Let $\dot{x}(t) = Ax(t)$ where $A$ is an n by n real matrix and $x(t)$ is a n dimensional vector. Let also $\dot{y}(t) = By(t) + Cx(t)$ where y is a vector and $B$,$C$ are n by n matrices. Suppose that the n by n matrix $D$ is such that $DA - BD = C$. Finnaly, suppose that $y(0) = Dx(0)$.
Prove that $y(t) = Dx(t)$ for all $t \geq 0$.
My attempt so far: If we can show that $\dot{y}(t) = D\dot{x}(t)$ for all $t \geq 0$ the result will follow by the fundamental theorem of calculus. By replacing $C$ with $DA - BD$ in the equation $\dot{y}(t) = By(t) + Cx(t)$ and then by replacing $Ax(t)$ with $\dot{x}(t)$ it follows from the initial condition $y(0) = Dx(0)$ that $\dot{y}(0) = D\dot{x}(0)$. Unfortunately, I don't know how to proceed to show that this final eq is true for all $t$. There is no mention that the matrix $D$ be invertible, but perhaps there is a typo and this is required? Any help is appreciated!
No, there are no additional assumptions on $D$.
The usual trick is to let $z(t)=y(t)-Dx(t)$ and try to prove $z(t)=0$ for all $t$. We know $z(0)=0$ and we calculate the differential equation of $z$: \begin{align} \dot{z}(t)&=\dot{y}(t)-D\dot{x}(t)\\ &=By(t)+Cx(t)-DAx(t)\\ &=By(t)-BDx(t)\\ &=B(y(t)-Dx(t))\\ &=Bz(t) \end{align} So $z(t)=0$ for all $t$.