I am studying for a comprehensive exam in non-linear ODE's and I have this in my book: $$\ddot{\xi}+c\bigg[x_1+\xi-\dfrac{\lambda}{a-x_1-\xi}\bigg] = 0$$ then it goes straight to
$$\ddot{\xi}+c\bigg[1-\dfrac{\lambda}{(a-x_1)^2}\bigg]\xi = 0.$$
How do they obtain this?
Thank you for any help or comments!
$\bf{ADD}:$ The original differential equation is $$\ddot{x}+c\bigg[x-\dfrac{\lambda}{a-x}\bigg] = 0$$ and $x = x_1+\xi$, where $x_1$ is a constant.
I'll describe the situation in more general terms, since you won't have the same equation on your exam. Write the original ODE as $$\ddot{x}+F(x)=0$$ This ODE may have an equilibrium solution $x\equiv x_1$, provided that the equation $F(x_1)=0$ has a solution. (In your example, it does.) We linearize the ODE around this equilibrium (as achille hui said) by letting $x=x_1+\xi$, so that $\xi$ is "small", and the approximation $F(x_1+\xi)\approx F'(x_1)\xi$ is reasonable. This leads to the linear ODE $$ \ddot{\xi } +F'(x_1)\xi =0$$ In your example $F(x)= c\bigg[x-\dfrac{\lambda}{a-x}\bigg]$, thus $F'(x_1)=c\bigg[1-\dfrac{\lambda}{(a-x_1)^2}\bigg]$.