Consider the differential equation

$\dot{x}(t)=\alpha x(t)+\beta u(t),$

with $\alpha ,\beta >0$ positive constant scalars and initial condition $x(0)={x}_{0}$. Function $u(\cdot )$ is known. The Laplace solution for this equation should be

$x(t)={e}^{\alpha (t-{t}_{0})}{x}_{0}+\beta {\int}_{0}^{t}{e}^{\alpha (t-\tau )}u(\tau )d\tau .$

However, if now I take the time derivative, I get

$\dot{x}(t)=\alpha {e}^{\alpha (t-{t}_{0})}{x}_{0}+\beta u(t).$

Comparing to the first equation, this leaves me with

$x(t)={e}^{\alpha (t-{t}_{0})}{x}_{0},$

which looks inconsistent with the Laplace solution. Anybody has a clue on what's going on here?

$\dot{x}(t)=\alpha x(t)+\beta u(t),$

with $\alpha ,\beta >0$ positive constant scalars and initial condition $x(0)={x}_{0}$. Function $u(\cdot )$ is known. The Laplace solution for this equation should be

$x(t)={e}^{\alpha (t-{t}_{0})}{x}_{0}+\beta {\int}_{0}^{t}{e}^{\alpha (t-\tau )}u(\tau )d\tau .$

However, if now I take the time derivative, I get

$\dot{x}(t)=\alpha {e}^{\alpha (t-{t}_{0})}{x}_{0}+\beta u(t).$

Comparing to the first equation, this leaves me with

$x(t)={e}^{\alpha (t-{t}_{0})}{x}_{0},$

which looks inconsistent with the Laplace solution. Anybody has a clue on what's going on here?