Solving a 2nd order ODE: $\frac{d^2}{dx^2}y(x)=\left(C+(1+x^2)^{-1}\right)y(x)$. I would like to solve the following ode:
$$\frac{d^2}{dx^2}y(x)=\left(C+(1+x^2)^{-1}\right)y(x),\quad x\in\mathbb{R};$$
with boundary condition $y(0)=1$. $C$ is just some constant.
I am very stuck with this. Does anyone have any suggestions of how to proceed?
 A: I am afraid that a closed form solution could not exist and that, provided a second  boundary condition, numerical method would be required.
Even if $C=0$ the solution is far away to be simple since given by
$$y=\, _2F_1\left(-\frac{\sqrt{5}+1}{4} ,\frac{\sqrt{5}-1}{4}
   ;\frac{1}{2};-x^2\right)+c_1\, x \,\, _2F_1\left(-\frac{\sqrt{5}-1}{4},\frac{\sqrt{5}+1}{4};\frac{3}{2};-x^2\right)$$ where appear hypergeometric functions.
A: Hint:
Assume $C\neq0$ for the key case:
$\dfrac{d^2y}{dx^2}=(C+(1+x^2)^{-1})y$
$(x^2+1)\dfrac{d^2y}{dx^2}-(Cx^2+C+1)y=0$
Let $t=x^2+1$ ,
Then $\dfrac{dy}{dx}=\dfrac{dy}{dt}\dfrac{dt}{dx}=2x\dfrac{dy}{dt}$
$\dfrac{d^2y}{dx^2}=\dfrac{d}{dx}\left(2x\dfrac{dy}{dt}\right)=2x\dfrac{d}{dx}\left(\dfrac{dy}{dt}\right)+2\dfrac{dy}{dt}=2x\dfrac{d}{dt}\left(\dfrac{dy}{dt}\right)\dfrac{dt}{dx}+2\dfrac{dy}{dt}=2x\dfrac{d^2y}{dt^2}2x+2\dfrac{dy}{dt}=4x^2\dfrac{d^2y}{dt^2}+2\dfrac{dy}{dt}=4(t-1)\dfrac{d^2y}{dt^2}+2\dfrac{dy}{dt}$
$\therefore t\left(4(t-1)\dfrac{d^2y}{dt^2}+2\dfrac{dy}{dt}\right)-(Ct+1)y=0$
$4t(t-1)\dfrac{d^2y}{dt^2}+2t\dfrac{dy}{dt}-(Ct+1)y=0$
Which relates to Heun's Confluent Equation.
