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Consider the ODE $$ \begin{cases} x(0)=(x_1(0),x_2(0))\neq 0\\ x'(t)=\begin{pmatrix}x_1(t)+x_2(t)& x_1(t)+x_2(t)\\ -x_1(t)& x_2(t)\end{pmatrix}^{-1}v(t), \end{cases} $$ with some known function $v(t)$ (note that the matrix is invertible as long as $x\not=0$).

Is there any smart observation on the problem that would suggest a certain kind of numerical solver? (E.g. implicit vs. explicit)

Bananach
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    Can this be rewritten as $\frac{d}{dt}\left( \begin{matrix} {{({{x}{1}}+{{x}{2}})}^{2}} \ x_{2}^{2}-x_{1}^{2} \ \end{matrix} \right)=2v(t)$? – Paul Sep 12 '16 at 09:31
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    In general it does not matter that much if a ODE $x' = f(x,t)$ is non-linear or not when solving it numerically (as long as it's not stiff). A standard RK solver will probably work perfectly well. If you analytically invert the matrix you can write it on the form $$\frac{d}{dx}\pmatrix{x_1\x_2} = \frac{1}{(x_1+x_2)^2}\pmatrix{x_2 & -(x_1+x_2)\ x_1 & (x_1+x_2)}v$$ You can solve it numerically on this form or you can go further analytically to get Paul's nice observation above which almost makes the problem trivial. – Winther Sep 12 '16 at 10:58

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