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?