Internal problem ID [5437]
Internal file name [OUTPUT/4928_Tuesday_February_06_2024_10_14_22_PM_25156565/index.tex]
Book: Schaums Outline. Theory and problems of Differential Equations, 1st edition. Frank Ayres.
McGraw Hill 1952
Section: Chapter 19. Linear equations with variable coefficients (Misc. types). Supplemetary
problems. Page 132
Problem number: 30.
ODE order: 3.
ODE degree: 1.
The type(s) of ODE detected by this program : "higher_order_missing_y"
Maple gives the following as the ode type
[[_3rd_order, _missing_y]]
\[ \boxed {\left (2 x^{3}-1\right ) y^{\prime \prime \prime }-6 x^{2} y^{\prime \prime }+6 y^{\prime } x=0} \] Since \(y\) is missing from the ode then we can use the substitution \(y^{\prime } = v \left (x \right )\) to reduce the order by one. The ODE becomes \begin {align*} \left (2 x^{3}-1\right ) v^{\prime \prime }\left (x \right )-6 x^{2} v^{\prime }\left (x \right )+6 x v \left (x \right ) = 0 \end {align*}
In normal form the ode \begin {align*} \left (2 x^{3}-1\right ) v^{\prime \prime }\left (x \right )-6 x^{2} v^{\prime }\left (x \right )+6 x v \left (x \right )&=0 \tag {1} \end {align*}
Becomes \begin {align*} v^{\prime \prime }\left (x \right )+p \left (x \right ) v^{\prime }\left (x \right )+q \left (x \right ) v \left (x \right )&=0 \tag {2} \end {align*}
Where \begin {align*} p \left (x \right )&=-\frac {6 x^{2}}{2 x^{3}-1}\\ q \left (x \right )&=\frac {6 x}{2 x^{3}-1} \end {align*}
Applying change of variables on the depndent variable \(v \left (x \right ) = v \left (x \right ) x^{n}\) to (2) gives the following ode where the dependent variables is \(v \left (x \right )\) and not \(v \left (x \right )\). \begin {align*} v^{\prime \prime }\left (x \right )+\left (\frac {2 n}{x}+p \right ) v^{\prime }\left (x \right )+\left (\frac {n \left (n -1\right )}{x^{2}}+\frac {n p}{x}+q \right ) v \left (x \right )&=0 \tag {3} \end {align*}
Let the coefficient of \(v \left (x \right )\) above be zero. Hence \begin {align*} \frac {n \left (n -1\right )}{x^{2}}+\frac {n p}{x}+q&=0 \tag {4} \end {align*}
Substituting the earlier values found for \(p \left (x \right )\) and \(q \left (x \right )\) into (4) gives \begin {align*} \frac {n \left (n -1\right )}{x^{2}}-\frac {6 n x}{2 x^{3}-1}+\frac {6 x}{2 x^{3}-1}&=0 \tag {5} \end {align*}
Solving (5) for \(n\) gives \begin {align*} n&=1 \tag {6} \end {align*}
Substituting this value in (3) gives \begin {align*} v^{\prime \prime }\left (x \right )+\left (\frac {2}{x}-\frac {6 x^{2}}{2 x^{3}-1}\right ) v^{\prime }\left (x \right )&=0 \\ v^{\prime \prime }\left (x \right )+\frac {\left (-2 x^{3}-2\right ) v^{\prime }\left (x \right )}{2 x^{4}-x}&=0 \tag {7} \\ \end {align*}
