Internal problem ID [9838]
Internal file name [OUTPUT/8783_Monday_June_06_2022_05_27_25_AM_8436101/index.tex]
Book: Differential Gleichungen, E. Kamke, 3rd ed. Chelsea Pub. NY, 1948
Section: Chapter 3, linear third order
Problem number: 1514.
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.
\[ \boxed {x^{3} y^{\prime \prime \prime }+6 x^{2} y^{\prime \prime }+\left (a \,x^{3}-12\right ) y=0} \] 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 )+6 x^{2} \left (\frac {d}{d x}y^{\prime }\right )+\left (a \,x^{3}-12\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 {6}{x}, P_{3}\left (x \right )=0, P_{4}\left (x \right )=\frac {a \,x^{3}-12}{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}}}=6 \\ {} & \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}}}=0 \\ {} & \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}}}=-12 \\ {} & \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..3 \\ {} & {} & 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^{2}\cdot \left (\frac {d}{d x}y^{\prime }\right )\hspace {3pt}\textrm {to series expansion}\hspace {3pt} \\ {} & {} & x^{2}\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} \\ {} & \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 (-2+r \right ) \left (r +3\right ) \left (r +2\right ) x^{r}+a_{1} \left (-1+r \right ) \left (r +4\right ) \left (r +3\right ) x^{1+r}+a_{2} r \left (r +5\right ) \left (r +4\right ) x^{r +2}+\left (\moverset {\infty }{\munderset {k =3}{\sum }}\left (a_{k} \left (k +r -2\right ) \left (r +3+k \right ) \left (r +2+k \right )+a_{k -3} a \right ) x^{k +r}\right )=0 \\ \bullet & {} & a_{0}\textrm {cannot be 0 by assumption, giving the indicial equation}\hspace {3pt} \\ {} & {} & \left (-2+r \right ) \left (r +3\right ) \left (r +2\right )=0 \\ \bullet & {} & \textrm {Values of r that satisfy the indicial equation}\hspace {3pt} \\ {} & {} & r \in \left \{-3, -2, 2\right \} \\ \bullet & {} & \textrm {The coefficients of each power of}\hspace {3pt} x \hspace {3pt}\textrm {must be 0}\hspace {3pt} \\ {} & {} & \left [a_{1} \left (-1+r \right ) \left (r +4\right ) \left (r +3\right )=0, a_{2} r \left (r +5\right ) \left (r +4\right )=0\right ] \\ \bullet & {} & \textrm {Solve for the dependent coefficient(s)}\hspace {3pt} \\ {} & {} & \left \{a_{1}=0, a_{2}=0\right \} \\ \bullet & {} & \textrm {Each term in the series must be 0, giving the recursion relation}\hspace {3pt} \\ {} & {} & a_{k} \left (k +r -2\right ) \left (r +3+k \right ) \left (r +2+k \right )+a_{k -3} a =0 \\ \bullet & {} & \textrm {Shift index using}\hspace {3pt} k \mathrm {->}k +3 \\ {} & {} & a_{k +3} \left (k +1+r \right ) \left (r +6+k \right ) \left (r +5+k \right )+a_{k} a =0 \\ \bullet & {} & \textrm {Recursion relation that defines series solution to ODE}\hspace {3pt} \\ {} & {} & a_{k +3}=-\frac {a_{k} a}{\left (k +1+r \right ) \left (r +6+k \right ) \left (r +5+k \right )} \\ \bullet & {} & \textrm {Recursion relation for}\hspace {3pt} r =-3 \\ {} & {} & a_{k +3}=-\frac {a_{k} a}{\left (k -2\right ) \left (k +3\right ) \left (2+k \right )} \\ \bullet & {} & \textrm {Solution for}\hspace {3pt} r =-3 \\ {} & {} & \left [y=\moverset {\infty }{\munderset {k =0}{\sum }}a_{k} x^{k -3}, a_{k +3}=-\frac {a_{k} a}{\left (k -2\right ) \left (k +3\right ) \left (2+k \right )}, a_{1}=0, a_{2}=0\right ] \\ \bullet & {} & \textrm {Recursion relation for}\hspace {3pt} r =-2 \\ {} & {} & a_{k +3}=-\frac {a_{k} a}{\left (k -1\right ) \left (k +4\right ) \left (k +3\right )} \\ \bullet & {} & \textrm {Solution for}\hspace {3pt} r =-2 \\ {} & {} & \left [y=\moverset {\infty }{\munderset {k =0}{\sum }}a_{k} x^{k -2}, a_{k +3}=-\frac {a_{k} a}{\left (k -1\right ) \left (k +4\right ) \left (k +3\right )}, a_{1}=0, a_{2}=0\right ] \\ \bullet & {} & \textrm {Recursion relation for}\hspace {3pt} r =2 \\ {} & {} & a_{k +3}=-\frac {a_{k} a}{\left (k +3\right ) \left (8+k \right ) \left (7+k \right )} \\ \bullet & {} & \textrm {Solution for}\hspace {3pt} r =2 \\ {} & {} & \left [y=\moverset {\infty }{\munderset {k =0}{\sum }}a_{k} x^{2+k}, a_{k +3}=-\frac {a_{k} a}{\left (k +3\right ) \left (8+k \right ) \left (7+k \right )}, a_{1}=0, a_{2}=0\right ] \\ \bullet & {} & \textrm {Combine solutions and rename parameters}\hspace {3pt} \\ {} & {} & \left [y=\left (\moverset {\infty }{\munderset {k =0}{\sum }}b_{k} x^{k -3}\right )+\left (\moverset {\infty }{\munderset {k =0}{\sum }}c_{k} x^{k -2}\right )+\left (\moverset {\infty }{\munderset {k =0}{\sum }}d_{k} x^{k +2}\right ), b_{k +3}=-\frac {b_{k} a}{\left (k -2\right ) \left (k +3\right ) \left (k +2\right )}, b_{1}=0, b_{2}=0, c_{k +3}=-\frac {c_{k} a}{\left (k -1\right ) \left (k +4\right ) \left (k +3\right )}, c_{1}=0, c_{2}=0, d_{k +3}=-\frac {d_{k} a}{\left (k +3\right ) \left (8+k \right ) \left (7+k \right )}, d_{1}=0, d_{2}=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`
✓ Solution by Maple
Time used: 0.031 (sec). Leaf size: 134
dsolve(x^3*diff(diff(diff(y(x),x),x),x)+6*x^2*diff(diff(y(x),x),x)+(a*x^3-12)*y(x)=0,y(x), singsol=all)
\[ y \left (x \right ) = \frac {-c_{2} \left (\left (-i-\sqrt {3}\right ) \left (-a^{4}\right )^{\frac {2}{3}}+i x \,a^{3}\right ) {\mathrm e}^{-\frac {\left (-a^{4}\right )^{\frac {1}{3}} \left (1+i \sqrt {3}\right ) x}{2 a}}+c_{3} \left (\left (-i+\sqrt {3}\right ) \left (-a^{4}\right )^{\frac {2}{3}}+i x \,a^{3}\right ) {\mathrm e}^{\frac {\left (-a^{4}\right )^{\frac {1}{3}} \left (i \sqrt {3}-1\right ) x}{2 a}}+c_{1} \left (x \,a^{3}+2 \left (-a^{4}\right )^{\frac {2}{3}}\right ) {\mathrm e}^{\frac {\left (-a^{4}\right )^{\frac {1}{3}} x}{a}}}{x^{3}} \]
✓ Solution by Mathematica
Time used: 0.366 (sec). Leaf size: 97
DSolve[(-12 + a*x^3)*y[x] + 6*x^2*y''[x] + x^3*Derivative[3][y][x] == 0,y[x],x,IncludeSingularSolutions -> True]
\[ y(x)\to \frac {c_1 e^{-\sqrt [3]{a} x} \left (\sqrt [3]{a} x+2\right )+c_2 e^{\sqrt [3]{-1} \sqrt [3]{a} x} \left (\sqrt [3]{a} x+2 (-1)^{2/3}\right )+c_3 e^{-(-1)^{2/3} \sqrt [3]{a} x} \left (\sqrt [3]{a} x-2 \sqrt [3]{-1}\right )}{x^3} \]