§7.2 Qualitative behavior of solutions to differential equations

Introduction

Slope fields

Equilibrium solutions and stability

For the initial value problem $\frac{dy}{dt} = t-2$ with $y(0) = 1$, the tangent line to the solution at $t = 0$ is

For $\frac{dy}{dt} = -\frac{1}{2}(y-4)$, the solution $y(t)$ increases when

The differential equation $\frac{dy}{dt} = -\frac{1}{2}y(y-4)$ has equilibrium solutions

Verify that $y(t) = 4+2e^{-t/2}$ solves $\frac{dy}{dt} = -\frac{1}{2}(y-4)$. Compute $y(0)$.

Every solution of $\frac{dy}{dt} = -\frac{1}{2}(y-4)$ has the form $y = 4+Ce^{-t/2}$. Find the constant $C$ for the solution with $y(0) = 2$.

If $y(t)$ models a population with positive initial value under $\frac{dy}{dt} = -\frac{1}{2}y(y-4)$, the population eventually tends to $4$.