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Chapter 2, Section 2.4, Question 18 Describe how solutions appear to behave as $t$ increases and how their behavior depends on the initial value $y_0$ when $t = 0$. $y' = y(7 - ty)$ For $y_0 > 0$, solutions decrease until they intersect the curve $y = \frac{7}{t}$, then they increase. For $y_0 = 0$, $y = 0$ is an equilibrium solution. For $y_0 < 0$, solutions decrease without bound as $t \to \infty$ For $y_0 > 0$, solutions increase without bound as $t \to \infty$ For $y_0 = 0$, $y = 0$ is an equilibrium solution. For $y_0 < 0$, solutions decrease without bound as $t \to \infty$

          Chapter 2, Section 2.4, Question 18
Describe how solutions appear to behave as $t$ increases and how their behavior depends on the initial value $y_0$ when $t = 0$.
$y' = y(7 - ty)$
For $y_0 > 0$, solutions decrease until they intersect the curve $y = \frac{7}{t}$, then they increase.
For $y_0 = 0$, $y = 0$ is an equilibrium solution.
For $y_0 < 0$, solutions decrease without bound as $t \to \infty$
For $y_0 > 0$, solutions increase without bound as $t \to \infty$
For $y_0 = 0$, $y = 0$ is an equilibrium solution.
For $y_0 < 0$, solutions decrease without bound as $t \to \infty$

Chapter 2, Section 2.4, Question 18
Describe how solutions appear to behave as t increases and how their behavior depends on the initial value y0 when t = 0.
y' = y(7 - ty)
For y0 > 0, solutions decrease until they intersect the curve y = (7)/(t), then they increase.
For y0 = 0, y = 0 is an equilibrium solution.
For y0 < 0, solutions decrease without bound as t →∞
For y0 > 0, solutions increase without bound as t →∞
For y0 = 0, y = 0 is an equilibrium solution.
For y0 < 0, solutions decrease without bound as t →∞

Added by Travis H.

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