Using the substitution \begin {align*} u \left (x \right ) = v^{\prime }\left (x \right ) \end {align*}
Then (7) becomes \begin {align*} u^{\prime }\left (x \right )+\frac {\left (-2 x^{3}-2\right ) u \left (x \right )}{2 x^{4}-x} = 0 \tag {8} \\ \end {align*}
The above is now solved for \(u \left (x \right )\). In canonical form the ODE is \begin {align*} u' &= F(x,u)\\ &= f( x) g(u)\\ &= \frac {2 u \left (x^{3}+1\right )}{x \left (2 x^{3}-1\right )} \end {align*}
Where \(f(x)=\frac {2 x^{3}+2}{x \left (2 x^{3}-1\right )}\) and \(g(u)=u\). Integrating both sides gives \begin {align*} \frac {1}{u} \,du &= \frac {2 x^{3}+2}{x \left (2 x^{3}-1\right )} \,d x\\ \int { \frac {1}{u} \,du} &= \int {\frac {2 x^{3}+2}{x \left (2 x^{3}-1\right )} \,d x}\\ \ln \left (u \right )&=\ln \left (2 x^{3}-1\right )-2 \ln \left (x \right )+c_{1}\\ u&={\mathrm e}^{\ln \left (2 x^{3}-1\right )-2 \ln \left (x \right )+c_{1}}\\ &=c_{1} {\mathrm e}^{\ln \left (2 x^{3}-1\right )-2 \ln \left (x \right )} \end {align*}
Which simplifies to \[ u \left (x \right ) = c_{1} \left (2 x -\frac {1}{x^{2}}\right ) \] Now that \(u \left (x \right )\) is known, then \begin {align*} v^{\prime }\left (x \right )&= u \left (x \right )\\ v \left (x \right )&= \int u \left (x \right )d x +c_{2}\\ &= c_{1} \left (x^{2}+\frac {1}{x}\right )+c_{2} \end {align*}
Hence \begin {align*} v \left (x \right )&= v \left (x \right ) x^{n}\\ &= \left (c_{1} \left (x^{2}+\frac {1}{x}\right )+c_{2} \right ) x\\ &= c_{1} x^{3}+c_{2} x +c_{1}\\ \end {align*}
But since \(y^{\prime } = v \left (x \right )\) then we now need to solve the ode \(y^{\prime } = \left (c_{1} \left (x^{2}+\frac {1}{x}\right )+c_{2} \right ) x\). Integrating both sides gives \begin {align*} y &= \int { c_{1} x^{3}+c_{2} x +c_{1}\,\mathop {\mathrm {d}x}}\\ &= \frac {x \left (c_{1} x^{3}+2 c_{2} x +4 c_{1} \right )}{4}+c_{3} \end {align*}
Summary
The solution(s) found are the following \begin{align*} \tag{1} y &= \frac {x \left (c_{1} x^{3}+2 c_{2} x +4 c_{1} \right )}{4}+c_{3} \\ \end{align*}
Verification of solutions
\[ y = \frac {x \left (c_{1} x^{3}+2 c_{2} x +4 c_{1} \right )}{4}+c_{3} \] Verified OK.
