An equilibrium solution in differential equations is defined as a constant solution where the derivative equals zero. For the differential equation dv/dt = 9.8 - (v/5), the equilibrium solution is v(t) = 49, as it satisfies the condition f(v_0) = 0. This means that at v = 49, the rate of change dv/dt is indeed zero. Therefore, the assertion that equilibrium solutions are always zero is incorrect; they can be any constant value that satisfies the equation.
PREREQUISITESStudents in introductory differential equations courses, educators teaching mathematical concepts, and anyone interested in understanding the behavior of dynamic systems through equilibrium analysis.
ehabmozart said: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??