Internal problem ID [9730]
Internal file name [OUTPUT/8672_Monday_June_06_2022_04_46_21_AM_69464698/index.tex]
Book: Differential Gleichungen, E. Kamke, 3rd ed. Chelsea Pub. NY, 1948
Section: Chapter 2, linear second order
Problem number: 1403.
ODE order: 2.
ODE degree: 1.
The type(s) of ODE detected by this program : "unknown"
Maple gives the following as the ode type
[[_2nd_order, _with_linear_symmetries]]
Unable to solve or complete the solution.
\[ \boxed {y^{\prime \prime }+\left (\frac {1-\operatorname {a1} -\operatorname {b1}}{x -\operatorname {c1}}+\frac {1-\operatorname {a2} -\operatorname {b2}}{x -\operatorname {c2}}+\frac {1-\operatorname {a3} -\operatorname {b3}}{x -\operatorname {c3}}\right ) y^{\prime }+\frac {\left (\frac {\operatorname {a1} \operatorname {b1} \left (\operatorname {c1} -\operatorname {c3} \right ) \left (\operatorname {c1} -\operatorname {c2} \right )}{x -\operatorname {c1}}+\frac {\operatorname {a2} \operatorname {b2} \left (\operatorname {c2} -\operatorname {c1} \right ) \left (\operatorname {c2} -\operatorname {c3} \right )}{x -\operatorname {c2}}+\frac {\operatorname {a3} \operatorname {b3} \left (\operatorname {c3} -\operatorname {c2} \right ) \left (\operatorname {c3} -\operatorname {c1} \right )}{x -\operatorname {c3}}\right ) y}{\left (x -\operatorname {c1} \right ) \left (x -\operatorname {c2} \right ) \left (x -\operatorname {c3} \right )}=0} \]
Maple trace
`Methods for second order ODEs: --- Trying classification methods --- 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 <- No Liouvillian solutions exists -> Trying a solution in terms of special functions: -> Bessel -> elliptic -> Legendre -> Whittaker -> hyper3: Equivalence to 1F1 under a power @ Moebius -> hypergeometric -> heuristic approach -> hyper3: Equivalence to 2F1, 1F1 or 0F1 under a power @ Moebius -> Mathieu -> Equivalence to the rational form of Mathieu ODE under a power @ Moebius trying a solution in terms of MeijerG functions -> Heun: Equivalence to the GHE or one of its 4 confluent cases under a power @ Moebius <- Heun successful: received ODE is equivalent to the HeunG ODE, case a <> 0, e <> 0, g <> 0, c = 0 `
✓ Solution by Maple
Time used: 0.672 (sec). Leaf size: 298
dsolve(diff(diff(y(x),x),x) = -((1-a1-b1)/(x-c1)+(1-a2-b2)/(x-c2)+(1-a3-b3)/(x-c3))*diff(y(x),x)-1/(x-c1)/(x-c2)/(x-c3)*(a1*b1*(c1-c3)*(c1-c2)/(x-c1)+a2*b2*(c2-c1)*(c2-c3)/(x-c2)+a3*b3*(c3-c2)*(c3-c1)/(x-c3))*y(x),y(x), singsol=all)
\[ y \left (x \right ) = \left (x -\operatorname {c2} \right )^{\operatorname {a2}} \left (x -\operatorname {c3} \right )^{\operatorname {b3}} \left (c_{1} \operatorname {HeunG}\left (\frac {\operatorname {c1} -\operatorname {c3}}{\operatorname {c1} -\operatorname {c2}}, \frac {\left (\left (-\operatorname {a3} -2 \operatorname {b1} -\operatorname {b2} +2\right ) \operatorname {c1} +\left (\operatorname {a3} +\operatorname {b1} -1\right ) \operatorname {c2} +\operatorname {c3} \left (\operatorname {b1} +\operatorname {b2} -1\right )\right ) \operatorname {a1} -\left (\operatorname {b1} -1\right ) \left (\operatorname {a2} +\operatorname {b3} \right ) \operatorname {c1} +\left (\operatorname {b1} \operatorname {b3} -\operatorname {a2} \operatorname {b2} +\operatorname {b3} \left (-1+\operatorname {a3} \right )\right ) \operatorname {c2} +\operatorname {c3} \left (\operatorname {b1} \operatorname {a2} +\left (\operatorname {b2} -1\right ) \operatorname {a2} -\operatorname {a3} \operatorname {b3} \right )}{\operatorname {c1} -\operatorname {c2}}, \operatorname {a1} +\operatorname {b3} +\operatorname {a2} , 2-\operatorname {a3} -\operatorname {b1} -\operatorname {b2} , \operatorname {a1} -\operatorname {b1} +1, \operatorname {a2} -\operatorname {b2} +1, \frac {-x +\operatorname {c1}}{\operatorname {c1} -\operatorname {c2}}\right ) \left (x -\operatorname {c1} \right )^{\operatorname {a1}}+c_{2} \operatorname {HeunG}\left (\frac {\operatorname {c1} -\operatorname {c3}}{\operatorname {c1} -\operatorname {c2}}, \frac {\left (\left (-2 \operatorname {a1} -\operatorname {a3} -\operatorname {b2} +2\right ) \operatorname {c1} +\left (\operatorname {a1} +\operatorname {a3} -1\right ) \operatorname {c2} +\operatorname {c3} \left (\operatorname {a1} +\operatorname {b2} -1\right )\right ) \operatorname {b1} -\left (\operatorname {a2} +\operatorname {b3} \right ) \left (\operatorname {a1} -1\right ) \operatorname {c1} +\left (\operatorname {a1} \operatorname {b3} -\operatorname {a2} \operatorname {b2} +\operatorname {b3} \left (-1+\operatorname {a3} \right )\right ) \operatorname {c2} +\operatorname {c3} \left (\operatorname {a1} \operatorname {a2} +\left (\operatorname {b2} -1\right ) \operatorname {a2} -\operatorname {a3} \operatorname {b3} \right )}{\operatorname {c1} -\operatorname {c2}}, \operatorname {b1} +\operatorname {b3} +\operatorname {a2} , 2-\operatorname {a1} -\operatorname {a3} -\operatorname {b2} , -\operatorname {a1} +\operatorname {b1} +1, \operatorname {a2} -\operatorname {b2} +1, \frac {-x +\operatorname {c1}}{\operatorname {c1} -\operatorname {c2}}\right ) \left (x -\operatorname {c1} \right )^{\operatorname {b1}}\right ) \]
✓ Solution by Mathematica
Time used: 17.454 (sec). Leaf size: 293
DSolve[y''[x] == -((((a1*b1*(c1 - c2)*(c1 - c3))/(-c1 + x) + (a2*b2*(-c1 + c2)*(c2 - c3))/(-c2 + x) + (a3*b3*(-c1 + c3)*(-c2 + c3))/(-c3 + x))*y[x])/((-c1 + x)*(-c2 + x)*(-c3 + x))) - ((1 - a1 - b1)/(-c1 + x) + (1 - a2 - b2)/(-c2 + x) + (1 - a3 - b3)/(-c3 + x))*y'[x],y[x],x,IncludeSingularSolutions -> True]
\[ y(x)\to (x-\text {c2})^{\text {a2}} (x-\text {c3})^{\text {b3}} \left (c_1 (x-\text {c1})^{\text {a1}} \text {HeunG}\left [\frac {\text {c1}-\text {c3}}{\text {c1}-\text {c2}},\frac {\text {a1} (-(\text {c1} (\text {a3}+2 \text {b1}+\text {b2}-2))+\text {c2} (\text {a3}+\text {b1}-1)+\text {c3} (\text {b1}+\text {b2}-1))+\text {a2} (-\text {b1} \text {c1}+(\text {b1}-1) \text {c3}+\text {b2} (\text {c3}-\text {c2})+\text {c1})+\text {b3} (\text {c2} (\text {a3}+\text {b1}-1)-\text {a3} \text {c3}-\text {b1} \text {c1}+\text {c1})}{\text {c1}-\text {c2}},-\text {a3}-\text {b1}-\text {b2}+2,\text {a1}+\text {a2}+\text {b3},\text {a1}-\text {b1}+1,\text {a2}-\text {b2}+1,\frac {\text {c1}-x}{\text {c1}-\text {c2}}\right ]+c_2 (x-\text {c1})^{\text {b1}} \text {HeunG}\left [\frac {\text {c1}-\text {c3}}{\text {c1}-\text {c2}},\frac {\text {a2} (-\text {a1} \text {c1}+(\text {a1}-1) \text {c3}+\text {b2} (\text {c3}-\text {c2})+\text {c1})+\text {b1} (-(\text {c1} (2 \text {a1}+\text {a3}+\text {b2}-2))+\text {c2} (\text {a1}+\text {a3}-1)+\text {c3} (\text {a1}+\text {b2}-1))+\text {b3} (\text {c2} (\text {a1}+\text {a3}-1)-\text {a1} \text {c1}-\text {a3} \text {c3}+\text {c1})}{\text {c1}-\text {c2}},-\text {a1}-\text {a3}-\text {b2}+2,\text {a2}+\text {b1}+\text {b3},-\text {a1}+\text {b1}+1,\text {a2}-\text {b2}+1,\frac {\text {c1}-x}{\text {c1}-\text {c2}}\right ]\right ) \]