Internal problem ID [9812]
Internal file name [OUTPUT/8755_Monday_June_06_2022_05_24_05_AM_2547530/index.tex]
Book: Differential Gleichungen, E. Kamke, 3rd ed. Chelsea Pub. NY, 1948
Section: Chapter 3, linear third order
Problem number: 1486.
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.
\[ \boxed {\left (2 x -1\right ) y^{\prime \prime \prime }-8 y^{\prime } x +8 y=0} \] Unable to solve this ODE.
\[ \begin {array}{lll} & {} & \textrm {Let's solve}\hspace {3pt} \\ {} & {} & \left (2 x -1\right ) \left (\frac {d}{d x}y^{\prime \prime }\right )-8 y^{\prime } x +8 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}=\frac {1}{2}\hspace {3pt}\textrm {is a regular singular point}\hspace {3pt} \\ {} & \circ & \textrm {Define functions}\hspace {3pt} \\ {} & {} & \left [P_{2}\left (x \right )=0, P_{3}\left (x \right )=-\frac {8 x}{2 x -1}, P_{4}\left (x \right )=\frac {8}{2 x -1}\right ] \\ {} & \circ & \left (x -\frac {1}{2}\right )\cdot P_{2}\left (x \right )\textrm {is analytic at}\hspace {3pt} x =\frac {1}{2} \\ {} & {} & \left (\left (x -\frac {1}{2}\right )\cdot P_{2}\left (x \right )\right )\bigg | {\mstack {}{_{x \hiderel {=}\frac {1}{2}}}}=0 \\ {} & \circ & \left (x -\frac {1}{2}\right )^{2}\cdot P_{3}\left (x \right )\textrm {is analytic at}\hspace {3pt} x =\frac {1}{2} \\ {} & {} & \left (\left (x -\frac {1}{2}\right )^{2}\cdot P_{3}\left (x \right )\right )\bigg | {\mstack {}{_{x \hiderel {=}\frac {1}{2}}}}=0 \\ {} & \circ & \left (x -\frac {1}{2}\right )^{3}\cdot P_{4}\left (x \right )\textrm {is analytic at}\hspace {3pt} x =\frac {1}{2} \\ {} & {} & \left (\left (x -\frac {1}{2}\right )^{3}\cdot P_{4}\left (x \right )\right )\bigg | {\mstack {}{_{x \hiderel {=}\frac {1}{2}}}}=0 \\ {} & \circ & x =\frac {1}{2}\textrm {is a regular singular point}\hspace {3pt} \\ & {} & \textrm {Check to see if}\hspace {3pt} x_{0}=\frac {1}{2}\hspace {3pt}\textrm {is a regular singular point}\hspace {3pt} \\ {} & {} & x_{0}=\frac {1}{2} \\ \bullet & {} & \textrm {Change variables using}\hspace {3pt} x =u +\frac {1}{2}\hspace {3pt}\textrm {so that the regular singular point is at}\hspace {3pt} u =0 \\ {} & {} & 2 u \left (\frac {d}{d u}\frac {d^{2}}{d u^{2}}y \left (u \right )\right )+\left (-8 u -4\right ) \left (\frac {d}{d u}y \left (u \right )\right )+8 y \left (u \right )=0 \\ \bullet & {} & \textrm {Assume series solution for}\hspace {3pt} y \left (u \right ) \\ {} & {} & y \left (u \right )=\moverset {\infty }{\munderset {k =0}{\sum }}a_{k} u^{k +r} \\ \square & {} & \textrm {Rewrite ODE with series expansions}\hspace {3pt} \\ {} & \circ & \textrm {Convert}\hspace {3pt} u^{m}\cdot \left (\frac {d}{d u}y \left (u \right )\right )\hspace {3pt}\textrm {to series expansion for}\hspace {3pt} m =0..1 \\ {} & {} & u^{m}\cdot \left (\frac {d}{d u}y \left (u \right )\right )=\moverset {\infty }{\munderset {k =0}{\sum }}a_{k} \left (k +r \right ) u^{k +r -1+m} \\ {} & \circ & \textrm {Shift index using}\hspace {3pt} k \mathrm {->}k +1-m \\ {} & {} & u^{m}\cdot \left (\frac {d}{d u}y \left (u \right )\right )=\moverset {\infty }{\munderset {k =-1+m}{\sum }}a_{k +1-m} \left (k +1-m +r \right ) u^{k +r} \\ {} & \circ & \textrm {Convert}\hspace {3pt} u \cdot \left (\frac {d}{d u}\frac {d^{2}}{d u^{2}}y \left (u \right )\right )\hspace {3pt}\textrm {to series expansion}\hspace {3pt} \\ {} & {} & u \cdot \left (\frac {d}{d u}\frac {d^{2}}{d u^{2}}y \left (u \right )\right )=\moverset {\infty }{\munderset {k =0}{\sum }}a_{k} \left (k +r \right ) \left (k +r -1\right ) \left (k +r -2\right ) u^{k +r -2} \\ {} & \circ & \textrm {Shift index using}\hspace {3pt} k \mathrm {->}k +2 \\ {} & {} & u \cdot \left (\frac {d}{d u}\frac {d^{2}}{d u^{2}}y \left (u \right )\right )=\moverset {\infty }{\munderset {k =-2}{\sum }}a_{k +2} \left (k +2+r \right ) \left (k +1+r \right ) \left (k +r \right ) u^{k +r} \\ & {} & \textrm {Rewrite ODE with series expansions}\hspace {3pt} \\ {} & {} & 2 a_{0} r \left (-1+r \right ) \left (-2+r \right ) u^{-2+r}+\left (2 a_{1} \left (1+r \right ) r \left (-1+r \right )-4 a_{0} r \right ) u^{-1+r}+\left (\moverset {\infty }{\munderset {k =0}{\sum }}\left (2 a_{k +2} \left (k +2+r \right ) \left (k +1+r \right ) \left (k +r \right )-4 a_{k +1} \left (k +1+r \right )-8 a_{k} \left (k +r -1\right )\right ) u^{k +r}\right )=0 \\ \bullet & {} & a_{0}\textrm {cannot be 0 by assumption, giving the indicial equation}\hspace {3pt} \\ {} & {} & 2 r \left (-1+r \right ) \left (-2+r \right )=0 \\ \bullet & {} & \textrm {Values of r that satisfy the indicial equation}\hspace {3pt} \\ {} & {} & r \in \left \{0, 1, 2\right \} \\ \bullet & {} & \textrm {Each term in the series must be 0, giving the recursion relation}\hspace {3pt} \\ {} & {} & 2 a_{k +2} \left (k +2+r \right ) \left (k +1+r \right ) \left (k +r \right )+\left (-8 a_{k}-4 a_{k +1}\right ) k +\left (-8 a_{k}-4 a_{k +1}\right ) r +8 a_{k}-4 a_{k +1}=0 \\ \bullet & {} & \textrm {Recursion relation that defines series solution to ODE}\hspace {3pt} \\ {} & {} & a_{k +2}=\frac {2 \left (2 a_{k} k +a_{k +1} k +2 a_{k} r +a_{k +1} r -2 a_{k}+a_{k +1}\right )}{\left (k +2+r \right ) \left (k +1+r \right ) \left (k +r \right )} \\ \bullet & {} & \textrm {Recursion relation for}\hspace {3pt} r =0 \\ {} & {} & a_{k +2}=\frac {2 \left (2 a_{k} k +a_{k +1} k -2 a_{k}+a_{k +1}\right )}{\left (k +2\right ) \left (k +1\right ) k} \\ \bullet & {} & \textrm {Series not valid for}\hspace {3pt} r =0\hspace {3pt}\textrm {, division by}\hspace {3pt} 0\hspace {3pt}\textrm {in the recursion relation at}\hspace {3pt} k =0 \\ {} & {} & a_{k +2}=\frac {2 \left (2 a_{k} k +a_{k +1} k -2 a_{k}+a_{k +1}\right )}{\left (k +2\right ) \left (k +1\right ) k} \\ \bullet & {} & \textrm {Recursion relation for}\hspace {3pt} r =1 \\ {} & {} & a_{k +2}=\frac {2 \left (2 a_{k} k +a_{k +1} k +2 a_{k +1}\right )}{\left (k +3\right ) \left (k +2\right ) \left (k +1\right )} \\ \bullet & {} & \textrm {Solution for}\hspace {3pt} r =1 \\ {} & {} & \left [y \left (u \right )=\moverset {\infty }{\munderset {k =0}{\sum }}a_{k} u^{k +1}, a_{k +2}=\frac {2 \left (2 a_{k} k +a_{k +1} k +2 a_{k +1}\right )}{\left (k +3\right ) \left (k +2\right ) \left (k +1\right )}, -4 a_{0}=0\right ] \\ \bullet & {} & \textrm {Revert the change of variables}\hspace {3pt} u =x -\frac {1}{2} \\ {} & {} & \left [y=\moverset {\infty }{\munderset {k =0}{\sum }}a_{k} \left (x -\frac {1}{2}\right )^{k +1}, a_{k +2}=\frac {2 \left (2 a_{k} k +a_{k +1} k +2 a_{k +1}\right )}{\left (k +3\right ) \left (k +2\right ) \left (k +1\right )}, -4 a_{0}=0\right ] \\ \bullet & {} & \textrm {Recursion relation for}\hspace {3pt} r =2 \\ {} & {} & a_{k +2}=\frac {2 \left (2 a_{k} k +a_{k +1} k +2 a_{k}+3 a_{k +1}\right )}{\left (k +4\right ) \left (k +3\right ) \left (k +2\right )} \\ \bullet & {} & \textrm {Solution for}\hspace {3pt} r =2 \\ {} & {} & \left [y \left (u \right )=\moverset {\infty }{\munderset {k =0}{\sum }}a_{k} u^{k +2}, a_{k +2}=\frac {2 \left (2 a_{k} k +a_{k +1} k +2 a_{k}+3 a_{k +1}\right )}{\left (k +4\right ) \left (k +3\right ) \left (k +2\right )}, 12 a_{1}-8 a_{0}=0\right ] \\ \bullet & {} & \textrm {Revert the change of variables}\hspace {3pt} u =x -\frac {1}{2} \\ {} & {} & \left [y=\moverset {\infty }{\munderset {k =0}{\sum }}a_{k} \left (x -\frac {1}{2}\right )^{k +2}, a_{k +2}=\frac {2 \left (2 a_{k} k +a_{k +1} k +2 a_{k}+3 a_{k +1}\right )}{\left (k +4\right ) \left (k +3\right ) \left (k +2\right )}, 12 a_{1}-8 a_{0}=0\right ] \\ \bullet & {} & \textrm {Combine solutions and rename parameters}\hspace {3pt} \\ {} & {} & \left [y=\left (\moverset {\infty }{\munderset {k =0}{\sum }}a_{k} \left (x -\frac {1}{2}\right )^{k +1}\right )+\left (\moverset {\infty }{\munderset {k =0}{\sum }}b_{k} \left (x -\frac {1}{2}\right )^{k +2}\right ), a_{k +2}=\frac {2 \left (2 k a_{k}+k a_{k +1}+2 a_{k +1}\right )}{\left (k +3\right ) \left (k +2\right ) \left (k +1\right )}, -4 a_{0}=0, b_{k +2}=\frac {2 \left (2 k b_{k}+k b_{k +1}+2 b_{k}+3 b_{k +1}\right )}{\left (k +4\right ) \left (k +3\right ) \left (k +2\right )}, 12 b_{1}-8 b_{0}=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) =`, [exp(2*x), x]
✓ Solution by Maple
Time used: 0.016 (sec). Leaf size: 50
dsolve((2*x-1)*diff(diff(diff(y(x),x),x),x)-8*x*diff(y(x),x)+8*y(x)=0,y(x), singsol=all)
\[ y \left (x \right ) = c_{1} x +c_{2} {\mathrm e}^{2 x}-\frac {c_{3} x \,{\mathrm e}^{-1} \operatorname {expIntegral}_{1}\left (-1+2 x \right )}{2}+\frac {c_{3} \operatorname {expIntegral}_{1}\left (-2+4 x \right ) {\mathrm e}^{2 x -2}}{4}+\frac {c_{3} {\mathrm e}^{-2 x}}{4} \]
✓ Solution by Mathematica
Time used: 0.372 (sec). Leaf size: 63
DSolve[8*y[x] - 8*x*y'[x] + (-1 + 2*x)*Derivative[3][y][x] == 0,y[x],x,IncludeSingularSolutions -> True]
\[ y(x)\to \frac {1}{4} \left (c_3 e^{2 x-2} \operatorname {ExpIntegralEi}(2-4 x)-\frac {2 c_3 x \operatorname {ExpIntegralEi}(1-2 x)}{e}+4 c_1 x-4 c_2 e^{2 x}-c_3 e^{-2 x}\right ) \]