Internal problem ID [6860]
Book: Collection of Kovacic problems
Section: section 1
Problem number: 128.
ODE order: 2.
ODE degree: 1.
CAS Maple gives this as type [[_2nd_order, _with_linear_symmetries]]
Solve \begin {gather*} \boxed {x^{2} \left (x^{2}+x +1\right ) y^{\prime \prime }-x \left (-2 x^{2}-4 x +1\right ) y^{\prime }+y=0} \end {gather*}
✓ Solution by Maple
Time used: 0.781 (sec). Leaf size: 269
dsolve(x^2*(1+x+x^2)*diff(y(x),x$2)-x*(1-4*x-2*x^2)*diff(y(x),x)+y(x)=0,y(x), singsol=all)
\[ y \relax (x ) = \frac {c_{1} {\mathrm e}^{-\frac {7 \sqrt {3}\, \arctan \left (\frac {\left (2 x +1\right ) \sqrt {3}}{3}\right )}{6}} x \left (i x \sqrt {3}-x -2\right )^{\frac {1}{4}-\frac {7 i \sqrt {3}}{12}} \left (i \sqrt {3}-2 x -1\right )^{\frac {1}{4}+\frac {7 i \sqrt {3}}{12}}}{\left (x^{2}+x +1\right )^{\frac {3}{4}}}+\frac {c_{2} {\mathrm e}^{-\frac {7 \sqrt {3}\, \arctan \left (\frac {\left (2 x +1\right ) \sqrt {3}}{3}\right )}{6}} \hypergeom \left (\left [\frac {3}{4}-\frac {\sqrt {\frac {-45 i \sqrt {3}-3}{1+i \sqrt {3}}}}{6}+\frac {7 i \sqrt {3}}{12}, \frac {3}{4}+\frac {\sqrt {\frac {-45 i \sqrt {3}-3}{1+i \sqrt {3}}}}{6}+\frac {7 i \sqrt {3}}{12}\right ], \left [1+\frac {2 \sqrt {\frac {-21 i \sqrt {3}+69}{\left (1+i \sqrt {3}\right )^{3}}}}{3}\right ], \frac {-4+2 i x \sqrt {3}-2 x}{\left (1+i \sqrt {3}\right ) \left (i \sqrt {3}-2 x -1\right )}\right ) x \left (i x \sqrt {3}-x -2\right )^{\frac {3}{4}+\frac {7 i \sqrt {3}}{12}} \left (i \sqrt {3}-2 x -1\right )^{-\frac {1}{4}-\frac {7 i \sqrt {3}}{12}}}{\left (x^{2}+x +1\right )^{\frac {3}{4}}} \]
✓ Solution by Mathematica
Time used: 0.639 (sec). Leaf size: 90
DSolve[x^2*(1+x+x^2)*y''[x]-x*(1-4*x-2*x^2)*y'[x]+y[x]==0,y[x],x,IncludeSingularSolutions -> True]
\begin{align*} y(x)\to \frac {x e^{-\frac {7 \text {ArcTan}\left (\frac {2 x+1}{\sqrt {3}}\right )}{\sqrt {3}}} \left (c_2 \int _1^x\frac {e^{\frac {7 \text {ArcTan}\left (\frac {2 K[1]+1}{\sqrt {3}}\right )}{\sqrt {3}}}}{K[1] \sqrt {K[1]^2+K[1]+1}}dK[1]+c_1\right )}{\sqrt {x^2+x+1}} \\ \end{align*}