Consider the following differential equation: d3x / dt3 + 6d2x / dt2 + 13dx / dt
ID: 1854389 • Letter: C
Question
Consider the following differential equation: d3x / dt3 + 6d2x / dt2 + 13dx / dt + 38x = f(x) where f(x) = 10e-5x Linearize the differential equation for x near 0.Explanation / Answer
If you have an autonomous equation dx/dy = f(x), where f is nonlinear, then you linearize by expanding f as a Taylor series but truncating after the linear term. So yes, it's about differentiating f (not integrating, which is what you'd like to do to solve the equation to get the solution x as a function of y). As with a Taylor series, you linearize about a particular value of x, call it p. This is crucial! You linearize ***about a particular point***. Typically you linearize about a fixed point (a.k.a. steady state or equilibrium point) of the system, i.e. where f(p) = 0 for some value(s) x=p. i.e. f(p+x) = f(p) + f'(p)*x + higher order terms so you get dy/dx = f(p) + f'(p) * x as your linearized system. In classes and textbooks you're often given that the equilibrium has been moved to p=0 for extra simplicity (it can always be translated to be at the origin without loss of generality) so that f(p)=0, and so you might see dy/dx = f'(p) * x written instead. To be a little more general, if x is in R^n (i.e. is a vector), then you take the Jacobian of f: R^n -> R^n instead of the usual 1D derivative. It's not as simple as just ignoring the nonlinear terms, especially if there are only nonlinear terms in your differential equation to begin with! Source(s): This is what I do for a living :) Also see on the web: http://planetmath.org/encyclopedia/Taylo… http://www.sosmath.com/diffeq/system/non…