linearize a nonlinear ode Could anyone suggest me how to linearize the following system of nonlinear odes (special attention to (2)
\begin{align}
    -cU'&=-U''+UV\tag{1}\\
 -cV'&=-k(k+1)V^{k-1}(V')^2+(k+1)V^k V''+UV-\mu V-\gamma V^2\tag{2}
\end{align}
?
Take $V=V(x)$, $U=U(x)$, $\mu$, $\gamma$ arbitrary constants. $k$ takes integer values.
 A: using 
\begin{align}
V(x) = V_0 + \delta V,\\
U(x) = U_0 + \delta U.
\end{align}
we find the linearized equations are
$$
-c\delta U' = -\delta U'' + U_0\delta V + V_0\delta U + U_0V_0,\\
-c\delta V' = \left(U_0V_0 - \mu V_0 -\gamma V_0^2\right) + \left(U_0 -\mu -2V_0\right)\delta V\\
+(k+1)V_0^k\delta V'' + V_0\delta U
$$
From the first equation the equilibrium point occurs at $U_0V_0=0$ and therefore the first linearized equation is 
$$
-c\delta U' = -\delta U'' + U_0\delta V + V_0\delta U.
$$
The second equation the eqluibrium points is given by
$$
U_0V_0 - \mu V_0 -\gamma V_0^2 =0
$$
which leads to the linearized equation 
$$
-c\delta V' = \left(U_0 -\mu -2V_0\right)\delta V +(k+1)V_0^k\delta V'' + V_0\delta U
$$
where the $V_0,U_0$ are constants. Though if you mean using 
$$
V(x) = \tilde{V}(x) + \delta V
$$
and linearize in that fashion then you can go through that yourself following the same convention with the exception that $\tilde{V}(x),\tilde{U}(x)$ are solutions of the original equation. 
$\textbf{Appendix}$
To further expand on how I arrived at the above linear form. lets take 
$$
cU' = -U'' + UV (1).
$$
we find the equilibrium points by setting the derivatives to zero as follows
$$
0 = -0 + UV\implies UV = U_0V_0 = 0
$$
The other equation will also yield an equilibrium equation which is set to zero. In the above I have been cheeky and not explicitly determined what the equilibrium points are, but in most cases I have come across you do not need to know that explicitly to form the linearised equations.
so continuing we have equations of the form $f = f_0 + \delta f$ (as above)
inserting into the Eq.(1) we find 
\begin{align}
-c\left(U_0'+\delta U'\right) = -\left(U_0''+\delta U''\right)\\
+\left(U_0+\delta U\right)\left(V_0+\delta V\right)
\end{align}
But the terms $f_0'$ are zero(since $f_0$ are constants) so we can further reduce.
$$
-c\delta U' = -\delta U''\\
+U_0V_0 +U_0\delta V + \delta U V_0 + \delta U\delta V
$$
Now we linearise by neglecting terms of $O(\delta f^2)$ and higher
$$
-c\delta U' = -\delta U'' + U_0V_0 +U_0\delta V + \delta U V_0 
$$
Can you take it from here? and also do the second equation?
Analogously, the second equation would become
$$
-c\delta V'=(k+1)V_0^k\delta V''+U_0 V_0-\mu V_0-\gamma V_0^2+(U_0-\mu-2V_0)\delta V+\delta UV_0
$$
?
Note that in the first term of rhs, we have $(k+1)V_0^k\delta V''$ instead of $(k+1)(V_0+\delta V)^k\delta V''$ since we are ignoring the higher power terms in the binomial expansion.
