what is the general solution of the given differential equation

Question

$$ 2\cdot (x+1)\cdot y′′(x) −(x+1)\cdot y′(x) +2\cdot y(x) = 0 $$ This is the differential equation. then how can i calculate the general solution that is valid in any interval not including the singular point.

Answer 1

This differential equation $$2\cdot (x+1)\cdot y′′(x) −(x+1)\cdot y′(x) +2\cdot y(x) = 0$$ is indeed a very difficult one and I did not find any way helping to find its general solution. Hoping that it could help you in your search of a solution, I used a CAS and arrived to $$y(x)=(x-3) (x+1) \left(c_2 \text{Ei}\left(\frac{x+1}{2}\right)+c_1\right)-2 c_2 e^{\frac{x+1}{2}} (x-1)$$ which is defined for all values of $x$.

However, the writing of the original post was not very clear and the differential equation could have been $$ (x+1)^2\cdot y′′(x) −(x+1)\cdot y′(x) +2\cdot y(x) = 0$$ which is slightly easier and for which the solution write $$y(x)=(x+1) \left(c_1 \sin (\log (x+1))+c_2 \cos (\log (x+1))\right)$$ which is defined for all $x \geq -1$

Answer 2

A proof of the result reported by Claude Lebovici : $$y(x)=(x-3) (x+1) \left(c_2 \text{Ei}\left(\frac{x+1}{2}\right)+c_1\right)-2 c_2 e^{\frac{x+1}{2}} (x-1)$$ is shown below :