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Take a look at this nonlinear differential equation and linearizing it about $(x,\dot{x})=(0.5,1)$ $$ \ddot{x} + 2x^2\dot{x} + 3\dot{x}^2 + x = 0 $$ In the book I'm reading, the author solved it as letting $y=\dot{x}$ and $z=2x^2\dot{x}+3\dot{x}^2$ $$ \ddot{x} + 2x^2y + 3y^2 + x = 0 $$ And he computed $$ \frac{dz}{dx}\Big\vert_{0.5,1} = 4xy = 2, \qquad \frac{dz}{dy}\Big\vert_{0.5,1} = 2x^2+6y = 6.5 \\ (z-z_0) = 2(x-x_0) + 6.5(y-y_0) \\ z = 2x + 6.5y + 4 $$ Finally, he got the linearized ODE as $$ \ddot{x} + 6.5\dot{x} + 3x - 4 = 0 $$

Now, I'm trying to reach same result using Jacobain matrix. If I let $x_1=x, x_2=\dot{x}$, I get $$ \begin{align} \dot{x}_1 &= x_2 \\ \dot{x}_2 &= -2x_1^2x_2 - 3x_2^2 - x_1 \end{align} $$ The jacobain matrix about (0.5, 1) is $$ J= \begin{bmatrix} 0&1\\ -3&-6.5 \end{bmatrix} $$ I got this linearized ODE as $$ \ddot{x} +6.5\dot{x} + 3x = 0 $$ It is missing the constant -4. What is the problem?

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The linearization of $F(u)$ is $$ F(u_0)+J_F(u_0)(u-u_0)=J_F(u_0)u+F(u_0)-J_F(u_0)u_0. $$ In your second method you are missing the last two terms. ($u=[x,\dot x]$)

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