Riccati ODE $y'(t)=(t^2+1)*y^2+t^2-1$, $y(0)=1$ I'm trying to solve the following initial value problem:
Riccati ODE:
$y'(t)=(t^2+1)*y^2+t^2-1$
$y(0)=1$
Sadly I'm out of ideas. Because it's $-1$ at the end of the rhs, I cannot use separation of variables. The Picard iteration led me nowhere (just to longer and longer polynomials). I recognize that it is a Riccati ODE and that I could transform it to a simpler one, if I knew a particular solution, but unfortunately I don't.
I attempted variation of parameters (even though it the ODE is not linear) by solving
$y'(t)=(t^2+1)\cdot y^2$, with $y_h(t)=-\frac{3}{c(t)+t(t^2+3)}$ and attempted to find a $c(t)$ such that $y_h(t)$ satisfies the the original initial value problem but that just led to an even more complicated Riccati problem. I also tried WolframAlpha.
Am I just missing the right guess at a particular solution here?
 A: Let $y=-\dfrac{1}{(t^2+1)u}\dfrac{du}{dt}$ ,
Then $\dfrac{dy}{dt}=-\dfrac{1}{(t^2+1)u}\dfrac{d^2u}{dt^2}+\dfrac{1}{(t^2+1)u^2}\left(\dfrac{du}{dt}\right)^2+\dfrac{2t}{(t^2+1)^2u}\dfrac{du}{dt}$
$\therefore-\dfrac{1}{(t^2+1)u}\dfrac{d^2u}{dt^2}+\dfrac{1}{(t^2+1)u^2}\left(\dfrac{du}{dt}\right)^2+\dfrac{2t}{(t^2+1)^2u}\dfrac{du}{dt}=\dfrac{1}{(t^2+1)u^2}\left(\dfrac{du}{dt}\right)^2+t^2-1$
$\dfrac{1}{(t^2+1)u}\dfrac{d^2u}{dt^2}-\dfrac{2t}{(t^2+1)^2u}\dfrac{du}{dt}+t^2-1=0$
$(t^2+1)\dfrac{d^2u}{dt^2}-2t\dfrac{du}{dt}+(t^2+1)^2(t^2-1)u=0$
Let $r=t^2$ ,
Then $\dfrac{du}{dt}=\dfrac{du}{dr}\dfrac{dr}{dt}=2t\dfrac{du}{dr}$
$\dfrac{d^2u}{dt^2}=\dfrac{d}{dt}\left(2t\dfrac{du}{dr}\right)=2t\dfrac{d}{dt}\left(\dfrac{du}{dr}\right)+2\dfrac{du}{dr}=2t\dfrac{d}{dr}\left(\dfrac{du}{dr}\right)\dfrac{dr}{dt}+2\dfrac{du}{dr}=2t\dfrac{d^2u}{dr^2}2t+2\dfrac{du}{dr}=4t^2\dfrac{d^2u}{dr^2}+2\dfrac{du}{dr}=4r\dfrac{d^2u}{dr^2}+2\dfrac{du}{dr}$
$\therefore(r+1)\left(4r\dfrac{d^2u}{dr^2}+2\dfrac{du}{dr}\right)-4r\dfrac{du}{dr}+(r+1)^2(r-1)u=0$
$4r(r+1)\dfrac{d^2u}{dr^2}-2(r-1)\dfrac{du}{dr}+(r+1)^2(r-1)u=0$
