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Show that the zero solution of $$ \large \ddot{x}+bx^2 \dot{x}+kx=0$$ is asymptotically stable if $b>0$ and unstable if $b<0$. Does this depend on the sign of $k$?

I know for 1st order equation but how to find stability of 2nd order equation?

Can I convert it into system of 1st order equations as follows:

Let $y=x$ and $z=\dot x$. Then,

$\dot z=\ddot x=-bx^2 \dot x-kx=-by^2z-ky,$

i.e., $\dot z=-by^2z-ky, ......(1)$

and $ \dot y=z, .......(2)$

These are the two equations.

But how to find zero solutions and stability ?

Do we need to linearize this system again?

Help me

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Any constant solution obviously has $\dot x=0$, $\ddot x=0$ so that the equation $kx=0$ remains.

For $k<0$ you get a saddle point, thus $k>0$.

For the stability consider $$ \frac{d}{dt}\frac12(\dot x^2+x^2)=-bx^2\dot x^2 $$ which tells you in which direction the solutions cross the circles around the origin.

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