# Linearization of Second Order Differential Equations

by Ruby Tyra
Tags: linearization, pendulum
 P: 1 I'm having some difficulties figuring out how to linearize second order differential equations for a double pendulum. I have an equation that is in the form of $\theta_{1}''$$\normalsize = function$ [$\theta_{1}$,$\theta_{2}$,$\theta_{1}'$,$\theta_{2}'$] (The original equation is found at http://www.myphysicslab.com/dbl_pendulum.html, the equations inside the orange rectangle.) I was told to replace that function by a linear function of all four variables but I don't know where to start with that since the original equation is much more complex than the simple pendulum example we were given. Thank you!
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 Quote by Ruby Tyra I'm having some difficulties figuring out how to linearize second order differential equations for a double pendulum. I have an equation that is in the form of $\theta_{1}''$$\normalsize = function$ [$\theta_{1}$,$\theta_{2}$,$\theta_{1}'$,$\theta_{2}'$] (The original equation is found at http://www.myphysicslab.com/dbl_pendulum.html, the equations inside the orange rectangle.) I was told to replace that function by a linear function of all four variables but I don't know where to start with that since the original equation is much more complex than the simple pendulum example we were given. Thank you!
If f = f(x,y,z,w) and you want to linearize, use the chain rule:

$$f(x,y,z,w) = f(x_0,y_0,z_0,w_0)+\frac{\partial{f}}{\partial{x}}(x-x_0)+\frac{\partial{f}}{\partial{y}}(y-y_0)+\frac{\partial{f}}{\partial{z}}(z-z_0)+\frac{\partial{f}}{\partial{w}}(w-w_0)$$

where the partials are evaluated at $$x_0,y_0,z_0,w_0$$
 PF Patron Sci Advisor Thanks Emeritus P: 38,429 Note that what Chestermiller is saying is essentially the same as replacing the function by a Taylor polynomial in all variables, then dropping all but the linear terms. And that, in turn, is the same as replacing the "surface" by its "tangent plane".

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