How to prove this variation-of-constants formula?

Question

If $A ,B\in \mathbb{R}^{n\times n},v\in\mathbb{R}^n$,then from the variation-of-constants formula $$\exp((A+B)\tau)v=\exp(A\tau)v+\int_0^\tau \exp(As)B\exp((A+B)(\tau-s))vds$$.

In my opinion,this formula can be seen as the soluion of the following ode: $$ \frac{d v}{dt}=(A+B)v,$$ Thus, we can get $$ \frac{d}{dt}(\exp(-At)v)=\exp(-At)Bv,$$ Then, $$ \exp((A+B)\tau)v=\exp(A\tau)v+\int_0^\tau \exp(A(\tau-s))Bvds$$ So, how to get the first identity?

Answer 1

The solution to $\dot{x} = Ax + Bu$, $x(0) = v$ is given by $x(t) = e^{At} v + \int_0^t e^{As} B u(t-s) ds$.

The solution to $\dot{w} = Aw + Bw$, $w(0) = v$ is given by $w(t) = e^{(A+B)t}v$ and also by $w(t) = e^{At} v + \int_0^t e^{As} B w(t-s) ds$, substituting gives the desired result.