Consider a solution of the linear system $$ \dot x=by$$ $$\dot y =ax$$ Then $u:=yx^{-1}$ solves your Riccati equation $$\dot u= a- ubu$$.

rmk. Note that the solution of the linear system is an exponential, defined for all $t\in\mathbb{R}$; while the solution of the Riccati equation is bounded to the interval where $x$ is invertible (possibly not the whole $\mathbb{R}$) Also, if you have an initial data $u(0)=u_0$, you can take $x(0)=I$ and $y(0)=u_0$ as initial data for the linear system.

Consider a solution of the linear system $$ \dot x=by$$ $$\dot y =ax$$ Then $u:=yx^{-1}$ solves your Riccati equation $$\dot u= a- ubu\, .$$

rmk. Note that the solution of the linear system is an exponential, defined for all $t\in\mathbb{R}$; while the solution of the Riccati equation is bounded to the interval where $x$ is invertible (possibly not the whole $\mathbb{R}$) Also, if you have an initial data $u(0)=u_0$, you can take $x(0)=I$ and $y(0)=u_0$ as initial data for the linear system.

Consider the linear system $$ \dot x=ay$$ $$\dot y =bx$$ Then $u:=yx^{-1}$ solves your Riccati equation $$\dot u= a- ubu$$.

Consider a solution of the linear system $$ \dot x=by$$ $$\dot y =ax$$ Then $u:=yx^{-1}$ solves your Riccati equation $$\dot u= a- ubu$$.

rmk. Note that the solution of the linear system is an exponential, defined for all $t\in\mathbb{R}$; while the solution of the Riccati equation is bounded to the interval where $x$ is invertible (possibly not the whole $\mathbb{R}$) Also, if you have an initial data $u(0)=u_0$, you can take $x(0)=I$ and $y(0)=u_0$ as initial data for the linear system.