\[ \begin {array}{lll} & {} & \textrm {Let's solve}\hspace {3pt} \\ {} & {} & \left (2 x^{3}-1\right ) y^{\prime \prime \prime }-6 x^{2} y^{\prime \prime }+6 y^{\prime } x =0 \\ \bullet & {} & \textrm {Highest derivative means the order of the ODE is}\hspace {3pt} 3 \\ {} & {} & y^{\prime \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 {6 x^{2}}{2 x^{3}-1}, P_{3}\left (x \right )=\frac {6 x}{2 x^{3}-1}, P_{4}\left (x \right )=0\right ] \\ {} & \circ & \left (x +\frac {2^{\frac {2}{3}}}{4}+\frac {\mathrm {I} \sqrt {3}\, 2^{\frac {2}{3}}}{4}\right )\cdot P_{2}\left (x \right )\textrm {is analytic at}\hspace {3pt} x =-\frac {2^{\frac {2}{3}}}{4}-\frac {\mathrm {I} \sqrt {3}\, 2^{\frac {2}{3}}}{4} \\ {} & {} & \left (\left (x +\frac {2^{\frac {2}{3}}}{4}+\frac {\mathrm {I} \sqrt {3}\, 2^{\frac {2}{3}}}{4}\right )\cdot P_{2}\left (x \right )\right )\bigg | {\mstack {}{_{x \hiderel {=}-\frac {2^{\frac {2}{3}}}{4}-\frac {\mathrm {I} \sqrt {3}\, 2^{\frac {2}{3}}}{4}}}}=0 \\ {} & \circ & \left (x +\frac {2^{\frac {2}{3}}}{4}+\frac {\mathrm {I} \sqrt {3}\, 2^{\frac {2}{3}}}{4}\right )^{2}\cdot P_{3}\left (x \right )\textrm {is analytic at}\hspace {3pt} x =-\frac {2^{\frac {2}{3}}}{4}-\frac {\mathrm {I} \sqrt {3}\, 2^{\frac {2}{3}}}{4} \\ {} & {} & \left (\left (x +\frac {2^{\frac {2}{3}}}{4}+\frac {\mathrm {I} \sqrt {3}\, 2^{\frac {2}{3}}}{4}\right )^{2}\cdot P_{3}\left (x \right )\right )\bigg | {\mstack {}{_{x \hiderel {=}-\frac {2^{\frac {2}{3}}}{4}-\frac {\mathrm {I} \sqrt {3}\, 2^{\frac {2}{3}}}{4}}}}=0 \\ {} & \circ & \left (x +\frac {2^{\frac {2}{3}}}{4}+\frac {\mathrm {I} \sqrt {3}\, 2^{\frac {2}{3}}}{4}\right )^{3}\cdot P_{4}\left (x \right )\textrm {is analytic at}\hspace {3pt} x =-\frac {2^{\frac {2}{3}}}{4}-\frac {\mathrm {I} \sqrt {3}\, 2^{\frac {2}{3}}}{4} \\ {} & {} & \left (\left (x +\frac {2^{\frac {2}{3}}}{4}+\frac {\mathrm {I} \sqrt {3}\, 2^{\frac {2}{3}}}{4}\right )^{3}\cdot P_{4}\left (x \right )\right )\bigg | {\mstack {}{_{x \hiderel {=}-\frac {2^{\frac {2}{3}}}{4}-\frac {\mathrm {I} \sqrt {3}\, 2^{\frac {2}{3}}}{4}}}}=0 \\ {} & \circ & x =-\frac {2^{\frac {2}{3}}}{4}-\frac {\mathrm {I} \sqrt {3}\, 2^{\frac {2}{3}}}{4}\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 {2^{\frac {2}{3}}}{4}-\frac {\mathrm {I} \sqrt {3}\, 2^{\frac {2}{3}}}{4} \\ \bullet & {} & \textrm {Change variables using}\hspace {3pt} x =u -\frac {2^{\frac {2}{3}}}{4}-\frac {\mathrm {I} \sqrt {3}\, 2^{\frac {2}{3}}}{4}\hspace {3pt}\textrm {so that the regular singular point is at}\hspace {3pt} u =0 \\ {} & {} & \left (2 u^{3}-\frac {3 u^{2} 2^{\frac {2}{3}}}{2}-\frac {3 \,\mathrm {I} u^{2} \sqrt {3}\, 2^{\frac {2}{3}}}{2}-\frac {3 u 2^{\frac {1}{3}}}{2}+\frac {3 \,\mathrm {I} u 2^{\frac {1}{3}} \sqrt {3}}{2}\right ) \left (\frac {d^{3}}{d u^{3}}y \left (u \right )\right )+\left (-6 u^{2}+3 u 2^{\frac {2}{3}}+3 \,\mathrm {I} u \sqrt {3}\, 2^{\frac {2}{3}}+\frac {3 \,2^{\frac {1}{3}}}{2}-\frac {3 \,\mathrm {I} \,2^{\frac {1}{3}} \sqrt {3}}{2}\right ) \left (\frac {d^{2}}{d u^{2}}y \left (u \right )\right )+\left (6 u -\frac {3 \,2^{\frac {2}{3}}}{2}-\frac {3 \,\mathrm {I} \sqrt {3}\, 2^{\frac {2}{3}}}{2}\right ) \left (\frac {d}{d u}y \left (u \right )\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^{2}}{d u^{2}}y \left (u \right )\right )\hspace {3pt}\textrm {to series expansion for}\hspace {3pt} m =0..2 \\ {} & {} & u^{m}\cdot \left (\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 ) u^{k +r -2+m} \\ {} & \circ & \textrm {Shift index using}\hspace {3pt} k \mathrm {->}k +2-m \\ {} & {} & u^{m}\cdot \left (\frac {d^{2}}{d u^{2}}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^{3}}{d u^{3}}y \left (u \right )\right )\hspace {3pt}\textrm {to series expansion for}\hspace {3pt} m =1..3 \\ {} & {} & u^{m}\cdot \left (\frac {d^{3}}{d u^{3}}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^{3}}{d u^{3}}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 \,2^{\frac {1}{3}} \left (\mathrm {I} \sqrt {3}-1\right ) \left (-3+r \right ) \left (r -1\right ) r a_{0} u^{-2+r}}{2}+\left (\frac {3 \,2^{\frac {1}{3}} \left (\mathrm {I} \sqrt {3}-1\right ) \left (-2+r \right ) r \left (1+r \right ) a_{1}}{2}-\frac {3 \,2^{\frac {2}{3}} \left (1+\mathrm {I} \sqrt {3}\right ) r \left (r^{2}-5 r +5\right ) a_{0}}{2}\right ) u^{r -1}+\left (\moverset {\infty }{\munderset {k =0}{\sum }}\left (\frac {3 \,2^{\frac {1}{3}} \left (\mathrm {I} \sqrt {3}-1\right ) \left (k +r -1\right ) \left (k +1+r \right ) \left (k +2+r \right ) a_{k +2}}{2}-\frac {3 \,2^{\frac {2}{3}} \left (1+\mathrm {I} \sqrt {3}\right ) \left (k +1+r \right ) \left (\left (k +1\right )^{2}+2 \left (k +1\right ) r +r^{2}-5 k -5 r \right ) a_{k +1}}{2}+2 a_{k} \left (k +r \right ) \left (k +r -2\right ) \left (k +r -4\right )\right ) u^{k +r}\right )=0 \\ \bullet & {} & a_{0}\textrm {cannot be 0 by assumption, giving the indicial equation}\hspace {3pt} \\ {} & {} & \frac {3 \,2^{\frac {1}{3}} \left (\mathrm {I} \sqrt {3}-1\right ) \left (-3+r \right ) \left (r -1\right ) r}{2}=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} \\ {} & {} & \frac {3 \,2^{\frac {1}{3}} \left (\mathrm {I} \sqrt {3}-1\right ) \left (k +r -1\right ) \left (k +1+r \right ) \left (k +2+r \right ) a_{k +2}}{2}-\frac {3 a_{k +1} \left (1+\mathrm {I} \sqrt {3}\right ) \left (k +1+r \right ) \left (k^{2}+\left (2 r -3\right ) k +r^{2}-3 r +1\right ) 2^{\frac {2}{3}}}{2}+2 a_{k} \left (k +r \right ) \left (k +r -2\right ) \left (k +r -4\right )=0 \\ \bullet & {} & \textrm {Recursion relation that defines series solution to ODE}\hspace {3pt} \\ {} & {} & a_{k +2}=\frac {\left (-12 k^{2} r a_{k}-12 k \,r^{2} a_{k}+48 k r a_{k}+3 \,2^{\frac {2}{3}} k^{3} a_{k +1}+3 \,2^{\frac {2}{3}} r^{3} a_{k +1}-6 \,2^{\frac {2}{3}} k^{2} a_{k +1}-6 \,2^{\frac {2}{3}} r^{2} a_{k +1}+9 \,2^{\frac {2}{3}} k^{2} r a_{k +1}+9 \,2^{\frac {2}{3}} k \,r^{2} a_{k +1}-12 \,2^{\frac {2}{3}} k r a_{k +1}+3 \,\mathrm {I} \,2^{\frac {2}{3}} \sqrt {3}\, a_{k +1}-6 \,2^{\frac {2}{3}} k a_{k +1}-6 \,2^{\frac {2}{3}} r a_{k +1}+9 \,\mathrm {I} \,2^{\frac {2}{3}} \sqrt {3}\, k^{2} r a_{k +1}+9 \,\mathrm {I} \,2^{\frac {2}{3}} \sqrt {3}\, k \,r^{2} a_{k +1}-12 \,\mathrm {I} \,2^{\frac {2}{3}} \sqrt {3}\, k r a_{k +1}+3 \,\mathrm {I} \,2^{\frac {2}{3}} \sqrt {3}\, k^{3} a_{k +1}+3 \,\mathrm {I} \,2^{\frac {2}{3}} \sqrt {3}\, r^{3} a_{k +1}-6 \,\mathrm {I} \,2^{\frac {2}{3}} \sqrt {3}\, k^{2} a_{k +1}-6 \,\mathrm {I} \,2^{\frac {2}{3}} \sqrt {3}\, r^{2} a_{k +1}-6 \,\mathrm {I} \,2^{\frac {2}{3}} \sqrt {3}\, k a_{k +1}-6 \,\mathrm {I} \,2^{\frac {2}{3}} \sqrt {3}\, r a_{k +1}-4 k^{3} a_{k}-4 r^{3} a_{k}+24 k^{2} a_{k}+24 r^{2} a_{k}-32 k a_{k}-32 r a_{k}+3 \,2^{\frac {2}{3}} a_{k +1}\right ) 2^{\frac {2}{3}}}{6 \left (3 \,\mathrm {I} \sqrt {3}\, k \,r^{2}-\mathrm {I} \sqrt {3}\, r +\mathrm {I} \sqrt {3}\, k^{3}+2 \,\mathrm {I} \sqrt {3}\, r^{2}+2 \,\mathrm {I} \sqrt {3}\, k^{2}+3 \,\mathrm {I} \sqrt {3}\, k^{2} r +\mathrm {I} \sqrt {3}\, r^{3}-\mathrm {I} \sqrt {3}\, k -2 \,\mathrm {I} \sqrt {3}-k^{3}-3 k^{2} r -3 k \,r^{2}-r^{3}+4 \,\mathrm {I} \sqrt {3}\, k r -2 k^{2}-4 k r -2 r^{2}+k +r +2\right )} \\ \bullet & {} & \textrm {Recursion relation for}\hspace {3pt} r =0 \\ {} & {} & a_{k +2}=\frac {\left (3 \,2^{\frac {2}{3}} k^{3} a_{k +1}-6 \,2^{\frac {2}{3}} k^{2} a_{k +1}+3 \,\mathrm {I} \,2^{\frac {2}{3}} \sqrt {3}\, a_{k +1}-6 \,2^{\frac {2}{3}} k a_{k +1}+3 \,\mathrm {I} \,2^{\frac {2}{3}} \sqrt {3}\, k^{3} a_{k +1}-6 \,\mathrm {I} \,2^{\frac {2}{3}} \sqrt {3}\, k^{2} a_{k +1}-6 \,\mathrm {I} \,2^{\frac {2}{3}} \sqrt {3}\, k a_{k +1}-4 k^{3} a_{k}+24 k^{2} a_{k}-32 k a_{k}+3 \,2^{\frac {2}{3}} a_{k +1}\right ) 2^{\frac {2}{3}}}{6 \left (2+\mathrm {I} \sqrt {3}\, k^{3}+2 \,\mathrm {I} \sqrt {3}\, k^{2}-\mathrm {I} \sqrt {3}\, k -2 \,\mathrm {I} \sqrt {3}-k^{3}-2 k^{2}+k \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 +2}=\frac {\left (3 \,2^{\frac {2}{3}} k^{3} a_{k +1}-6 \,2^{\frac {2}{3}} k^{2} a_{k +1}+3 \,\mathrm {I} \,2^{\frac {2}{3}} \sqrt {3}\, a_{k +1}-6 \,2^{\frac {2}{3}} k a_{k +1}+3 \,\mathrm {I} \,2^{\frac {2}{3}} \sqrt {3}\, k^{3} a_{k +1}-6 \,\mathrm {I} \,2^{\frac {2}{3}} \sqrt {3}\, k^{2} a_{k +1}-6 \,\mathrm {I} \,2^{\frac {2}{3}} \sqrt {3}\, k