\[\frac{\mathrm dx}{\mathrm dt} = -4\pi^2x+x^2\]
\[x_0 = \pi^2,\qquad v_0=0\]

\[\frac{\mathrm dx}{\mathrm dt} = -4\pi^2x+x^2\]
\[x_0 = \pi^2,\qquad v_0=0\]

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Is \(\pi\) the circle constant?

yes

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