm {->}k +3-m \\ {} & {} & x^{m}\cdot \left (\frac {d}{d x}y^{\prime \prime }\right )=\moverset {\infty }{\munderset {k =-3+m}{\sum }}a_{k +3-m} \left (k +3-m +r \right ) \left (k +2-m +r \right ) \left (k +1-m +r \right ) x^{k +r} \\ & {} & \textrm {Rewrite ODE with series expansions}\hspace {3pt} \\ {} & {} & a_{0} r \left (r +1\right ) \left (-1+r \right ) x^{-2+r}+a_{1} \left (r +1\right ) \left (r +2\right ) r \,x^{-1+r}+\left (\moverset {\infty }{\munderset {k =0}{\sum }}\left (a_{k +2} \left (r +2+k \right ) \left (r +3+k \right ) \left (k +r +1\right )+a_{k} \left (r +2+k \right ) \left (k +r -2\right ) \left (r +3+k \right )\right ) x^{k +r}\right )=0 \\ \bullet & {} & a_{0}\textrm {cannot be 0 by assumption, giving the indicial equation}\hspace {3pt} \\ {} & {} & r \left (r +1\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} \\ {} & {} & \left (\left (a_{k}+a_{k +2}\right ) k +\left (a_{k}+a_{k +2}\right ) r -2 a_{k}+a_{k +2}\right ) \left (r +3+k \right ) \left (r +2+k \right )=0 \\ \bullet & {} & \textrm {Recursion relation that defines series solution to ODE}\hspace {3pt} \\ {} & {} & a_{k +2}=-\frac {a_{k} \left (k +r -2\right )}{k +r +1} \\ \bullet & {} & \textrm {Recursion relation for}\hspace {3pt} r =-1 \\ {} & {} & a_{k +2}=-\frac {a_{k} \left (k -3\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 {a_{k} \left (k -3\right )}{k}, 0=0\right ] \\ \bullet & {} & \textrm {Recursion relation for}\hspace {3pt} r =0\hspace {3pt}\textrm {; series terminates at}\hspace {3pt} k =2 \\ {} & {} & a_{k +2}=-\frac {a_{k} \left (k -2\right )}{k +1} \\ \bullet & {} & \textrm {Solution for}\hspace {3pt} r =0 \\ {} & {} & \left [y=\moverset {\infty }{\munderset {k =0}{\sum }}a_{k} x^{k}, a_{k +2}=-\frac {a_{k} \left (k -2\right )}{k +1}, 0=0\right ] \\ \bullet & {} & \textrm {Recursion relation for}\hspace {3pt} r =1 \\ {} & {} & a_{k +2}=-\frac {a_{k} \left (k -1\right )}{k +2} \\ \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 {a_{k} \left (k -1\right )}{k +2}, 6 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 {a_{k} \left (k -3\right )}{k}, 0=0, b_{k +2}=-\frac {b_{k} \left (k -2\right )}{k +1}, 0=0, c_{k +2}=-\frac {c_{k} \left (k -1\right )}{k +2}, 6 c_{1}=0\right ] \end {array} \]
Maple trace
`Methods for third 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: 3; missing the dependent variable trying Louvillian solutions for 3rd order ODEs, imprimitive case Louvillian solutions for 3rd order ODEs, imprimitive case: input is reducible, switching to DFactorsols checking if the LODE is of Euler type expon. solutions partially successful. Result(s) =`, [(x^2+1)^(1/2)*x, 2*x^2+1]
✓ Solution by Maple
Time used: 0.032 (sec). Leaf size: 60
dsolve((x^2+1)*x*diff(diff(diff(y(x),x),x),x)+3*(2*x^2+1)*diff(diff(y(x),x),x)-12*y(x)=0,y(x), singsol=all)
\[ y \left (x \right ) = \frac {3 \,\operatorname {arctanh}\left (\frac {1}{\sqrt {x^{2}+1}}\right ) \sqrt {x^{2}+1}\, c_{2} x^{2}+c_{1} \sqrt {x^{2}+1}\, x^{2}+2 c_{3} x^{3}-3 c_{2} x^{2}+c_{3} x -c_{2}}{x} \]
✓ Solution by Mathematica
Time used: 0.483 (sec). Leaf size: 69
DSolve[-12*y[x] + 3*(1 + 2*x^2)*y''[x] + x*(1 + x^2)*Derivative[3][y][x] == 0,y[x],x,IncludeSingularSolutions -> True]
\[ y(x)\to \frac {1}{6} \left (-3 c_3 x \sqrt {x^2+1} \text {arctanh}\left (\sqrt {x^2+1}\right )+c_1 \left (4 x^2+2\right )+2 c_2 x \sqrt {x^2+1}+3 c_3 x+\frac {c_3}{x}\right ) \]