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


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