91Ó°ÊÓ

First, let's rewrite the differential equation: $$\frac{d}{dt}u(t) - 4 u(t) = 3$$ Now that we have the differential equation in the standard form, we can move to the next step.

An equilibrium solution refers to a steady-state constant value of \(u(t)\) when the rate of change is 0. Mathematically, we find equilibrium solutions by setting \(u'(t) = 0\) and solving for \(u(t)\): $$0 - 4 u(t) = 3 \implies -4u(t) = 3 \implies u(t) = -\frac{3}{4}$$ The equilibrium solution for the given DE is \(u(t) = -\frac{3}{4}\).

To determine the stability of the equilibrium solution, we look at the behavior of solutions near the equilibrium. If the equilibrium solution is stable, neighboring solutions will also approach this value as \(t \to \infty\). In contrast, if the equilibrium is unstable, neighboring solutions will diverge away from the equilibrium value as \(t \to \infty\). We can determine the behavior by looking at the change in the solutions at different points around the equilibrium point \(u=-\frac{3}{4}\). We analyze the terms inside the DE: $$\frac{d}{dt}u(t) - 4 u(t)$$ For \(u(t) > -\frac{3}{4}\): $$\frac{d}{dt} u(t) - 4 u(t) > 0$$ So, \(\frac{d}{dt}u(t) > 4u(t)\), and the solution increases. For \(u(t) < -\frac{3}{4}\): $$\frac{d}{dt} u(t) - 4 u(t) < 0$$ So, \(\frac{d}{dt}u(t) < 4u(t)\), and the solution decreases. From the above analysis, we can conclude that the equilibrium solution \(u(t)=-\frac{3}{4}\) is an unstable equilibrium.

Now we will sketch the direction field for the given DE, for \(t\geq 0\). 1. Draw a horizontal line at \(u = -\frac{3}{4}\) to represent the equilibrium solution. 2. For values of \(u\) greater than \(-\frac{3}{4}\), indicate that the solutions are increasing by sketching arrows pointing upward. 3. For values of \(u\) less than \(-\frac{3}{4}\), indicate that the solutions are decreasing by sketching arrows pointing downward. The arrows indicate whether the solution is increasing or decreasing on either side of the equilibrium solution.