My problem
Given $$x(t)=x_Te^{-a(T-t)}-\int_t^T e^{-a(v-t)} b\,dv$$ , $$x_T=x(T)=constant$$ Show that x(t) is a solution to $$\dot{x}(t)=ax(t)+b $$ a and b are constants
My attempt to solution $$\dot{x}(t)=ax_Te^{-a(T-t)}-\int_t^T ae^{-a(v-t)} b\,dv=a(x_Te^{-a(T-t)}-\int_t^T e^{-a(v-t)} b\,dv)=ax(t)$$
I don't see where I can get the b from.
Thank you for help.
You may use $$ \partial_t\int\limits^t_T e^{-a(v-t)}bdv = \partial_t\left(e^{at}\int\limits^t_T e^{-av}bdv\right) = \int\limits^t_T e^{-av}bdv\left(\partial_te^{at}\right)+e^{at}\left(\partial_t\int\limits^t_T e^{-av}bdv\right)\\ = ae^{at}\int\limits^t_T e^{-av}bdv+e^{at}be^{-at} = ae^{at}\int\limits^t_T e^{-av}bdv + b $$ where we used the product rule and $\partial_t\int\limits^t_u f(x)dx = f(t)$. Also note: $-\int\limits^t_u f(x)dx = \int\limits_t^u f(x)dx$.