Mathematics

Mathematics

Problem

The differential equation dydx=(4−y)(y+1)\dfrac{dy}{dx}=(4-y)(y+1) has equilibrium solutions at y=−1y=-1 and y=4y=4. Two solution curves satisfy y(0)=0y(0)=0 and y(0)=3y(0)=3, respectively. Which of the following statements best compares the long-term behavior of these two solutions as xx increases? a) Both solutions decrease and approach y=−1y=-1 as x→∞x\to\infty. b) Both solutions increase without bound as x→∞x\to\infty. c) Both solutions increase and approach y=4y=4 as x→∞x\to\infty. d) The solution with y(0)=0y(0)=0 approaches y=−1y=-1, while the solution with y(0)=3y(0)=3 approaches y=4y=4.
Intuitive explanation

This is about stability of equilibrium solutions and how initial conditions determine which equilibrium a solution approaches. The key is to analyze the sign of dydx\frac{dy}{dx} in each region between and outside the equilibria. When the derivative is positive, yy increases toward the nearest equilibrium; when negative, it decreases. By checking the sign of (4−y)(y+1)(4-y)(y+1) at your initial values, you'll see which equilibrium each solution is "trapped" between and therefore must approach as x→∞x \to \infty.

Intuitive explanation

This is about stability of equilibrium solutions and how initial conditions determine which equilibrium a solution approaches. The key is to analyze the sign of dydx\frac{dy}{dx} in each region between and outside the equilibria. When the derivative is positive, yy increases toward the nearest equilibrium; when negative, it decreases. By checking the sign of (4−y)(y+1)(4-y)(y+1) at your initial values, you'll see which equilibrium each solution is "trapped" between and therefore must approach as x→∞x \to \infty.

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