Internal problem ID [11338]
Internal file name [OUTPUT/10325_Friday_January_27_2023_02_36_58_AM_31623902/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 62. Summary. Page 144
Problem number: Ex 4.
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, _fully, _exact, _linear]]
Unable to solve or complete the solution.
\[ \boxed {\left (x^{3}+1\right ) y^{\prime \prime \prime }+9 x^{2} y^{\prime \prime }+18 x y^{\prime }+6 y=0} \] Unable to solve this ODE.
\[ \begin {array}{lll} & {} & \textrm {Let's solve}\hspace {3pt} \\ {} & {} & \left (x^{3}+1\right ) \left (\frac {d}{d x}y^{\prime \prime }\right )+9 \left (\frac {d}{d x}y^{\prime }\right ) x^{2}+18 x y^{\prime }+6 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 {9 x^{2}}{x^{3}+1}, P_{3}\left (x \right )=\frac {18 x}{x^{3}+1}, P_{4}\left (x \right )=\frac {6}{x^{3}+1}\right ] \\ {} & \circ & \left (x -\frac {1}{2}+\frac {\mathrm {I} \sqrt {3}}{2}\right )\cdot P_{2}\left (x \right )\textrm {is analytic at}\hspace {3pt} x =\frac {1}{2}-\frac {\mathrm {I} \sqrt {3}}{2} \\ {} & {} & \left (\left (x -\frac {1}{2}+\frac {\mathrm {I} \sqrt {3}}{2}\right )\cdot P_{2}\left (x \right )\right )\bigg | {\mstack {}{_{x \hiderel {=}\frac {1}{2}-\frac {\mathrm {I} \sqrt {3}}{2}}}}=0 \\ {} & \circ & \left (x -\frac {1}{2}+\frac {\mathrm {I} \sqrt {3}}{2}\right )^{2}\cdot P_{3}\left (x \right )\textrm {is analytic at}\hspace {3pt} x =\frac {1}{2}-\frac {\mathrm {I} \sqrt {3}}{2} \\ {} & {} & \left (\left (x -\frac {1}{2}+\frac {\mathrm {I} \sqrt {3}}{2}\right )^{2}\cdot P_{3}\left (x \right )\right )\bigg | {\mstack {}{_{x \hiderel {=}\frac {1}{2}-\frac {\mathrm {I} \sqrt {3}}{2}}}}=0 \\ {} & \circ & \left (x -\frac {1}{2}+\frac {\mathrm {I} \sqrt {3}}{2}\right )^{3}\cdot P_{4}\left (x \right )\textrm {is analytic at}\hspace {3pt} x =\frac {1}{2}-\frac {\mathrm {I} \sqrt {3}}{2} \\ {} & {} & \left (\left (x -\frac {1}{2}+\frac {\mathrm {I} \sqrt {3}}{2}\right )^{3}\cdot P_{4}\left (x \right )\right )\bigg | {\mstack {}{_{x \hiderel {=}\frac {1}{2}-\frac {\mathrm {I} \sqrt {3}}{2}}}}=0 \\ {} & \circ & x =\frac {1}{2}-\frac {\mathrm {I} \sqrt {3}}{2}\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}=\frac {1}{2}-\frac {\mathrm {I} \sqrt {3}}{2} \\ \bullet & {} & \textrm {Change variables using}\hspace {3pt} x =u +\frac {1}{2}-\frac {\mathrm {I} \sqrt {3}}{2}\hspace {3pt}\textrm {so that the regular singular point is at}\hspace {3pt} u =0 \\ {} & {} & \left (u^{3}+\frac {3 u^{2}}{2}-\frac {3 \,\mathrm {I} u^{2} \sqrt {3}}{2}-\frac {3 u}{2}-\frac {3 \,\mathrm {I} u \sqrt {3}}{2}\right ) \left (\frac {d}{d u}\frac {d^{2}}{d u^{2}}y \left (u \right )\right )+\left (9 u^{2}+9 u -9 \,\mathrm {I} u \sqrt {3}-\frac {9}{2}-\frac {9 \,\mathrm {I} \sqrt {3}}{2}\right ) \left (\frac {d}{d u}\frac {d}{d u}y \left (u \right )\right )+\left (18 u +9-9 \,\mathrm {I} \sqrt {3}\right ) \left (\frac {d}{d u}y \left (u \right )\right )+6 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} \\ {} & {} & -\frac {3 \left (1+\mathrm {I} \sqrt {3}\right ) \left (r +1\right ) \left (r -1\right ) r a_{0} u^{-2+r}}{2}+\left (-\frac {3 \left (1+\mathrm {I} \sqrt {3}\right ) \left (r +2\right ) r \left (r +1\right ) a_{1}}{2}-\frac {3 \left (\mathrm {I} \sqrt {3}-1\right ) \left (r +2\right ) r \left (r +1\right ) a_{0}}{2}\right ) u^{r -1}+\left (\moverset {\infty }{\munderset {k =0}{\sum }}\left (-\frac {3 \left (1+\mathrm {I} \sqrt {3}\right ) \left (r +3+k \right ) \left (r +1+k \right ) \left (r +2+k \right ) a_{k +2}}{2}-\frac {3 \left (\mathrm {I} \sqrt {3}-1\right ) \left (r +3+k \right ) \left (r +1+k \right ) \left (r +2+k \right ) a_{k +1}}{2}+a_{k} \left (r +1+k \right ) \left (r +3+k \right ) \left (r +2+k \right )\right ) u^{k +r}\right )=0 \\ \bullet & {} & a_{0}\textrm {cannot be 0 by assumption, giving the indicial equation}\hspace {3pt} \\ {} & {} & -\frac {3 \left (1+\mathrm {I} \sqrt {3}\right ) \left (r +1\right ) \left (r -1\right ) r}{2}=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} \\ {} & {} & \frac {3 \left (r +2+k \right ) \left (r +1+k \right ) \left (r +3+k \right ) \left (-\mathrm {I} \left (a_{k +1}+a_{k +2}\right ) \sqrt {3}+\frac {2 a_{k}}{3}+a_{k +1}-a_{k +2}\right )}{2}=0 \\ \bullet & {} & \textrm {Recursion relation that defines series solution to ODE}\hspace {3pt} \\ {} & {} & a_{k +2}=-\frac {3 \,\mathrm {I} \sqrt {3}\, a_{k +1}-2 a_{k}-3 a_{k +1}}{3 \left (1+\mathrm {I} \sqrt {3}\right )} \\ \bullet & {} & \textrm {Recursion relation for}\hspace {3pt} r =-1 \\ {} & {} & a_{k +2}=-\frac {3 \,\mathrm {I} \sqrt {3}\, a_{k +1}-2 a_{k}-3 a_{k +1}}{3 \left (1+\mathrm {I} \sqrt {3}\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 {3 \,\mathrm {I} \sqrt {3}\, a_{k +1}-2 a_{k}-3 a_{k +1}}{3 \left (1+\mathrm {I} \sqrt {3}\right )}, 0=0\right ] \\ \bullet & {} & \textrm {Revert the change of variables}\hspace {3pt} u =x -\frac {1}{2}+\frac {\mathrm {I} \sqrt {3}}{2} \\ {} & {} & \left [y=\moverset {\infty }{\munderset {k =0}{\sum }}a_{k} \left (x -\frac {1}{2}+\frac {\mathrm {I} \sqrt {3}}{2}\right )^{k -1}, a_{k +2}=-\frac {3 \,\mathrm {I} \sqrt {3}\, a_{k +1}-2 a_{k}-3 a_{k +1}}{3 \left (1+\mathrm {I} \sqrt {3}\right )}, 0=0\right ] \\ \bullet & {} & \textrm {Recursion relation for}\hspace {3pt} r =0 \\ {} & {} & a_{k +2}=-\frac {3 \,\mathrm {I} \sqrt {3}\, a_{k +1}-2 a_{k}-3 a_{k +1}}{3 \left (1+\mathrm {I} \sqrt {3}\right )} \\ \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 {3 \,\mathrm {I} \sqrt {3}\, a_{k +1}-2 a_{k}-3 a_{k +1}}{3 \left (1+\mathrm {I} \sqrt {3}\right )}, 0=0\right ] \\ \bullet & {} & \textrm {Revert the change of variables}\hspace {3pt} u =x -\frac {1}{2}+\frac {\mathrm {I} \sqrt {3}}{2} \\ {} & {} & \left [y=\moverset {\infty }{\munderset {k =0}{\sum }}a_{k} \left (x -\frac {1}{2}+\frac {\mathrm {I} \sqrt {3}}{2}\right )^{k}, a_{k +2}=-\frac {3 \,\mathrm {I} \sqrt {3}\, a_{k +1}-2 a_{k}-3 a_{k +1}}{3 \left (1+\mathrm {I} \sqrt {3}\right )}, 0=0\right ] \\ \bullet & {} & \textrm {Recursion relation for}\hspace {3pt} r =1 \\ {} & {} & a_{k +2}=-\frac {3 \,\mathrm {I} \sqrt {3}\, a_{k +1}-2 a_{k}-3 a_{k +1}}{3 \left (1+\mathrm {I} \sqrt {3}\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 {3 \,\mathrm {I} \sqrt {3}\, a_{k +1}-2 a_{k}-3 a_{k +1}}{3 \left (1+\mathrm {I} \sqrt {3}\right )}, -9 \left (1+\mathrm {I} \sqrt {3}\right ) a_{1}-9 \left (\mathrm {I} \sqrt {3}-1\right ) a_{0}=0\right ] \\ \bullet & {} & \textrm {Revert the change of variables}\hspace {3pt} u =x -\frac {1}{2}+\frac {\mathrm {I} \sqrt {3}}{2} \\ {} & {} & \left [y=\moverset {\infty }{\munderset {k =0}{\sum }}a_{k} \left (x -\frac {1}{2}+\frac {\mathrm {I} \sqrt {3}}{2}\right )^{k +1}, a_{k +2}=-\frac {3 \,\mathrm {I} \sqrt {3}\, a_{k +1}-2 a_{k}-3 a_{k +1}}{3 \left (1+\mathrm {I} \sqrt {3}\right )}, -9 \left (1+\mathrm {I} \sqrt {3}\right ) a_{1}-9 \left (\mathrm {I} \sqrt {3}-1\right ) 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}+\frac {\mathrm {I} \sqrt {3}}{2}\right )^{k -1}\right )+\left (\moverset {\infty }{\munderset {k =0}{\sum }}b_{k} \left (x -\frac {1}{2}+\frac {\mathrm {I} \sqrt {3}}{2}\right )^{k}\right )+\left (\moverset {\infty }{\munderset {k =0}{\sum }}c_{k} \left (x -\frac {1}{2}+\frac {\mathrm {I} \sqrt {3}}{2}\right )^{k +1}\right ), a_{k +2}=-\frac {3 \,\mathrm {I} \sqrt {3}\, a_{k +1}-2 a_{k}-3 a_{k +1}}{3 \left (1+\mathrm {I} \sqrt {3}\right )}, 0=0, b_{k +2}=-\frac {3 \,\mathrm {I} \sqrt {3}\, b_{k +1}-2 b_{k}-3 b_{k +1}}{3 \left (1+\mathrm {I} \sqrt {3}\right )}, 0=0, c_{k +2}=-\frac {3 \,\mathrm {I} \sqrt {3}\, c_{k +1}-2 c_{k}-3 c_{k +1}}{3 \left (1+\mathrm {I} \sqrt {3}\right )}, -9 \left (1+\mathrm {I} \sqrt {3}\right ) c_{1}-9 \left (\mathrm {I} \sqrt {3}-1\right ) c_{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 <- high order exact linear fully integrable successful`
✓ Solution by Maple
Time used: 0.0 (sec). Leaf size: 30
dsolve((1+x^3)*diff(y(x),x$3)+9*x^2*diff(y(x),x$2)+18*x*diff(y(x),x)+6*y(x)=0,y(x), singsol=all)
\[ y \left (x \right ) = \frac {c_{1} x^{2}+c_{2} x +c_{3}}{\left (1+x \right ) \left (x^{2}-x +1\right )} \]
✓ Solution by Mathematica
Time used: 0.015 (sec). Leaf size: 31
DSolve[(1+x^3)*y'''[x]+9*x^2*y''[x]+18*x*y'[x]+6*y[x]==0,y[x],x,IncludeSingularSolutions -> True]
\[ y(x)\to \frac {c_3 x^2+2 c_2 x+2 c_1}{2 x^3+2} \]