Internal problem ID [9845]
Internal file name [OUTPUT/8790_Monday_June_06_2022_05_29_35_AM_13190768/index.tex]
Book: Differential Gleichungen, E. Kamke, 3rd ed. Chelsea Pub. NY, 1948
Section: Chapter 3, linear third order
Problem number: 1521.
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.
\[ \boxed {\left (x +1\right ) x^{3} y^{\prime \prime \prime }-\left (4 x +2\right ) x^{2} y^{\prime \prime }+\left (10 x +4\right ) x y^{\prime }-4 \left (3 x +1\right ) y=0} \] Unable to solve this ODE.
\[ \begin {array}{lll} & {} & \textrm {Let's solve}\hspace {3pt} \\ {} & {} & \left (x +1\right ) x^{3} \left (\frac {d}{d x}y^{\prime \prime }\right )-\left (4 x +2\right ) x^{2} \left (\frac {d}{d x}y^{\prime }\right )+\left (10 x +4\right ) x y^{\prime }-4 \left (3 x +1\right ) 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 {2 \left (2 x +1\right )}{\left (x +1\right ) x}, P_{3}\left (x \right )=\frac {2 \left (5 x +2\right )}{\left (x +1\right ) x^{2}}, P_{4}\left (x \right )=-\frac {4 \left (3 x +1\right )}{\left (x +1\right ) x^{3}}\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}}}=-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} \\ {} & {} & \left (x +1\right ) x^{3} \left (\frac {d}{d x}y^{\prime \prime }\right )-2 \left (2 x +1\right ) x^{2} \left (\frac {d}{d x}y^{\prime }\right )+2 \left (5 x +2\right ) x y^{\prime }+\left (-12 x -4\right ) 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^{4}-3 u^{3}+3 u^{2}-u \right ) \left (\frac {d}{d u}\frac {d^{2}}{d u^{2}}y \left (u \right )\right )+\left (-4 u^{3}+10 u^{2}-8 u +2\right ) \left (\frac {d}{d u}\frac {d}{d u}y \left (u \right )\right )+\left (10 u^{2}-16 u +6\right ) \left (\frac {d}{d u}y \left (u \right )\right )+\left (-12 u +8\right ) 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 y \left (u \right )\hspace {3pt}\textrm {to series expansion for}\hspace {3pt} m =0..1 \\ {} & {} & u^{m}\cdot y \left (u \right )=\moverset {\infty }{\munderset {k =0}{\sum }}a_{k} u^{k +r +m} \\ {} & \circ & \textrm {Shift index using}\hspace {3pt} k \mathrm {->}k -m \\ {} & {} & u^{m}\cdot y \left (u \right )=\moverset {\infty }{\munderset {k =m}{\sum }}a_{k -m} u^{k +r} \\ {} & \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..2 \\ {} & {} & 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..3 \\ {} & {} & 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..4 \\ {} & {} & 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+r \right ) \left (-4+r \right ) u^{r -2}+\left (-a_{1} \left (1+r \right ) r \left (r -3\right )+a_{0} r \left (-5+3 r \right ) \left (-4+r \right )\right ) u^{-1+r}+\left (-a_{2} \left (2+r \right ) \left (1+r \right ) \left (r -2\right )+a_{1} \left (1+r \right ) \left (-2+3 r \right ) \left (r -3\right )-a_{0} \left (3 r^{3}-19 r^{2}+32 r -8\right )\right ) u^{r}+\left (\moverset {\infty }{\munderset {k =1}{\sum }}\left (-a_{k +2} \left (k +2+r \right ) \left (k +1+r \right ) \left (k +r -2\right )+a_{k +1} \left (k +1+r \right ) \left (3 k -2+3 r \right ) \left (k +r -3\right )-a_{k} \left (3 k^{3}+9 k^{2} r +9 k \,r^{2}+3 r^{3}-19 k^{2}-38 k r -19 r^{2}+32 k +32 r -8\right )+a_{k -1} \left (k +r -4\right ) \left (k +r -3\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+r \right ) \left (-4+r \right )=0 \\ \bullet & {} & \textrm {Values of r that satisfy the indicial equation}\hspace {3pt} \\ {} & {} & r \in \left \{0, 1, 4\right \} \\ \bullet & {} & \textrm {The coefficients of each power of}\hspace {3pt} u \hspace {3pt}\textrm {must be 0}\hspace {3pt} \\ {} & {} & \left [-a_{1} \left (1+r \right ) r \left (r -3\right )+a_{0} r \left (-5+3 r \right ) \left (-4+r \right )=0, -a_{2} \left (2+r \right ) \left (1+r \right ) \left (r -2\right )+a_{1} \left (1+r \right ) \left (-2+3 r \right ) \left (r -3\right )-a_{0} \left (3 r^{3}-19 r^{2}+32 r -8\right )=0\right ] \\ \bullet & {} & \textrm {Each term in the series must be 0, giving the recursion relation}\hspace {3pt} \\ {} & {} & -a_{k +2} \left (k +2+r \right ) \left (k +1+r \right ) \left (k +r -2\right )+a_{k +1} \left (k +1+r \right ) \left (3 k -2+3 r \right ) \left (k +r -3\right )-a_{k} \left (3 k^{3}+9 k^{2} r +9 k \,r^{2}+3 r^{3}-19 k^{2}-38 k r -19 r^{2}+32 k +32 r -8\right )+a_{k -1} \left (k +r -4\right ) \left (k +r -3\right )^{2}=0 \\ \bullet & {} & \textrm {Shift index using}\hspace {3pt} k \mathrm {->}k +1 \\ {} & {} & -a_{k +3} \left (k +3+r \right ) \left (k +2+r \right ) \left (k +r -1\right )+a_{k +2} \left (k +2+r \right ) \left (3 k +1+3 r \right ) \left (k +r -2\right )-a_{k +1} \left (3 \left (k +1\right )^{3}+9 \left (k +1\right )^{2} r +9 \left (k +1\right ) r^{2}+3 r^{3}-19 \left (k +1\right )^{2}-38 \left (k +1\right ) r -19 r^{2}+32 k +24+32 r \right )+a_{k} \left (k +r -3\right ) \left (k +r -2\right )^{2}=0 \\ \bullet & {} & \textrm {Recursion relation that defines series solution to ODE}\hspace {3pt} \\ {} & {} & a_{k +3}=\frac {k^{3} a_{k}-3 k^{3} a_{k +1}+3 k^{3} a_{k +2}+3 k^{2} r a_{k}-9 k^{2} r a_{k +1}+9 k^{2} r a_{k +2}+3 k \,r^{2} a_{k}-9 k \,r^{2} a_{k +1}+9 k \,r^{2} a_{k +2}+r^{3} a_{k}-3 r^{3} a_{k +1}+3 r^{3} a_{k +2}-7 k^{2} a_{k}+10 k^{2} a_{k +1}+k^{2} a_{k +2}-14 k r a_{k}+20 k r a_{k +1}+2 k r a_{k +2}-7 r^{2} a_{k}+10 r^{2} a_{k +1}+r^{2} a_{k +2}+16 k a_{k}-3 k a_{k +1}-12 k a_{k +2}+16 r a_{k}-3 r a_{k +1}-12 r a_{k +2}-12 a_{k}-8 a_{k +1}-4 a_{k +2}}{\left (k +3+r \right ) \left (k +2+r \right ) \left (k +r -1\right )} \\ \bullet & {} & \textrm {Recursion relation for}\hspace {3pt} r =0 \\ {} & {} & a_{k +3}=\frac {k^{3} a_{k}-3 k^{3} a_{k +1}+3 k^{3} a_{k +2}-7 k^{2} a_{k}+10 k^{2} a_{k +1}+k^{2} a_{k +2}+16 k a_{k}-3 k a_{k +1}-12 k a_{k +2}-12 a_{k}-8 a_{k +1}-4 a_{k +2}}{\left (k +3\right ) \left (k +2\right ) \left (k -1\right )} \\ \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 =1 \\ {} & {} & a_{k +3}=\frac {k^{3} a_{k}-3 k^{3} a_{k +1}+3 k^{3} a_{k +2}-7 k^{2} a_{k}+10 k^{2} a_{k +1}+k^{2} a_{k +2}+16 k a_{k}-3 k a_{k +1}-12 k a_{k +2}-12 a_{k}-8 a_{k +1}-4 a_{k +2}}{\left (k +3\right ) \left (k +2\right ) \left (k -1\right )} \\ \bullet & {} & \textrm {Recursion relation for}\hspace {3pt} r =1 \\ {} & {} & a_{k +3}=\frac {k^{3} a_{k}-3 k^{3} a_{k +1}+3 k^{3} a_{k +2}-4 k^{2} a_{k}+k^{2} a_{k +1}+10 k^{2} a_{k +2}+5 k a_{k}+8 k a_{k +1}-k a_{k +2}-2 a_{k}-4 a_{k +1}-12 a_{k +2}}{\left (k +4\right ) \left (k +3\right ) k} \\ \bullet & {} & \textrm {Series not valid for}\hspace {3pt} r =1\hspace {3pt}\textrm {, division by}\hspace {3pt} 0\hspace {3pt}\textrm {in the recursion relation at}\hspace {3pt} k =0 \\ {} & {} & a_{k +3}=\frac {k^{3} a_{k}-3 k^{3} a_{k +1}+3 k^{3} a_{k +2}-4 k^{2} a_{k}+k^{2} a_{k +1}+10 k^{2} a_{k +2}+5 k a_{k}+8 k a_{k +1}-k a_{k +2}-2 a_{k}-4 a_{k +1}-12 a_{k +2}}{\left (k +4\right ) \left (k +3\right ) k} \\ \bullet & {} & \textrm {Recursion relation for}\hspace {3pt} r =4 \\ {} & {} & a_{k +3}=\frac {k^{3} a_{k}-3 k^{3} a_{k +1}+3 k^{3} a_{k +2}+5 k^{2} a_{k}-26 k^{2} a_{k +1}+37 k^{2} a_{k +2}+8 k a_{k}-67 k a_{k +1}+140 k a_{k +2}+4 a_{k}-52 a_{k +1}+156 a_{k +2}}{\left (k +7\right ) \left (k +6\right ) \left (k +3\right )} \\ \bullet & {} & \textrm {Solution for}\hspace {3pt} r =4 \\ {} & {} & \left [y \left (u \right )=\moverset {\infty }{\munderset {k =0}{\sum }}a_{k} u^{k +4}, a_{k +3}=\frac {k^{3} a_{k}-3 k^{3} a_{k +1}+3 k^{3} a_{k +2}+5 k^{2} a_{k}-26 k^{2} a_{k +1}+37 k^{2} a_{k +2}+8 k a_{k}-67 k a_{k +1}+140 k a_{k +2}+4 a_{k}-52 a_{k +1}+156 a_{k +2}}{\left (k +7\right ) \left (k +6\right ) \left (k +3\right )}, -20 a_{1}=0, -60 a_{2}+50 a_{1}-8 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 +4}, a_{k +3}=\frac {k^{3} a_{k}-3 k^{3} a_{k +1}+3 k^{3} a_{k +2}+5 k^{2} a_{k}-26 k^{2} a_{k +1}+37 k^{2} a_{k +2}+8 k a_{k}-67 k a_{k +1}+140 k a_{k +2}+4 a_{k}-52 a_{k +1}+156 a_{k +2}}{\left (k +7\right ) \left (k +6\right ) \left (k +3\right )}, -20 a_{1}=0, -60 a_{2}+50 a_{1}-8 a_{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) =`, [x^2]
✓ Solution by Maple
Time used: 0.031 (sec). Leaf size: 28
dsolve((x+1)*x^3*diff(diff(diff(y(x),x),x),x)-(4*x+2)*x^2*diff(diff(y(x),x),x)+(10*x+4)*x*diff(y(x),x)-4*(3*x+1)*y(x)=0,y(x), singsol=all)
\[ y \left (x \right ) = x \left (\ln \left (x \right )^{2} c_{3} x +c_{2} x \ln \left (x \right )+x^{2} c_{3} +c_{1} x +c_{3} \right ) \]
✓ Solution by Mathematica
Time used: 0.054 (sec). Leaf size: 29
DSolve[-4*(1 + 3*x)*y[x] + x*(4 + 10*x)*y'[x] - x^2*(2 + 4*x)*y''[x] + x^3*(1 + x)*Derivative[3][y][x] == 0,y[x],x,IncludeSingularSolutions -> True]
\[ y(x)\to x^2 \left (c_3 \left (x+\frac {1}{x}+\log ^2(x)\right )+c_2 \log (x)+c_1\right ) \]