# Differential Equation of Type Ricatti as part of Short Rate Model

I currently despair of the following solution of a differiental equation (Ricatti Type) as part of a short rate model:

$$B_t=\frac{1}{2}aB^2+bB-1$$ First I am "guessing" a particular solution $$c \in \mathbb{R}$$ as its derivative is zero. That is straightforward and I get e.g. $$-\frac{b}{a}+\frac{\sqrt{b^2+2a}}{a}=s$$

I call $$\sqrt{b^2+2a} = k$$ from now on.

For a general solution y I set $$y=s+u$$ and insert y into the initial equation.

$$s'+u'=\frac{1}{2}a(s+u)^2+b(s+u)-1$$ As s'= 0: $$u'=\frac{1}{2}as^2+\frac{1}{2}au^2+asu+bs+bu-1$$ We already know that $$u'=\frac{1}{2}as^2+bs-1=0$$ thus we are left with $$u'=\frac{1}{2}au^2+asu+bu$$ We now can set $$z=\frac{1}{u}$$ Therefore, $$u'=-\frac{z'}{z^2}$$ Inserting this into the equation delivers: $$z_t+z(as+b)-\frac{1}{2}a$$

which is a linear DGL with the solution: $$z=\frac{-\frac{1}{2}a \int\limits_{t}^{T} exp(k(T-t)) d\tau+C}{exp(k(T-t))}$$

Furthermore (see definition of k above): $$as+b=k$$

Now I have to resubstitute to find y (which is B) with: $$y=\frac{1}{z}+s$$ So I have: $$y=\frac{exp(k(T-t))}{-\frac{1}{2}a \int\limits_{t}^{T} exp(k(T-t)) d\tau+C}+s$$ And I now that $$B(T,T)=y(T,T)=0$$ in order to find C.

Does anyone already spot any mistake I might have made, because no matter what I do I never obtain the solution proposed, being: $$B(t,T)=\frac{2(exp(k(T-t))-1)}{(b+k)(exp(k(T-t))-1)+2k}$$ Or if no mistake is spotted, can somebody carry me over the finish line. I tries everything, also by inserting numbers, but still...something is off.

Thank you very much in advance for any help! Kev