Is an equilibrium solution always zero in differential equations?

[ehabmozart]

I am still in the first few pages of my differential course. I just passed through the term equilibrium solution. Apparently, it meant for me as when the solution of the DE is zero. For example, dv/dt= 9.8-(v/5) ... has an equilibrium solution of 49 where dv.dt is actually zero. Is this the case or not??

 

[pasmith]

I am still in the first few pages of my differential course. I just passed through the term equilibrium solution. Apparently, it meant for me as when the solution of the DE is zero. For example, dv/dt= 9.8-(v/5) ... has an equilibrium solution of 49 where dv.dt is actually zero. Is this the case or not??

An equilibrium solution is a solution which is a constant. Since the derivative of a constant is zero, a constant solution $v(t) = v_0$ of $\dot v(t) = f(v(t))$ must satisfy $f(v_0) = 0$.

Thus any equilibrium solution of $\dot v = 9.8 - \frac15 v$ must satisfy $9.8 - \frac15 v = 0$, ie. $v(t) = 49$.