59.2.9 problem 9

Internal problem ID [10002]
Book : Collection of Kovacic problems
Section : section 2. Solution found using all possible Kovacic cases
Problem number : 9
Date solved : Monday, January 27, 2025 at 06:16:37 PM
CAS classification : [[_2nd_order, _with_linear_symmetries]]

\begin{align*} x^{2} \left (-x^{2}+2\right ) y^{\prime \prime }-x \left (4 x^{2}+3\right ) y^{\prime }+\left (-2 x^{2}+2\right ) y&=0 \end{align*}

✓ Solution by Maple

Time used: 0.089 (sec). Leaf size: 47

dsolve(x^2*(2-x^2)*diff(y(x), x$2) - x*(3+4*x^2)*diff(y(x), x) + (2-2*x^2)*y(x) = 0,y(x), singsol=all)
 

\[ y = c_{1} x^{2} \operatorname {hypergeom}\left (\left [\frac {3}{2}, 2\right ], \left [\frac {7}{4}\right ], \frac {x^{2}}{2}\right )+\frac {c_{2} \sqrt {x}\, \left (x^{2}+1\right )}{\left (x^{2}-2\right ) \left (-2 x^{2}+4\right )^{{3}/{4}}} \]

✓ Solution by Mathematica

Time used: 0.606 (sec). Leaf size: 133

DSolve[x^2*(2-x^2)*D[y[x],{x,2}] - x*(3+4*x^2)*D[y[x],x] + (2-2*x^2)*y[x] == 0,y[x],x,IncludeSingularSolutions -> True]
 

\[ y(x)\to \left (x^2+1\right ) \exp \left (\int _1^x-\frac {1-2 K[1]^2}{4 K[1]-2 K[1]^3}dK[1]-\frac {1}{2} \int _1^x\frac {4 K[2]^2+3}{K[2] \left (K[2]^2-2\right )}dK[2]\right ) \left (c_2 \int _1^x\frac {\exp \left (-2 \int _1^{K[3]}-\frac {1-2 K[1]^2}{4 K[1]-2 K[1]^3}dK[1]\right )}{\left (K[3]^2+1\right )^2}dK[3]+c_1\right ) \]