Does differentiating an integro-differential equation results in equivalent stability of the solution?

I have a dynamical system in the form of an integro-differential equation which I want to analyze in terms of stability. To demonstrate my problem consider the following integro-differential equation:

$$\dot{x}(t)=ax(t)+b\int_0^tx(\tau)\text{d}\tau,$$

where $$\dot{x}(t)$$ denotes the time derivative of $$x(t)$$. If we derive the above equation and try $$x(t)=\text{e}^{\lambda t}$$ we get:

$$\ddot{x}(t)-a\dot{x}(t)-bx(t)=\text{e}^{\lambda t}\left(\lambda^2-a\lambda-b\right) = 0$$ which we can solve for $$\lambda$$. However if we first use the substitution $$x(t)=\text{e}^{\lambda t}$$ to the original equation we get:

$$\lambda\text{e}^{\lambda t}=a\text{e}^{\lambda t}+b\left[\frac{1}{\lambda}\text{e}^{\lambda \tau}\right]^t_0=a\text{e}^{\lambda t}+\left[\frac{b}{\lambda}\text{e}^{\lambda t}-\frac{b}{\lambda}\right]$$ in which case we end up with:

$$\text{e}^{\lambda t}\left(\lambda-\frac{b}{\lambda}-a\right) = \frac{b}{\lambda}.$$

For me it seams like differentiating an integro-differential equation modifies it's stability properties. Does this mean that in case of stability analysis one cannot differentiate? Or these are equivalent in terms of stability and I am just missing an obvious point? Thank you for any comments, ideas!

• The integrated form of your equation contains the condition that $\dot x(0)= a x(0)$, which is lost in the second order differential equation. So if you are looking for exponentials solving the original equation, the only choice will be $\lambda =a$, and then you must have had $b=0$. And for $b=0$ both equations determining $\lambda$ coincide. – Dirk Werner Sep 30 '18 at 20:14

$$\begin{cases} \dot{p}(t) = bx(t) ,& p(0) = 0 \\ \dot{x}(t) = a x(t) + p (t),& x(0) = ? \end{cases}$$
The initial condition that $$p(0)$$ comes from the requirement that $$p(t) = b\int_0^t x(s) ds$$.
Ignoring the prescribed initial data, to do stability analysis of such a system with two unknowns, your hypothesized exponential solution should look like $$(p(t),x(t)) = (p(0),x(0)) e^{\lambda t}$$, where both $$\lambda$$ and the eigenvector need to be solved for.
A priori there is no reason that an eigenvector can be chosen so that $$p(0) = 0$$, matching your initial conditions. In fact, as pointed out by Dirk Werner, an eigenvector with $$p(0) = 0$$ is only possible for this equation when $$b = 0$$.
However, you can still use the information from the stability analysis of the system (or equivalently, the linear second order equation you derived). You just need to solve first and impose the initial condition $$p(0)$$ later. Denoting by $$\lambda_\pm$$ the roots of $$\lambda^2 - a \lambda - b$$, you have that solutions to your integral-differential equation can always take the form $$x(t) = x_+ e^{\lambda_+ t} + x_- e^{\lambda_- t}$$ By Dirk Werner's analysis, when $$b\neq 0$$ the eigenvectors must have a $$p$$ component, implying that a $$p(0) = 0$$ solution necessarily has a non-trivial projection into both eigenspaces. This means that as long as one of the roots $$\lambda_{\pm}$$ is positive, your equation is unstable.