The Riccati equation
$$
y'=a(x)\,y^2+b(x)\,y+c(x)
$$
is transformed into the second order linear equation
$$
u''-\Bigl(\frac{a'}{a}+b\Bigr)u'+a\,c\,u=0
$$ in the unknown $u(x)$ by the change
$$
y=-\frac{1}{a(x)}\,\frac{u'}{u}.
$$
If the solution of this equation is $u=C_1u_1+C_2u_2$, then the solution of the original equation is
$$
y=-\frac{1}{a(x)}\,\frac{C_1u'_1+C_2u'_2}{C_1u_1+C_2u_2}=-\frac{1}{a(x)}\,\frac{u'_1+K\,u'_2}{u_1+K\,u_2}\ ,
$$
where $K=C_2/C_1$ if $C_1\ne0$; a similar formula holds if $C_2\ne0$. So, the solution of the Riccati equation depends only on one constant.
Hint:
Let $r=\sinh(x)$ ,
Then $\dfrac{du}{dx}=\dfrac{du}{dr}\dfrac{dr}{dx}=\cosh(x)\dfrac{du}{dr}$
$\dfrac{d^2u}{dx^2}=\dfrac{d}{dx}\left(\cosh(x)\dfrac{du}{dr}\right)=\cosh(x)\dfrac{d}{dx}\left(\dfrac{du}{dr}\right)+\sinh(x)\dfrac{du}{dr}=\cosh(x)\dfrac{d}{dr}\left(\dfrac{du}{dr}\right)\dfrac{dr}{dx}+\sinh(x)\dfrac{du}{dr}=\cosh(x)\dfrac{d^2u}{dr^2}\cosh(x)+\sinh(x)\dfrac{du}{dr}=\cosh^2(x)\dfrac{d^2u}{dr^2}+\sinh(x)\dfrac{du}{dr}$
$\therefore\sinh^2(x)\left(\cosh^2(x)\dfrac{d^2u}{dr^2}+\sinh(x)\dfrac{du}{dr}\right)+\cosh^2(x)\sinh(x)\dfrac{du}{dr}+(\lambda-\sinh^2(x))u=0$
$\sinh^2(x)\cosh^2(x)\dfrac{d^2u}{dr^2}+\sinh(x)(\sinh^2(x)+\cosh^2(x))\dfrac{du}{dr}+(\lambda-\sinh^2(x))u=0$
$r^2(r^2+1)\dfrac{d^2u}{dr^2}+r(2r^2+1)\dfrac{du}{dr}+(\lambda-r^2)u=0$
Let $s=r^2$ ,
Then $\dfrac{du}{dr}=\dfrac{du}{ds}\dfrac{ds}{dr}=2r\dfrac{du}{ds}$
$\dfrac{d^2u}{dr^2}=\dfrac{d}{dr}\left(2r\dfrac{du}{ds}\right)=2r\dfrac{d}{dr}\left(\dfrac{du}{ds}\right)+2\dfrac{du}{ds}=2r\dfrac{d}{ds}\left(\dfrac{du}{ds}\right)\dfrac{ds}{dr}+2\dfrac{du}{ds}=2r\dfrac{d^2u}{ds^2}2r+2\dfrac{du}{ds}=4r^2\dfrac{d^2u}{ds^2}+2\dfrac{du}{ds}$
$\therefore r^2(r^2+1)\left(4r^2\dfrac{d^2u}{ds^2}+2\dfrac{du}{ds}\right)+2r^2(2r^2+1)\dfrac{du}{ds}+(\lambda-r^2)u=0$
$4r^4(r^2+1)\dfrac{d^2u}{ds^2}+2r^2(3r^2+2)\dfrac{du}{ds}+(\lambda-r^2)u=0$
$4s^2(s+1)\dfrac{d^2u}{ds^2}+2s(3s+2)\dfrac{du}{ds}+(\lambda-s)u=0$
$\dfrac{d^2u}{ds^2}+\dfrac{3s+2}{2s(s+1)}\dfrac{du}{ds}+\dfrac{\lambda-s}{4s^2(s+1)}u=0$
$\dfrac{d^2u}{ds^2}+\left(\dfrac{1}{s}+\dfrac{1}{2(s+1)}\right)\dfrac{du}{ds}+\left(\dfrac{\lambda}{4s^2}-\dfrac{\lambda+1}{4s(s+1)}\right)u=0$
Let $u=s^av$ ,
Then $\dfrac{du}{ds}=s^a\dfrac{dv}{ds}+as^{a-1}v$
$\dfrac{d^2u}{ds^2}=s^a\dfrac{d^2v}{ds^2}+as^{a-1}\dfrac{dv}{ds}+as^{a-1}\dfrac{dv}{ds}+a(a-1)s^{a-2}v=s^a\dfrac{d^2v}{ds^2}+2as^{a-1}\dfrac{dv}{ds}+a(a-1)s^{a-2}v$
$\therefore s^a\dfrac{d^2v}{ds^2}+2as^{a-1}\dfrac{dv}{ds}+a(a-1)s^{a-2}v+\left(\dfrac{1}{s}+\dfrac{1}{2(s+1)}\right)\left(s^a\dfrac{dv}{ds}+as^{a-1}v\right)+\left(\dfrac{\lambda}{4s^2}-\dfrac{\lambda+1}{4s(s+1)}\right)s^av=0$
$\dfrac{d^2v}{ds^2}+\dfrac{2a}{s}\dfrac{dv}{ds}+\dfrac{a(a-1)}{s^2}v+\left(\dfrac{1}{s}+\dfrac{1}{2(s+1)}\right)\dfrac{dv}{ds}+\left(\dfrac{a}{s^2}+\dfrac{a}{2s(s+1)}\right)v+\left(\dfrac{\lambda}{4s^2}-\dfrac{\lambda+1}{4s(s+1)}\right)v=0$
$\dfrac{d^2v}{ds^2}+\left(\dfrac{2a+1}{s}+\dfrac{1}{2(s+1)}\right)\dfrac{dv}{ds}+\left(\dfrac{4a^2+\lambda}{4s^2}+\dfrac{2a-\lambda-1}{4s(s+1)}\right)v=0$
Choose $4a^2+\lambda=0$ , i.e. $a=\pm i\dfrac{\sqrt\lambda}{2}$ , the ODE becomes
$\dfrac{d^2v}{ds^2}+\left(\dfrac{\pm i\sqrt\lambda+1}{s}+\dfrac{1}{2(s+1)}\right)\dfrac{dv}{ds}+\dfrac{\pm i\sqrt\lambda-\lambda-1}{4s(s+1)}v=0$
Which reduces to Gaussian hypergeometric equation.
Best Answer
In the same spirit as @Tavish, consider that $$f(t)=(A^2 +b)\sinh (At+B)+aA \cosh(At+B)=0$$ needing to be true for all $t$, use the Taylor series around $t=0$ $$f(t)=\left(a A \cosh (B)+\left(A^2+b\right) \sinh (B)\right)+A \left(a A \sinh(B)+\left(A^2+b\right) \cosh (B)\right)t+O\left(t^2\right)$$
So, from the first coefficient $$b=-\text{csch}(B) \left(a A \cosh (B)+A^2 \sinh (B)\right)$$ Plug in the second coefficient and you are left with $$a A^2 \text{csch}(B)=0$$ Assuming $A\neq0$ then $a=0$ and then $b=-A^2$.