Recognizing linear ordinary differential equations

[fishingspree2]

Homework Statement

Is the following equation a linear ODE?
\frac{d^{2}R}{dt^{2}}=-\frac{k}{R^{2}} where k is a constant

Homework Equations

A linear ordinary differential equation can be written in the following form:
a_{n}\left ( x \right )\frac{{d}^{n}y}{{d}x^{n}}+a_{n-1}\left ( x \right )\frac{{d}^{n-1}y}{{d}x^{n-1}}+...+a_{1}\left ( x \right )\frac{{d}y}{{d}x}+a_{0}\left ( x \right )y=g\left ( x \right )

The Attempt at a Solution

Well, the correct answer in the textbook is: not a linear ODE. But there is something I don't really understand. If we multiply both sides by R^{2} then we have:
R^{2}\frac{d^{2}R}{dt^{2}}=-k

The right hand side is of the form g\left ( x \right )=-k, so this is good. Also, the \frac{{d}^{2}R}{{d}t^{2}} term is power of 1, which is also good. The problem relies in the a_{n}\left ( x \right ) term. The coefficient in front of the derivative must at most depend on the independant variable, in this case, t.

We have R^{2}. Well, in my opinion, we can't know whether it is linear or not because we don't know how R is explicitely defined. R could be a function of t in the following form: R=F\left ( t \right ). If this is the case, then R^{2} could be substitued by \left (F\left ( t \right ) \right )^{2}. Then we would clearly see that R^{2} depend only on the variable t and we could conclude that the given equation is linear.

Given my arguments, I don't understand why we can already tell that the equation is non-linear. Thank you very much!

 

[micromass]

Well, even if R=F(t), that would not make it linear. The problem is that F is an unknown function. That is: we don't know (a priori) the exact form of F.

The idea is that \alpha_i are known functions that depend on t. But R is not a known function. Thus it is a non-linear equation.