problem Ex 4

Internal problem ID [11325]
Internal ﬁle name [OUTPUT/10311_Tuesday_December_27_2022_04_06_15_AM_43338447/index.tex]

Book: An elementary treatise on diﬀerential equations by Abraham Cohen. DC heath publishers. 1906
Section: Chapter IX, Miscellaneous methods for solving equations of higher order than ﬁrst. Article 60. Exact equation. Integrating factor. Page 139
Problem number: Ex 4.
ODE order: 3.
ODE degree: 1.

\[ \boxed {\left (x^{3}-x \right ) y^{\prime \prime \prime }+\left (8 x^{2}-3\right ) y^{\prime \prime }+14 x y^{\prime }+4 y=0} \] Unable to solve this ODE.

36.4.1 Maple step by step solution

\[ \begin {array}{lll} & {} & \textrm {Let's solve}\hspace {3pt} \\ {} & {} & \left (x^{3}-x \right ) \left (\frac {d}{d x}y^{\prime \prime }\right )+\left (8 x^{2}-3\right ) \left (\frac {d}{d x}y^{\prime }\right )+14 x y^{\prime }+4 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 {8 x^{2}-3}{x \left (x^{2}-1\right )}, P_{3}\left (x \right )=\frac {14}{x^{2}-1}, P_{4}\left (x \right )=\frac {4}{x \left (x^{2}-1\right )}\right ] \\ {} & \circ & \left (x +1\right )\cdot P_{2}\left (x \right )\textrm {is analytic at}\hspace {3pt} x =-1 \\ {} & {} & \left (\left (x +1\right )\cdot P_{2}\left (x \right )\right )\bigg | {\mstack {}{_{x \hiderel {=}-1}}}=\frac {5}{2} \\ {} & \circ & \left (x +1\right )^{2}\cdot P_{3}\left (x \right )\textrm {is analytic at}\hspace {3pt} x =-1 \\ {} & {} & \left (\left (x +1\right )^{2}\cdot P_{3}\left (x \right )\right )\bigg | {\mstack {}{_{x \hiderel {=}-1}}}=0 \\ {} & \circ & \left (x +1\right )^{3}\cdot P_{4}\left (x \right )\textrm {is analytic at}\hspace {3pt} x =-1 \\ {} & {} & \left (\left (x +1\right )^{3}\cdot P_{4}\left (x \right )\right )\bigg | {\mstack {}{_{x \hiderel {=}-1}}}=0 \\ {} & \circ & x =-1\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}=-1 \\ \bullet & {} & \textrm {Multiply by denominators}\hspace {3pt} \\ {} & {} & x \left (x^{2}-1\right ) \left (\frac {d}{d x}y^{\prime \prime }\right )+\left (8 x^{2}-3\right ) \left (\frac {d}{d x}y^{\prime }\right )+14 x y^{\prime }+4 y=0 \\ \bullet & {} & \textrm {Change variables using}\hspace {3pt} x =u -1\hspace {3pt}\textrm {so that the regular singular point is at}\hspace {3pt} u =0 \\ {} & {} & \left (u^{3}-3 u^{2}+2 u \right ) \left (\frac {d}{d u}\frac {d^{2}}{d u^{2}}y \left (u \right )\right )+\left (8 u^{2}-16 u +5\right ) \left (\frac {d}{d u}\frac {d}{d u}y \left (u \right )\right )+\left (14 u -14\right ) \left (\frac {d}{d u}y \left (u \right )\right )+4 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^{m}\cdot \left (\frac {d}{d u}\frac {d}{d u}y \left (u \right )\right )\hspace {3pt}\textrm {to series expansion for}\hspace {3pt} m =0..2 \\ {} & {} & u^{m}\cdot \left (\frac {d}{d u}\frac {d}{d u}y \left (u \right )\right )=\moverset {\infty }{\munderset {k =0}{\sum }}a_{k} \left (k +r \right ) \left (k +r -1\right ) u^{k +r -2+m} \\ {} & \circ & \textrm {Shift index using}\hspace {3pt} k \mathrm {->}k +2-m \\ {} & {} & u^{m}\cdot \left (\frac {d}{d u}\frac {d}{d u}y \left (u \right )\right )=\moverset {\infty }{\munderset {k =-2+m}{\sum }}a_{k +2-m} \left (k +2-m +r \right ) \left (k +1-m +r \right ) u^{k +r} \\ {} & \circ & \textrm {Convert}\hspace {3pt} u^{m}\cdot \left (\frac {d}{d u}\frac {d^{2}}{d u^{2}}y \left (u \right )\right )\hspace {3pt}\textrm {to series expansion for}\hspace {3pt} m =1..3 \\ {} & {} & u^{m}\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 -3+m} \\ {} & \circ & \textrm {Shift index using}\hspace {3pt} k \mathrm {->}k +3-m \\ {} & {} & u^{m}\cdot \left (\frac {d}{d u}\frac {d^{2}}{d u^{2}}y \left (u \right )\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 ) u^{k +r} \\ & {} & \textrm {Rewrite ODE with series expansions}\hspace {3pt} \\ {} & {} & a_{0} r \left (1+2 r \right ) \left (-1+r \right ) u^{-2+r}+\left (a_{1} \left (r +1\right ) \left (2 r +3\right ) r -a_{0} r \left (3 r +4\right ) \left (r +1\right )\right ) u^{-1+r}+\left (\moverset {\infty }{\munderset {k =0}{\sum }}\left (a_{k +2} \left (2+k +r \right ) \left (2 k +5+2 r \right ) \left (k +r +1\right )-a_{k +1} \left (k +r +1\right ) \left (3 k +7+3 r \right ) \left (2+k +r \right )+a_{k} \left (k +r +1\right ) \left (2+k +r \right )^{2}\right ) u^{k +r}\right )=0 \\ \bullet & {} & a_{0}\textrm {cannot be 0 by assumption, giving the indicial equation}\hspace {3pt} \\ {} & {} & r \left (1+2 r \right ) \left (-1+r \right )=0 \\ \bullet & {} & \textrm {Values of r that satisfy the indicial equation}\hspace {3pt} \\ {} & {} & r \in \left \{0, 1, -\frac {1}{2}\right \} \\ \bullet & {} & \textrm {Each term in the series must be 0, giving the recursion relation}\hspace {3pt} \\ {} & {} & \left (2+k +r \right ) \left (k +r +1\right ) \left (\left (a_{k}-3 a_{k +1}+2 a_{k +2}\right ) k +\left (a_{k}-3 a_{k +1}+2 a_{k +2}\right ) r +2 a_{k}-7 a_{k +1}+5 a_{k +2}\right )=0 \\ \bullet & {} & \textrm {Recursion relation that defines series solution to ODE}\hspace {3pt} \\ {} & {} & a_{k +2}=-\frac {k a_{k}-3 k a_{k +1}+r a_{k}-3 r a_{k +1}+2 a_{k}-7 a_{k +1}}{2 k +5+2 r} \\ \bullet & {} & \textrm {Recursion relation for}\hspace {3pt} r =0 \\ {} & {} & a_{k +2}=-\frac {k a_{k}-3 k a_{k +1}+2 a_{k}-7 a_{k +1}}{2 k +5} \\ \bullet & {} & \textrm {Solution for}\hspace {3pt} r =0 \\ {} & {} & \left [y \left (u \right )=\moverset {\infty }{\munderset {k =0}{\sum }}a_{k} u^{k}, a_{k +2}=-\frac {k a_{k}-3 k a_{k +1}+2 a_{k}-7 a_{k +1}}{2 k +5}, 0=0\right ] \\ \bullet & {} & \textrm {Revert the change of variables}\hspace {3pt} u =x +1 \\ {} & {} & \left [y=\moverset {\infty }{\munderset {k =0}{\sum }}a_{k} \left (x +1\right )^{k}, a_{k +2}=-\frac {k a_{k}-3 k a_{k +1}+2 a_{k}-7 a_{k +1}}{2 k +5}, 0=0\right ] \\ \bullet & {} & \textrm {Recursion relation for}\hspace {3pt} r =1 \\ {} & {} & a_{k +2}=-\frac {k a_{k}-3 k a_{k +1}+3 a_{k}-10 a_{k +1}}{2 k +7} \\ \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 {k a_{k}-3 k a_{k +1}+3 a_{k}-10 a_{k +1}}{2 k +7}, 10 a_{1}-14 a_{0}=0\right ] \\ \bullet & {} & \textrm {Revert the change of variables}\hspace {3pt} u =x +1 \\ {} & {} & \left [y=\moverset {\infty }{\munderset {k =0}{\sum }}a_{k} \left (x +1\right )^{k +1}, a_{k +2}=-\frac {k a_{k}-3 k a_{k +1}+3 a_{k}-10 a_{k +1}}{2 k +7}, 10 a_{1}-14 a_{0}=0\right ] \\ \bullet & {} & \textrm {Recursion relation for}\hspace {3pt} r =-\frac {1}{2} \\ {} & {} & a_{k +2}=-\frac {k a_{k}-3 k a_{k +1}+\frac {3}{2} a_{k}-\frac {11}{2} a_{k +1}}{2 k +4} \\ \bullet & {} & \textrm {Solution for}\hspace {3pt} r =-\frac {1}{2} \\ {} & {} & \left [y \left (u \right )=\moverset {\infty }{\munderset {k =0}{\sum }}a_{k} u^{k -\frac {1}{2}}, a_{k +2}=-\frac {k a_{k}-3 k a_{k +1}+\frac {3}{2} a_{k}-\frac {11}{2} a_{k +1}}{2 k +4}, -\frac {a_{1}}{2}+\frac {5 a_{0}}{8}=0\right ] \\ \bullet & {} & \textrm {Revert the change of variables}\hspace {3pt} u =x +1 \\ {} & {} & \left [y=\moverset {\infty }{\munderset {k =0}{\sum }}a_{k} \left (x +1\right )^{k -\frac {1}{2}}, a_{k +2}=-\frac {k a_{k}-3 k a_{k +1}+\frac {3}{2} a_{k}-\frac {11}{2} a_{k +1}}{2 k +4}, -\frac {a_{1}}{2}+\frac {5 a_{0}}{8}=0\right ] \\ \bullet & {} & \textrm {Combine solutions and rename parameters}\hspace {3pt} \\ {} & {} & \left [y=\left (\moverset {\infty }{\munderset {k =0}{\sum }}a_{k} \left (x +1\right )^{k}\right )+\left (\moverset {\infty }{\munderset {k =0}{\sum }}b_{k} \left (x +1\right )^{k +1}\right )+\left (\moverset {\infty }{\munderset {k =0}{\sum }}c_{k} \left (x +1\right )^{k -\frac {1}{2}}\right ), a_{k +2}=-\frac {k a_{k}-3 k a_{k +1}+2 a_{k}-7 a_{k +1}}{2 k +5}, 0=0, b_{k +2}=-\frac {k b_{k}-3 k b_{k +1}+3 b_{k}-10 b_{k +1}}{2 k +7}, 10 b_{1}-14 b_{0}=0, c_{k +2}=-\frac {k c_{k}-3 k c_{k +1}+\frac {3}{2} c_{k}-\frac {11}{2} c_{k +1}}{2 k +4}, -\frac {c_{1}}{2}+\frac {5 c_{0}}{8}=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 
<- high order exact linear fully integrable successful`

\[ y \left (x \right ) = \frac {\frac {c_{3}}{\sqrt {1+x}\, \sqrt {-1+x}}+c_{1} +\frac {c_{2} \ln \left (x +\sqrt {x^{2}-1}\right )}{\sqrt {x^{2}-1}}}{x} \]

\[ y(x)\to \frac {-\frac {c_2}{\sqrt {x^2-1}}+\frac {c_3 \log \left (\sqrt {x^2-1}-x\right )}{\sqrt {x^2-1}}+c_1}{x} \]

36.4 problem Ex 4

36.4.1 Maple step by step solution