Internal problem ID [9840]
Internal file name [OUTPUT/8785_Monday_June_06_2022_05_27_43_AM_29828229/index.tex]
Book: Differential Gleichungen, E. Kamke, 3rd ed. Chelsea Pub. NY, 1948
Section: Chapter 3, linear third order
Problem number: 1516.
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.
\[ \boxed {x^{3} y^{\prime \prime \prime }+\left (x +3\right ) x^{2} y^{\prime \prime }+5 \left (x -6\right ) x y^{\prime }+\left (4 x +30\right ) y=0} \] Unable to solve this ODE.
\[ \begin {array}{lll} & {} & \textrm {Let's solve}\hspace {3pt} \\ {} & {} & x^{3} \left (\frac {d}{d x}y^{\prime \prime }\right )+\left (x +3\right ) x^{2} \left (\frac {d}{d x}y^{\prime }\right )+5 \left (x -6\right ) x y^{\prime }+\left (4 x +30\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}=0\hspace {3pt}\textrm {is a regular singular point}\hspace {3pt} \\ {} & \circ & \textrm {Define functions}\hspace {3pt} \\ {} & {} & \left [P_{2}\left (x \right )=\frac {x +3}{x}, P_{3}\left (x \right )=\frac {5 \left (x -6\right )}{x^{2}}, P_{4}\left (x \right )=\frac {2 \left (2 x +15\right )}{x^{3}}\right ] \\ {} & \circ & x \cdot P_{2}\left (x \right )\textrm {is analytic at}\hspace {3pt} x =0 \\ {} & {} & \left (x \cdot P_{2}\left (x \right )\right )\bigg | {\mstack {}{_{x \hiderel {=}0}}}=3 \\ {} & \circ & x^{2}\cdot P_{3}\left (x \right )\textrm {is analytic at}\hspace {3pt} x =0 \\ {} & {} & \left (x^{2}\cdot P_{3}\left (x \right )\right )\bigg | {\mstack {}{_{x \hiderel {=}0}}}=-30 \\ {} & \circ & x^{3}\cdot P_{4}\left (x \right )\textrm {is analytic at}\hspace {3pt} x =0 \\ {} & {} & \left (x^{3}\cdot P_{4}\left (x \right )\right )\bigg | {\mstack {}{_{x \hiderel {=}0}}}=30 \\ {} & \circ & x =0\textrm {is a regular singular point}\hspace {3pt} \\ & {} & \textrm {Check to see if}\hspace {3pt} x_{0}=0\hspace {3pt}\textrm {is a regular singular point}\hspace {3pt} \\ {} & {} & x_{0}=0 \\ \bullet & {} & \textrm {Assume series solution for}\hspace {3pt} y \\ {} & {} & y=\moverset {\infty }{\munderset {k =0}{\sum }}a_{k} x^{k +r} \\ \square & {} & \textrm {Rewrite ODE with series expansions}\hspace {3pt} \\ {} & \circ & \textrm {Convert}\hspace {3pt} x^{m}\cdot y\hspace {3pt}\textrm {to series expansion for}\hspace {3pt} m =0..1 \\ {} & {} & x^{m}\cdot y=\moverset {\infty }{\munderset {k =0}{\sum }}a_{k} x^{k +r +m} \\ {} & \circ & \textrm {Shift index using}\hspace {3pt} k \mathrm {->}k -m \\ {} & {} & x^{m}\cdot y=\moverset {\infty }{\munderset {k =m}{\sum }}a_{k -m} x^{k +r} \\ {} & \circ & \textrm {Convert}\hspace {3pt} x^{m}\cdot y^{\prime }\hspace {3pt}\textrm {to series expansion for}\hspace {3pt} m =1..2 \\ {} & {} & x^{m}\cdot y^{\prime }=\moverset {\infty }{\munderset {k =0}{\sum }}a_{k} \left (k +r \right ) x^{k +r -1+m} \\ {} & \circ & \textrm {Shift index using}\hspace {3pt} k \mathrm {->}k +1-m \\ {} & {} & x^{m}\cdot y^{\prime }=\moverset {\infty }{\munderset {k =-1+m}{\sum }}a_{k +1-m} \left (k +1-m +r \right ) x^{k +r} \\ {} & \circ & \textrm {Convert}\hspace {3pt} x^{m}\cdot \left (\frac {d}{d x}y^{\prime }\right )\hspace {3pt}\textrm {to series expansion for}\hspace {3pt} m =2..3 \\ {} & {} & x^{m}\cdot \left (\frac {d}{d x}y^{\prime }\right )=\moverset {\infty }{\munderset {k =0}{\sum }}a_{k} \left (k +r \right ) \left (k +r -1\right ) x^{k +r -2+m} \\ {} & \circ & \textrm {Shift index using}\hspace {3pt} k \mathrm {->}k +2-m \\ {} & {} & x^{m}\cdot \left (\frac {d}{d x}y^{\prime }\right )=\moverset {\infty }{\munderset {k =-2+m}{\sum }}a_{k +2-m} \left (k +2-m +r \right ) \left (k +1-m +r \right ) x^{k +r} \\ {} & \circ & \textrm {Convert}\hspace {3pt} x^{3}\cdot \left (\frac {d}{d x}y^{\prime \prime }\right )\hspace {3pt}\textrm {to series expansion}\hspace {3pt} \\ {} & {} & x^{3}\cdot \left (\frac {d}{d x}y^{\prime \prime }\right )=\moverset {\infty }{\munderset {k =0}{\sum }}a_{k} \left (k +r \right ) \left (k +r -1\right ) \left (k +r -2\right ) x^{k +r} \\ & {} & \textrm {Rewrite ODE with series expansions}\hspace {3pt} \\ {} & {} & a_{0} \left (r +6\right ) \left (r -5\right ) \left (-1+r \right ) x^{r}+\left (\moverset {\infty }{\munderset {k =1}{\sum }}\left (a_{k} \left (r +6+k \right ) \left (r -5+k \right ) \left (k +r -1\right )+a_{k -1} \left (k +1+r \right )^{2}\right ) x^{k +r}\right )=0 \\ \bullet & {} & a_{0}\textrm {cannot be 0 by assumption, giving the indicial equation}\hspace {3pt} \\ {} & {} & \left (r +6\right ) \left (r -5\right ) \left (-1+r \right )=0 \\ \bullet & {} & \textrm {Values of r that satisfy the indicial equation}\hspace {3pt} \\ {} & {} & r \in \left \{-6, 1, 5\right \} \\ \bullet & {} & \textrm {Each term in the series must be 0, giving the recursion relation}\hspace {3pt} \\ {} & {} & a_{k} \left (r +6+k \right ) \left (r -5+k \right ) \left (k +r -1\right )+a_{k -1} \left (k +1+r \right )^{2}=0 \\ \bullet & {} & \textrm {Shift index using}\hspace {3pt} k \mathrm {->}k +1 \\ {} & {} & a_{k +1} \left (r +7+k \right ) \left (r -4+k \right ) \left (k +r \right )+a_{k} \left (k +r +2\right )^{2}=0 \\ \bullet & {} & \textrm {Recursion relation that defines series solution to ODE}\hspace {3pt} \\ {} & {} & a_{k +1}=-\frac {a_{k} \left (k +r +2\right )^{2}}{\left (r +7+k \right ) \left (r -4+k \right ) \left (k +r \right )} \\ \bullet & {} & \textrm {Recursion relation for}\hspace {3pt} r =-6\hspace {3pt}\textrm {; series terminates at}\hspace {3pt} k =4 \\ {} & {} & a_{k +1}=-\frac {a_{k} \left (k -4\right )^{2}}{\left (k +1\right ) \left (-10+k \right ) \left (k -6\right )} \\ \bullet & {} & \textrm {Apply recursion relation for}\hspace {3pt} k =0 \\ {} & {} & a_{1}=-\frac {4 a_{0}}{15} \\ \bullet & {} & \textrm {Apply recursion relation for}\hspace {3pt} k =1 \\ {} & {} & a_{2}=-\frac {a_{1}}{10} \\ \bullet & {} & \textrm {Express in terms of}\hspace {3pt} a_{0} \\ {} & {} & a_{2}=\frac {2 a_{0}}{75} \\ \bullet & {} & \textrm {Apply recursion relation for}\hspace {3pt} k =2 \\ {} & {} & a_{3}=-\frac {a_{2}}{24} \\ \bullet & {} & \textrm {Express in terms of}\hspace {3pt} a_{0} \\ {} & {} & a_{3}=-\frac {a_{0}}{900} \\ \bullet & {} & \textrm {Apply recursion relation for}\hspace {3pt} k =3 \\ {} & {} & a_{4}=-\frac {a_{3}}{84} \\ \bullet & {} & \textrm {Express in terms of}\hspace {3pt} a_{0} \\ {} & {} & a_{4}=\frac {a_{0}}{75600} \\ \bullet & {} & \textrm {Terminating series solution of the ODE for}\hspace {3pt} r =-6\hspace {3pt}\textrm {. Use reduction of order to find the second linearly independent solution}\hspace {3pt} \\ {} & {} & y=a_{0}\cdot \left (1-\frac {4}{15} x +\frac {2}{75} x^{2}-\frac {1}{900} x^{3}+\frac {1}{75600} x^{4}\right ) \\ \bullet & {} & \textrm {Recursion relation for}\hspace {3pt} r =1 \\ {} & {} & a_{k +1}=-\frac {a_{k} \left (k +3\right )^{2}}{\left (8+k \right ) \left (-3+k \right ) \left (k +1\right )} \\ \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 =3 \\ {} & {} & a_{k +1}=-\frac {a_{k} \left (k +3\right )^{2}}{\left (8+k \right ) \left (-3+k \right ) \left (k +1\right )} \\ \bullet & {} & \textrm {Recursion relation for}\hspace {3pt} r =5 \\ {} & {} & a_{k +1}=-\frac {a_{k} \left (k +7\right )^{2}}{\left (12+k \right ) \left (k +1\right ) \left (k +5\right )} \\ \bullet & {} & \textrm {Solution for}\hspace {3pt} r =5 \\ {} & {} & \left [y=\moverset {\infty }{\munderset {k =0}{\sum }}a_{k} x^{k +5}, a_{k +1}=-\frac {a_{k} \left (k +7\right )^{2}}{\left (12+k \right ) \left (k +1\right ) \left (k +5\right )}\right ] \\ \bullet & {} & \textrm {Combine solutions and rename parameters}\hspace {3pt} \\ {} & {} & \left [y=a_{0}\cdot \left (1-\frac {4}{15} x +\frac {2}{75} x^{2}-\frac {1}{900} x^{3}+\frac {1}{75600} x^{4}\right )+\left (\moverset {\infty }{\munderset {k =0}{\sum }}b_{k} x^{k +5}\right ), b_{k +1}=-\frac {b_{k} \left (7+k \right )^{2}}{\left (12+k \right ) \left (k +1\right ) \left (k +5\right )}\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^4-84*x^3+2016*x^2-20160*x+75600)/x^6, exp(-x)*(x^8+28*x^7+450*x^6+5100*x^5+
✓ Solution by Maple
Time used: 0.047 (sec). Leaf size: 188
dsolve(x^3*diff(diff(diff(y(x),x),x),x)+(x+3)*x^2*diff(diff(y(x),x),x)+5*(x-6)*x*diff(y(x),x)+(4*x+30)*y(x)=0,y(x), singsol=all)
\[ y \left (x \right ) = \frac {c_{3} {\mathrm e}^{-x} \left (x^{8}+28 x^{7}+450 x^{6}+5100 x^{5}+42900 x^{4}+267120 x^{3}+1179360 x^{2}+3326400 x +4536000\right ) \operatorname {expIntegral}_{1}\left (-x \right )+c_{2} {\mathrm e}^{-x} \left (x^{8}+28 x^{7}+450 x^{6}+5100 x^{5}+42900 x^{4}+267120 x^{3}+1179360 x^{2}+3326400 x +4536000\right )+60 c_{3} \left (x^{4}-84 x^{3}+2016 x^{2}-20160 x +75600\right ) \ln \left (x \right )+c_{3} x^{7}+29 c_{3} x^{6}+480 x^{5} c_{3} +\left (c_{1} +5612 c_{3} \right ) x^{4}+\left (-84 c_{1} +40152 c_{3} \right ) x^{3}+\left (2016 c_{1} +654192 c_{3} \right ) x^{2}+\left (-20160 c_{1} -2761920 c_{3} \right ) x +75600 c_{1} +27367200 c_{3}}{x^{6}} \]
✗ Solution by Mathematica
Time used: 0.0 (sec). Leaf size: 0
DSolve[(30 + 4*x)*y[x] + 5*(-6 + x)*x*y'[x] + x^2*(3 + x)*y''[x] + x^3*Derivative[3][y][x] == 0,y[x],x,IncludeSingularSolutions -> True]
Timed out