# Consider the differential equation <mrow class="MJX-TeXAtom-ORD"> <mover> x

Consider the differential equation
$\stackrel{˙}{x}\left(t\right)=\alpha x\left(t\right)+\beta u\left(t\right),$
with $\alpha ,\beta >0$ positive constant scalars and initial condition $x\left(0\right)={x}_{0}$. Function $u\left(\cdot \right)$ is known. The Laplace solution for this equation should be
$x\left(t\right)={e}^{\alpha \left(t-{t}_{0}\right)}{x}_{0}+\beta {\int }_{0}^{t}{e}^{\alpha \left(t-\tau \right)}u\left(\tau \right)d\tau .$
However, if now I take the time derivative, I get
$\stackrel{˙}{x}\left(t\right)=\alpha {e}^{\alpha \left(t-{t}_{0}\right)}{x}_{0}+\beta u\left(t\right).$
Comparing to the first equation, this leaves me with
$x\left(t\right)={e}^{\alpha \left(t-{t}_{0}\right)}{x}_{0},$
which looks inconsistent with the Laplace solution. Anybody has a clue on what's going on here?
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Kayley Jackson
The mistake is that t is also inside the integral.
To get things right, write
$\frac{d}{dt}{\int }_{0}^{t}{e}^{\alpha \left(t-\tau \right)}u\left(\tau \right)d\tau =\frac{d}{dt}\left[{e}^{\alpha t}{\int }_{0}^{t}{e}^{-\alpha \tau }u\left(\tau \right)d\tau \right]$
and use the product rule.
This gives
$=\alpha {e}^{\alpha t}{\int }_{0}^{t}{e}^{-\alpha \tau }u\left(\tau \right)d\tau +{e}^{\alpha t}{e}^{-\alpha t}u\left(t\right)=\alpha {\int }_{0}^{t}{e}^{\alpha \left(t-\tau \right)}u\left(\tau \right)d\tau +u\left(t\right)$
which allows you to check that you have a solution of the ODE.