a_{k +1}-4 k^{3} a_{k}+24 k^{2} a_{k}-32 k a_{k}+3 \,2^{\frac {2}{3}} a_{k +1}\right ) 2^{\frac {2}{3}}}{6 \left (2+\mathrm {I} \sqrt {3}\, k^{3}+2 \,\mathrm {I} \sqrt {3}\, k^{2}-\mathrm {I} \sqrt {3}\, k -2 \,\mathrm {I} \sqrt {3}-k^{3}-2 k^{2}+k \right )} \\ \bullet & {} & \textrm {Recursion relation for}\hspace {3pt} r =1 \\ {} & {} & a_{k +2}=\frac {\left (3 \,2^{\frac {2}{3}} k^{3} a_{k +1}+3 \,2^{\frac {2}{3}} k^{2} a_{k +1}-12 a_{k}+3 \,\mathrm {I} \,2^{\frac {2}{3}} \sqrt {3}\, k^{2} a_{k +1}-9 \,2^{\frac {2}{3}} k a_{k +1}-6 \,\mathrm {I} \,2^{\frac {2}{3}} \sqrt {3}\, a_{k +1}+3 \,\mathrm {I} \,2^{\frac {2}{3}} \sqrt {3}\, k^{3} a_{k +1}-4 k^{3} a_{k}+12 k^{2} a_{k}+4 k a_{k}-6 \,2^{\frac {2}{3}} a_{k +1}-9 \,\mathrm {I} \,2^{\frac {2}{3}} \sqrt {3}\, k a_{k +1}\right ) 2^{\frac {2}{3}}}{6 \left (\mathrm {I} \sqrt {3}\, k^{3}+5 \,\mathrm {I} \sqrt {3}\, k^{2}+6 \,\mathrm {I} \sqrt {3}\, k -k^{3}-5 k^{2}-6 k \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 =0 \\ {} & {} & a_{k +2}=\frac {\left (3 \,2^{\frac {2}{3}} k^{3} a_{k +1}+3 \,2^{\frac {2}{3}} k^{2} a_{k +1}-12 a_{k}+3 \,\mathrm {I} \,2^{\frac {2}{3}} \sqrt {3}\, k^{2} a_{k +1}-9 \,2^{\frac {2}{3}} k a_{k +1}-6 \,\mathrm {I} \,2^{\frac {2}{3}} \sqrt {3}\, a_{k +1}+3 \,\mathrm {I} \,2^{\frac {2}{3}} \sqrt {3}\, k^{3} a_{k +1}-4 k^{3} a_{k}+12 k^{2} a_{k}+4 k a_{k}-6 \,2^{\frac {2}{3}} a_{k +1}-9 \,\mathrm {I} \,2^{\frac {2}{3}} \sqrt {3}\, k a_{k +1}\right ) 2^{\frac {2}{3}}}{6 \left (\mathrm {I} \sqrt {3}\, k^{3}+5 \,\mathrm {I} \sqrt {3}\, k^{2}+6 \,\mathrm {I} \sqrt {3}\, k -k^{3}-5 k^{2}-6 k \right )} \\ \bullet & {} & \textrm {Recursion relation for}\hspace {3pt} r =3 \\ {} & {} & a_{k +2}=\frac {\left (3 \,2^{\frac {2}{3}} k^{3} a_{k +1}+21 \,2^{\frac {2}{3}} k^{2} a_{k +1}+12 a_{k}+39 \,2^{\frac {2}{3}} k a_{k +1}+3 \,\mathrm {I} \,2^{\frac {2}{3}} \sqrt {3}\, k^{3} a_{k +1}-4 k^{3} a_{k}-12 k^{2} a_{k}+4 k a_{k}+12 \,2^{\frac {2}{3}} a_{k +1}+12 \,\mathrm {I} \,2^{\frac {2}{3}} \sqrt {3}\, a_{k +1}+21 \,\mathrm {I} \,2^{\frac {2}{3}} \sqrt {3}\, k^{2} a_{k +1}+39 \,\mathrm {I} \,2^{\frac {2}{3}} \sqrt {3}\, k a_{k +1}\right ) 2^{\frac {2}{3}}}{6 \left (\mathrm {I} \sqrt {3}\, k^{3}+11 \,\mathrm {I} \sqrt {3}\, k^{2}+38 \,\mathrm {I} \sqrt {3}\, k +40 \,\mathrm {I} \sqrt {3}-40-k^{3}-11 k^{2}-38 k \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 {\left (3 \,2^{\frac {2}{3}} k^{3} a_{k +1}+21 \,2^{\frac {2}{3}} k^{2} a_{k +1}+12 a_{k}+39 \,2^{\frac {2}{3}} k a_{k +1}+3 \,\mathrm {I} \,2^{\frac {2}{3}} \sqrt {3}\, k^{3} a_{k +1}-4 k^{3} a_{k}-12 k^{2} a_{k}+4 k a_{k}+12 \,2^{\frac {2}{3}} a_{k +1}+12 \,\mathrm {I} \,2^{\frac {2}{3}} \sqrt {3}\, a_{k +1}+21 \,\mathrm {I} \,2^{\frac {2}{3}} \sqrt {3}\, k^{2} a_{k +1}+39 \,\mathrm {I} \,2^{\frac {2}{3}} \sqrt {3}\, k a_{k +1}\right ) 2^{\frac {2}{3}}}{6 \left (\mathrm {I} \sqrt {3}\, k^{3}+11 \,\mathrm {I} \sqrt {3}\, k^{2}+38 \,\mathrm {I} \sqrt {3}\, k +40 \,\mathrm {I} \sqrt {3}-40-k^{3}-11 k^{2}-38 k \right )}, 18 \,2^{\frac {1}{3}} \left (\mathrm {I} \sqrt {3}-1\right ) a_{1}+\frac {9 \,2^{\frac {2}{3}} \left (1+\mathrm {I} \sqrt {3}\right ) a_{0}}{2}=0\right ] \\ \bullet & {} & \textrm {Revert the change of variables}\hspace {3pt} u =x +\frac {2^{\frac {2}{3}}}{4}+\frac {\mathrm {I} \sqrt {3}\, 2^{\frac {2}{3}}}{4} \\ {} & {} & \left [y=\moverset {\infty }{\munderset {k =0}{\sum }}a_{k} \left (x +\frac {2^{\frac {2}{3}}}{4}+\frac {\mathrm {I} \sqrt {3}\, 2^{\frac {2}{3}}}{4}\right )^{k +3}, a_{k +2}=\frac {\left (3 \,2^{\frac {2}{3}} k^{3} a_{k +1}+21 \,2^{\frac {2}{3}} k^{2} a_{k +1}+12 a_{k}+39 \,2^{\frac {2}{3}} k a_{k +1}+3 \,\mathrm {I} \,2^{\frac {2}{3}} \sqrt {3}\, k^{3} a_{k +1}-4 k^{3} a_{k}-12 k^{2} a_{k}+4 k a_{k}+12 \,2^{\frac {2}{3}} a_{k +1}+12 \,\mathrm {I} \,2^{\frac {2}{3}} \sqrt {3}\, a_{k +1}+21 \,\mathrm {I} \,2^{\frac {2}{3}} \sqrt {3}\, k^{2} a_{k +1}+39 \,\mathrm {I} \,2^{\frac {2}{3}} \sqrt {3}\, k a_{k +1}\right ) 2^{\frac {2}{3}}}{6 \left (\mathrm {I} \sqrt {3}\, k^{3}+11 \,\mathrm {I} \sqrt {3}\, k^{2}+38 \,\mathrm {I} \sqrt {3}\, k +40 \,\mathrm {I} \sqrt {3}-40-k^{3}-11 k^{2}-38 k \right )}, 18 \,2^{\frac {1}{3}} \left (\mathrm {I} \sqrt {3}-1\right ) a_{1}+\frac {9 \,2^{\frac {2}{3}} \left (1+\mathrm {I} \sqrt {3}\right ) a_{0}}{2}=0\right ] \end {array} \]
Maple trace Kovacic algorithm successful
`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 a quadrature checking if the LODE has constant coefficients checking if the LODE is of Euler type trying a symmetry of the form [xi=0, eta=F(x)] checking if the LODE is missing y -> Trying a Liouvillian solution using Kovacics algorithm A Liouvillian solution exists Reducible group (found an exponential solution) <- Kovacics algorithm successful <- missing the dependent variable successful`
✓ Solution by Maple
Time used: 0.015 (sec). Leaf size: 20
dsolve((2*x^3-1)*diff(y(x),x$3)-6*x^2*diff(y(x),x$2)+6*x*diff(y(x),x)=0,y(x), singsol=all)
\[ y \left (x \right ) = c_{3} x^{4}+c_{2} x^{2}+4 c_{3} x +c_{1} \]
✓ Solution by Mathematica
Time used: 1.357 (sec). Leaf size: 31
DSolve[(2*x^3-1)*y'''[x]-6*x^2*y''[x]+6*x*y'[x]==0,y[x],x,IncludeSingularSolutions -> True]
\[ y(x)\to -\frac {c_2 x^4}{4}+\frac {c_1 x^2}{2}-c_2 x+c_3 \]