Internal problem ID [9843]
Internal file name [OUTPUT/8788_Monday_June_06_2022_05_29_15_AM_29931924/index.tex]
Book: Differential Gleichungen, E. Kamke, 3rd ed. Chelsea Pub. NY, 1948
Section: Chapter 3, linear third order
Problem number: 1519.
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.
\[ \boxed {\left (x +3\right ) x^{2} y^{\prime \prime \prime }-3 x \left (x +2\right ) y^{\prime \prime }+6 \left (x +1\right ) y^{\prime }-6 y=0} \] Unable to solve this ODE.
\[ \begin {array}{lll} & {} & \textrm {Let's solve}\hspace {3pt} \\ {} & {} & \left (x +3\right ) x^{2} \left (\frac {d}{d x}y^{\prime \prime }\right )-3 x \left (x +2\right ) \left (\frac {d}{d x}y^{\prime }\right )+6 \left (x +1\right ) 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 {3 \left (x +2\right )}{x \left (x +3\right )}, P_{3}\left (x \right )=\frac {6 \left (x +1\right )}{\left (x +3\right ) x^{2}}, P_{4}\left (x \right )=-\frac {6}{\left (x +3\right ) x^{2}}\right ] \\ {} & \circ & \left (x +3\right )\cdot P_{2}\left (x \right )\textrm {is analytic at}\hspace {3pt} x =-3 \\ {} & {} & \left (\left (x +3\right )\cdot P_{2}\left (x \right )\right )\bigg | {\mstack {}{_{x \hiderel {=}-3}}}=-1 \\ {} & \circ & \left (x +3\right )^{2}\cdot P_{3}\left (x \right )\textrm {is analytic at}\hspace {3pt} x =-3 \\ {} & {} & \left (\left (x +3\right )^{2}\cdot P_{3}\left (x \right )\right )\bigg | {\mstack {}{_{x \hiderel {=}-3}}}=0 \\ {} & \circ & \left (x +3\right )^{3}\cdot P_{4}\left (x \right )\textrm {is analytic at}\hspace {3pt} x =-3 \\ {} & {} & \left (\left (x +3\right )^{3}\cdot P_{4}\left (x \right )\right )\bigg | {\mstack {}{_{x \hiderel {=}-3}}}=0 \\ {} & \circ & x =-3\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}=-3 \\ \bullet & {} & \textrm {Multiply by denominators}\hspace {3pt} \\ {} & {} & -6 y+\left (6 x +6\right ) y^{\prime }+\left (x +3\right ) x^{2} \left (\frac {d}{d x}y^{\prime \prime }\right )-3 x \left (x +2\right ) \left (\frac {d}{d x}y^{\prime }\right )=0 \\ \bullet & {} & \textrm {Change variables using}\hspace {3pt} x =u -3\hspace {3pt}\textrm {so that the regular singular point is at}\hspace {3pt} u =0 \\ {} & {} & \left (u^{3}-6 u^{2}+9 u \right ) \left (\frac {d}{d u}\frac {d^{2}}{d u^{2}}y \left (u \right )\right )+\left (-3 u^{2}+12 u -9\right ) \left (\frac {d}{d u}\frac {d}{d u}y \left (u \right )\right )+\left (6 u -12\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} \\ {} & {} & 9 a_{0} r \left (-1+r \right ) \left (-3+r \right ) u^{-2+r}+\left (9 a_{1} \left (1+r \right ) r \left (-2+r \right )-6 a_{0} r \left (-2+r \right ) \left (-3+r \right )\right ) u^{-1+r}+\left (\moverset {\infty }{\munderset {k =0}{\sum }}\left (9 a_{k +2} \left (k +2+r \right ) \left (k +1+r \right ) \left (k +r -1\right )-6 a_{k +1} \left (k +1+r \right ) \left (k +r -1\right ) \left (k +r -2\right )+a_{k} \left (k +r -1\right ) \left (k +r -2\right ) \left (r -3+k \right )\right ) u^{k +r}\right )=0 \\ \bullet & {} & a_{0}\textrm {cannot be 0 by assumption, giving the indicial equation}\hspace {3pt} \\ {} & {} & 9 r \left (-1+r \right ) \left (-3+r \right )=0 \\ \bullet & {} & \textrm {Values of r that satisfy the indicial equation}\hspace {3pt} \\ {} & {} & r \in \left \{0, 1, 3\right \} \\ \bullet & {} & \textrm {Each term in the series must be 0, giving the recursion relation}\hspace {3pt} \\ {} & {} & \left (k +r -1\right ) \left (\left (a_{k}-6 a_{k +1}+9 a_{k +2}\right ) k^{2}+\left (2 \left (a_{k}-6 a_{k +1}+9 a_{k +2}\right ) r -5 a_{k}+6 a_{k +1}+27 a_{k +2}\right ) k +\left (a_{k}-6 a_{k +1}+9 a_{k +2}\right ) r^{2}+\left (-5 a_{k}+6 a_{k +1}+27 a_{k +2}\right ) r +6 a_{k}+12 a_{k +1}+18 a_{k +2}\right )=0 \\ \bullet & {} & \textrm {Recursion relation that defines series solution to ODE}\hspace {3pt} \\ {} & {} & a_{k +2}=-\frac {k^{2} a_{k}-6 k^{2} a_{k +1}+2 k r a_{k}-12 k r a_{k +1}+r^{2} a_{k}-6 r^{2} a_{k +1}-5 k a_{k}+6 k a_{k +1}-5 r a_{k}+6 r a_{k +1}+6 a_{k}+12 a_{k +1}}{9 \left (k^{2}+2 k r +r^{2}+3 k +3 r +2\right )} \\ \bullet & {} & \textrm {Recursion relation for}\hspace {3pt} r =0\hspace {3pt}\textrm {; series terminates at}\hspace {3pt} k =2 \\ {} & {} & a_{k +2}=-\frac {k^{2} a_{k}-6 k^{2} a_{k +1}-5 k a_{k}+6 k a_{k +1}+6 a_{k}+12 a_{k +1}}{9 \left (k^{2}+3 k +2\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 {k^{2} a_{k}-6 k^{2} a_{k +1}-5 k a_{k}+6 k a_{k +1}+6 a_{k}+12 a_{k +1}}{9 \left (k^{2}+3 k +2\right )}, 0=0\right ] \\ \bullet & {} & \textrm {Revert the change of variables}\hspace {3pt} u =x +3 \\ {} & {} & \left [y=\moverset {\infty }{\munderset {k =0}{\sum }}a_{k} \left (x +3\right )^{k}, a_{k +2}=-\frac {k^{2} a_{k}-6 k^{2} a_{k +1}-5 k a_{k}+6 k a_{k +1}+6 a_{k}+12 a_{k +1}}{9 \left (k^{2}+3 k +2\right )}, 0=0\right ] \\ \bullet & {} & \textrm {Recursion relation for}\hspace {3pt} r =1\hspace {3pt}\textrm {; series terminates at}\hspace {3pt} k =1 \\ {} & {} & a_{k +2}=-\frac {k^{2} a_{k}-6 k^{2} a_{k +1}-3 k a_{k}-6 k a_{k +1}+2 a_{k}+12 a_{k +1}}{9 \left (k^{2}+5 k +6\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 {k^{2} a_{k}-6 k^{2} a_{k +1}-3 k a_{k}-6 k a_{k +1}+2 a_{k}+12 a_{k +1}}{9 \left (k^{2}+5 k +6\right )}, -18 a_{1}-12 a_{0}=0\right ] \\ \bullet & {} & \textrm {Revert the change of variables}\hspace {3pt} u =x +3 \\ {} & {} & \left [y=\moverset {\infty }{\munderset {k =0}{\sum }}a_{k} \left (x +3\right )^{k +1}, a_{k +2}=-\frac {k^{2} a_{k}-6 k^{2} a_{k +1}-3 k a_{k}-6 k a_{k +1}+2 a_{k}+12 a_{k +1}}{9 \left (k^{2}+5 k +6\right )}, -18 a_{1}-12 a_{0}=0\right ] \\ \bullet & {} & \textrm {Recursion relation for}\hspace {3pt} r =3 \\ {} & {} & a_{k +2}=-\frac {k^{2} a_{k}-6 k^{2} a_{k +1}+k a_{k}-30 k a_{k +1}-24 a_{k +1}}{9 \left (k^{2}+9 k +20\right )} \\ \bullet & {} & \textrm {Solution for}\hspace {3pt} r =3 \\ {} & {} & \left [y \left (u \right )=\moverset {\infty }{\munderset {k =0}{\sum }}a_{k} u^{k +3}, a_{k +2}=-\frac {k^{2} a_{k}-6 k^{2} a_{k +1}+k a_{k}-30 k a_{k +1}-24 a_{k +1}}{9 \left (k^{2}+9 k +20\right )}, 108 a_{1}=0\right ] \\ \bullet & {} & \textrm {Revert the change of variables}\hspace {3pt} u =x +3 \\ {} & {} & \left [y=\moverset {\infty }{\munderset {k =0}{\sum }}a_{k} \left (x +3\right )^{k +3}, a_{k +2}=-\frac {k^{2} a_{k}-6 k^{2} a_{k +1}+k a_{k}-30 k a_{k +1}-24 a_{k +1}}{9 \left (k^{2}+9 k +20\right )}, 108 a_{1}=0\right ] \\ \bullet & {} & \textrm {Combine solutions and rename parameters}\hspace {3pt} \\ {} & {} & \left [y=\left (\moverset {\infty }{\munderset {k =0}{\sum }}a_{k} \left (x +3\right )^{k}\right )+\left (\moverset {\infty }{\munderset {k =0}{\sum }}b_{k} \left (x +3\right )^{k +1}\right )+\left (\moverset {\infty }{\munderset {k =0}{\sum }}c_{k} \left (x +3\right )^{k +3}\right ), a_{k +2}=-\frac {k^{2} a_{k}-6 k^{2} a_{k +1}-5 k a_{k}+6 k a_{k +1}+6 a_{k}+12 a_{k +1}}{9 \left (k^{2}+3 k +2\right )}, 0=0, b_{k +2}=-\frac {k^{2} b_{k}-6 k^{2} b_{k +1}-3 k b_{k}-6 k b_{k +1}+2 b_{k}+12 b_{k +1}}{9 \left (k^{2}+5 k +6\right )}, -18 b_{1}-12 b_{0}=0, c_{k +2}=-\frac {k^{2} c_{k}-6 k^{2} c_{k +1}+k c_{k}-30 k c_{k +1}-24 c_{k +1}}{9 \left (k^{2}+9 k +20\right )}, 108 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 Equation is the LCLM of -1/(x+1)*y(x)+diff(y(x),x), -2/x*y(x)+diff(y(x),x), -3/x*y(x)+diff(y(x),x) trying differential order: 1; missing the dependent variable checking if the LODE is of Euler type <- LODE of Euler type successful Euler equation successful trying differential order: 1; missing the dependent variable checking if the LODE is of Euler type <- LODE of Euler type successful Euler equation successful trying differential order: 1; missing the dependent variable checking if the LODE is of Euler type <- LODE of Euler type successful Euler equation successful <- solving the LCLM ode successful `
✓ Solution by Maple
Time used: 0.015 (sec). Leaf size: 19
dsolve((x+3)*x^2*diff(diff(diff(y(x),x),x),x)-3*x*(x+2)*diff(diff(y(x),x),x)+6*(x+1)*diff(y(x),x)-6*y(x)=0,y(x), singsol=all)
\[ y \left (x \right ) = c_{2} x^{3}+c_{1} x^{2}+c_{3} x +c_{3} \]
✓ Solution by Mathematica
Time used: 0.024 (sec). Leaf size: 58
DSolve[-6*y[x] + 6*(1 + x)*y'[x] - 3*x*(2 + x)*y''[x] + x^2*(3 + x)*Derivative[3][y][x] == 0,y[x],x,IncludeSingularSolutions -> True]
\[ y(x)\to \frac {1}{8} \left (2 c_1 \left (x^3-3 x^2+3 x+3\right )-(x-1) \left (4 c_2 \left (x^2-2 x-1\right )+c_3 \left (-3 x^2+2 x+1\right )\right )\right ) \]