Solving state-space function with using of Runge-Kutta method

You have a system $dot{x} = f(x)$ with

$$ f(x) = A x = begin{pmatrix} 0 & 1 & 0 \ 0 & 0 & 1 \ -10 & -5 & -2 end{pmatrix} begin{pmatrix} x_1 \ x_2 \ x_3 end{pmatrix} = begin{pmatrix} x_2 \ x_3 \ -10 x_1 - 5 x_2 - 2 x_3 end{pmatrix} $$

The formula for Runge Kutta with step size $h$ is:

$$ begin{align} k_1 &= f(x(t)) \ k_2 &= f(x(t) + frac{h}{2}k_1) \ k_3 &= f(x(t) + frac{h}{2}k_2) \ k_4 &= f(x(t) + h k_3) \ x(t + h) &= x(t) + frac{h}{6}(k_1 + 2 k_2 + 2 k_3 + k_4) end{align} $$

So all you need to do is choose a step size $h$ and initial condition $x(0) = x_0$.

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For example use $h = 0.01$ and

$$ x_0 = begin{pmatrix} 5 \ -2 \ 3 end{pmatrix} $$

Insert this:

$$ begin{align} k_1 &= f(x_0) = begin{pmatrix} -2 \ 3 \ -46 end{pmatrix} \ k_2 &= f(x_0 + frac{h}{2}k_1) = begin{pmatrix} -1.985 \ 2.77 \ -45.515 end{pmatrix} \ k_3 &= f(x_0 + frac{h}{2}k_2) = begin{pmatrix} -1.98615 \ 2.772425 \ -45.51485 end{pmatrix} \ k_4 &= f(x_0 + h k_3) = begin{pmatrix} -1.97227575 \ 2.5448515 \ -45.02970925 end{pmatrix} \ end{align} $$

And so:

$$ x(0.01) = x(0) + frac{0.01}{6}(k_1 + 2 k_2 + 2 k_3 + k_4) = begin{pmatrix} 4.98014237375 \ -1.972283830833333 \ 2.544850984583333 end{pmatrix} $$

If you repeat this 500 times you get:

If you simulate even longer you can see this is an undamped oscillation which makes sense because the three eigenvalues of $A$ are $-2, sqrt{5}i, -sqrt{5}i$.