Apparent contradiction in first order differential equation 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?
 A: The mistake is that $t$ is also inside the integral.
To get things right, write 
$$
\frac d{dt} \int_0^t e^{\alpha(t-\tau)}u(\tau)d\tau
= \frac d{dt}\left[e^{\alpha t} \int_0^t e^{-\alpha \tau}u(\tau)d\tau\right]
$$ and use the product rule.
This gives
$$
= \alpha e^{\alpha t} \int_0^t e^{-\alpha \tau}u(\tau)d\tau
+ e^{\alpha t} e^{-\alpha t}u(t) = \alpha \int_0^t e^{\alpha(t-\tau)}u(\tau)d\tau + u(t)
$$
which allows you to check that you have a solution of the ODE.
A: The derivative is not correct.
\begin{align*}
\dot{x}(t) 
&= \alpha e^{\alpha (t-t_0)} + \beta \frac{\mathrm{d}}{\mathrm{dt}} \left(\int_0 ^t e^{\alpha (t-\tau)} u(\tau) \,\mathrm{d}\tau\right) \\
&= \alpha e^{\alpha (t-t_0)} + \beta\int_0 ^t \frac{\partial}{\partial t} \left( e^{\alpha (t-\tau)} u(\tau) \,\mathrm{d}\tau\right) + \beta u(t)  \\
&= \alpha e^{\alpha (t-t_0)} + \beta \alpha \int_0 ^t e^{\alpha (t-\tau)} u(\tau) \,\mathrm{d}\tau + \beta u(t)\\
&= \alpha\left(e^{\alpha (t-t_0)} + \beta \int_0 ^t e^{\alpha (t-\tau)} u(\tau) \,\mathrm{d}\tau\right) + \beta u(t)\end{align*}
Therefore
$$\dot{x}(t) = \alpha x(t) + \beta u(t)$$

Observation. Take a look at Leibniz Rule. It is useful when you can not use the trick mookid suggests.
