Consider the differential equation $$\dot{v}(t)+2\lambda v(t)=r(t) \tag{Eq. 1}$$ of a causal system with $$r(t) \not\equiv0$$
Using Laplace transform, $$sV(s)-V_0+2\lambda V(s)=R(s)$$ where $V_0$ is the value of $v(t)$ at $t=0$.
$$V(s)=V_0(\frac{1}{s+2\lambda})+R(s)(\frac{1}{s+2\lambda})$$ Using inverse Laplace transform, the solution to the differntial equation is therefore $$v(t)=V_0e^{-2\lambda t}+\int_{0}^{t}e^{-2\lambda (t-\tau)}r(\tau)d\tau$$ where $v_h(t) = V_0e^{-2\lambda t}$ is the homogeneous solution and $v_p(t) = \int_{0}^{t}e^{-2\lambda (t-\tau)}r(\tau)d\tau$ is the particular solution. Note that both $v_h(t)$ and $v_p(t)$ satisfy Eq. 1.
We know that $$ \frac{d}{dt}(f(t)*g(t)) = (\frac{d}{dt}f(t))*g(t) \tag{Eq.2}$$
If we let $f(t) = e^{-2\lambda (t-\tau)}$ and $g(t)=r(t)$, then $$\dot{v_p}(t)=\frac{d}{dt}(f(t)*g(t))=(-2\lambda e^{-2 \lambda t})*r(t)=-2\lambda\int_{0}^{t}e^{-2\lambda (t-\tau)}r(\tau)d\tau$$
Applying Eq. 1 on $v_p(t)$, $$\dot{v_p}(t)+2\lambda v_p(t)=r(t)$$ $$-2\lambda\int_{0}^{t}e^{-2\lambda (t-\tau)}r(\tau)d\tau + 2\lambda\int_{0}^{t}e^{-2\lambda (t-\tau)}r(\tau)d\tau=r(t)$$ $$0 = r(t)$$ which is a contradiction. Why is this so?
