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In particular, what could be a benefit of linearizing a system of first order non-linear ODEs? I have seen a number of papers derive linearized versions of non-linear systems of ODEs, but I can't seem to find much of an argument as to why those are better than the original non-linear system.

For example, given a system of 3 ODEs $\dot x=f_x(x,y,z) $ , $\ \dot y=f_y(x,y,z)$ , $\ \dot z=f_z(x,y,z)$, say you derive some linear system $\dot x= a_1(t)x+a_2(t)y+a_3(t)z$ , $\dot y= b_1(t)x+b_2(t)y+b_3(t)z$ , $\dot z= c_1(t)x+c_2(t)y+c_3(t)z$, and that this linear system provides a decent approximation to the nonlinear system.

To me, from a couple weeks of searching the internet, it seems there aren't really any benefits unless the linear system has certain properties (such as constant coefficients, as is the case with the Clohessy Wiltshire equations) that let you solve for some explicit solution. It seems to me that, without an explicit solution, all you have is another set of differential equations to plug into an integrator. At that point you might as well just use the original nonlinear system (i.e. use the integrator to approximate the solution to the original equation rather than an approximation of the original equation).

I was wondering if there are any advantages other than the possibility of an explicit solution. For example, could a linear system somehow improve computational efficiency? I have been browsing the internet for a little while now, but it has proven hard to sift through material about linearizing systems and solving linear ODEs etc.

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  • $\begingroup$ Even if the coefficients of the linear system are non-constant $X'=A(t)X$ you have a nice expression for the solution $X=\exp(\int A(t))$. Even if less explicit than if $A$ is constant, it is still an equation more manageable than the non-linear system. $\endgroup$ – user569098 Jun 27 '18 at 20:00
  • $\begingroup$ @LB_O I thought that was only true if $A(t)$ and $\int A(t)$ were commutative? math24.net/… $\endgroup$ – Miles Johnson Jun 27 '18 at 20:06
  • $\begingroup$ Which is larger than the class of constant matrices. $\endgroup$ – user569098 Jun 27 '18 at 20:14
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    $\begingroup$ When the matrix doesn't commute with its primitive, the equation is still somewhat manageable. Numerical analysis is not my area, but I would guess that also when trying to approximate the solution the linear system is likely to be more tractable in general. $\endgroup$ – user569098 Jun 27 '18 at 20:17
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When you linearize at an equilibrium point, you always get a linear system with constant coefficients. And that system (provided all eigenvalues have nonzero real part) will tell you whether the equilibrium is stable or not, and the phase portrait of the linear system will tell you approximately what the phase portrait of the nonlinear system looks like near the equilibrium.

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