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

Question

Consider the differential equation

\dot{x} (t) = α x (t) + β u (t),

with

α, β > 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^{α (t - t_{0})} x_{0} + β \int_{0}^{t} e^{α (t - τ)} u (τ) d τ .

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

\dot{x} (t) = α e^{α (t - t_{0})} x_{0} + β u (t) .

Comparing to the first equation, this leaves me with

x (t) = e^{α (t - t_{0})} x_{0},

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

Answer 1

The mistake is that t is also inside the integral.
To get things right, write

\frac{d}{d t} \int_{0}^{t} e^{α (t - τ)} u (τ) d τ = \frac{d}{d t} [e^{α t} \int_{0}^{t} e^{- α τ} u (τ) d τ]

and use the product rule.
This gives

= α e^{α t} \int_{0}^{t} e^{- α τ} u (τ) d τ + e^{α t} e^{- α t} u (t) = α \int_{0}^{t} e^{α (t - τ)} u (τ) d τ + u (t)

which allows you to check that you have a solution of the ODE.

Answers (1)