Particular solution of a Ricatti ODE satisfying initial condition

Question

I want to solve the following Ricatti ODE: $y′ = (y −t)^2 + 1, y(0) = 2$ but I am unable to find a particular solution that will satisfy the initial condition. I found solutions to the equation (say, $y = t$), but none of them satisfy $y(0)=2$. Any hints as to how to approach this problem?

Answer 1

Hint: $$y′ = (y −t)^2 + 1, y(0) = 2$$ $$(y-t)′ = (y −t)^2 $$ $$z′ = z^2 $$ Where $z=y-t$. The DE is separable.

---

Edit: $$\dfrac {z'}{z^2}=1$$ Integrate: $$-\dfrac 1 z =t+c$$ $$\dfrac 1 {y-t}=-t+C$$ Apply initial condition: $$y(0)=2 \implies C= \dfrac 12$$ The solution is therefore: $$\dfrac {1}{y-t}+t=\dfrac 12$$ $$y(t)=\dfrac {2}{1-2t}+t$$