§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

For an autonomous equation such as $\frac{dy}{dt} = -\frac{1}{2}y(y-4)$, the slope field has the property that

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

For $\frac{dy}{dt} = -\frac{1}{2}y(y-4)$, the equilibrium classifications are

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 $y(0) = 2$.

For $\frac{dy}{dt} = -\frac{1}{2}(y-4)$, compute the rate $\frac{dy}{dt}$ at the instant when $y = 6$.

For $\frac{dy}{dt} = -\frac{1}{2}y(y-4)$, the solution with $y(0) = 5$ stays above $4$ for all time while decreasing toward the equilibrium $4$.

For an autonomous equation $\frac{dy}{dt} = f(y)$ with equilibrium $\overline{y}$, the equilibrium is stable exactly when $f$ changes from positive to negative as $y$ increases through $\overline{y}$.

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 $4$.

For $\frac{dy}{dt} = -\frac{1}{2}(y-4)$, the solution satisfying $y(0) = 4$ is the constant function $y(t) = 4$.