Chapter 4: Problem 2

Find the equilibrium solutions and determine their stability for the following systems. For each case, draw representative solutions and phase lines. a. \(y^{\prime}=y^{2}-6 y-16\). b. \(y^{\prime}=\cos y\). c. \(y^{\prime}=y(y-2)(y+3)\). d. \(y^{\prime}=y^{2}(y+1)(y-4)\).

Short Answer

Expert verified

a. The equilibrium solutions for \(y^{\prime}=y^{2}-6 y-16\) are \(y=8\) (stable), \(y=-2\) (unstable). b. The equilibrium solutions for \(y^{\prime}=\cos y\) are \(y=\frac{\pi}{2}+n\pi\) (all stable). c. The equilibrium solutions for \(y^{\prime}=y(y-2)(y+3)\) are \(y=0\) (unstable), \(y=2\) (stable), \(y=-3\) (unstable). d. The equilibrium solutions for \(y^{\prime}=y^{2}(y+1)(y-4)\) are \(y=0\) (unstable), \(y=-1\) (stable), \(y=4\) (unstable).

Step by step solution

Finding the equilibrium points

To find the equilibrium solutions, set the derivative equal to zero and solve for \(y\). a. For \(y^{\prime}=y^{2}-6 y-16\), setting the equation equal to zero gives us \(y^{2}-6y-16=0\). After factoring, we get \((y-8)(y+2)=0\), yielding \(y=8\) and \(y=-2\). b. For \(y^{\prime}=\cos y\), setting the equation equal to zero means \(\cos y=0\). This happens when \(y=\frac{\pi}{2}+n\pi\), for any integer \(n\). c. For \(y^{\prime}=y(y-2)(y+3)\), setting \(y(y-2)(y+3)=0\) gives us \(y=0, 2, -3\). d. For \(y^{\prime}=y^{2}(y+1)(y-4)\), setting \(y^{2}(y+1)(y-4)=0\) gives \(y=0, -1, 4\).

Determining stability

The sign of the derivative between the equilibrium points determines whether an equilibrium solution is stable. a. For \(y^{\prime}=y^{2}-6 y-16\), the intervals are \(-\infty\) to -2, -2 to 8, and 8 to \(\infty\). Choose convenient points such as -3, 0, and 10, respectively. Plug these into the derivative and observe their sign. b. Use common test points within intervals for \(y^{\prime}=\cos y\) such as \(\frac{\pi}{4}\), \(\frac{3\pi}{4}\), etc. c. The intervals for \(y^{\prime}=y(y-2)(y+3)\) are \(-\infty\) to -3, -3 to 0, 0 to 2, and 2 to \(\infty\). d. The intervals for \(y^{\prime}=y^{2}(y+1)(y-4)\) are \(-\infty\) to -1, -1 to 0, 0 to 4, and 4 to \(\infty\).

Representative solutions and phase lines

Sketch the representative solutions and phase lines based on the equilibrium solutions and their stability. Arrange the equilibrium solutions on a line in ascending order and track the sign changes over the interval. Regions between equilibrium points where the derivative has the same sign belong to the same phase. Negative derivative values indicate decreasing solutions while positive values indicate increasing solutions.