I want to solve the initial value problem:
\begin{cases} \dfrac{dx}{dt} + 2x(t) = f(t), t > 0 \\ x(0) = x_0 \\ \end{cases}
I start off by finding the integrating factor, which is $ e^{2t}$. Multiplying both sides with the integrating factor and using the chain rule gives me following:
$ \Big(x(t) * e^{2t} \Big)' = f(t) * e^{2t} $
Now i set $t = \tau$ and integrate both sides, i get
$$ \int_0^t\Big(x(\tau) * e^{2\tau} \Big)' \ d\tau = \int_0^t\Big(f(\tau) * e^{2\tau} \Big) \ d\tau$$
From this i get that
$$ \left [x(\tau) e^{2\tau} \right]_{0}^{t} = \int_0^t\Big(f(\tau) * e^{2\tau} \Big) \ d\tau$$ and then from this we get that
$$ x(t) = e^{-2t}x_0 +\int_0^t\Big(f(\tau) * e^{-2(t-\tau)} \Big) \ d\tau $$ which is the solution.
I have to following questions:
- Why do we set $ t = \tau $?
Also, why do i get that $$ \left [x(\tau) e^{2\tau} \right]_{0}^{t} = \int_0^t\Big(f(\tau) * e^{2\tau} \Big) \ d\tau $$
The way i always learned to solve IVP with integrating factor is that
$$ \int\Big(x(t) * e^{2t} \Big)' = x(t) * e^{2t} $$ since the integral and the derivative cancel each other out, so why don't is the following wrong?
$$ x(t) * e^{2t} = \int f(t) * e^{2t}$$
Hopefully someone will understand what i mean. Thanks in advance!
Setting $t=\tau$ distinguishes the limits of integration from the variable of integration, which is a dummy variable.
The final indefinite integral would introduce a constant of integration and doesn't allow you to compute $x(t)$ without finding that constant. Using the definite integrals, the constant doesn't appear, it cancels out, and you have an explicit formula for $x(t)$.