Question

38–43. Equilibrium solutions A differential equation of the form y′(t)=f(y) is said to be autonomous (the function f depends only on y). The constant function y=y0 is an equilibrium solution of the equation provided f(y0)=0 (because then y'(t)=0 and the solution remains constant for all t). Note that equilibrium solutions correspond to horizontal lines in the direction field. Note also that for autonomous equations, the direction field is independent of t. Carry out the following analysis on the given equations.
b. Sketch the direction field, for t≥0.

y′(t) = 2y + 4

Accepted Answer

Identify the given autonomous differential equation: $$y'(t) = 2y + 4. $$Since it depends only on $$y$$, it fits the form $$y'(t) = f(y)$$ with $$f(y) = 2y + 4. $$Find the equilibrium solution(s) by setting $$f(y) = 0$$, which means solving $$2y + 4 = 0$$ for $$y. $$This gives the constant solution(s) where the slope is zero. Understand that the direction field consists of small line segments at various points $$(t, y)$$ with slope given by $$y'(t) = 2y + 4. $$Since the equation is autonomous, the slope depends only on $$y$$, not on $$t. To $$sketch the direction field for $$t \geq 0$$, choose several values of $$y ($$both above and below the equilibrium solution) and calculate the slope $$2y + 4 at $$each. Draw short line segments with these slopes at points along the vertical lines for different $$t$$ values. Note that at the equilibrium solution, the slope is zero, so the direction field will have horizontal line segments. For $$y$$ values greater than the equilibrium, the slope will be positive (lines slanting upward), and for $$y$$ values less than the equilibrium, the slope will be negative (lines slanting downward). This helps visualize the behavior of solutions over time.

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Concetti chiave

Autonomous Differential Equations

Equilibrium Solutions

Direction Fields (Slope Fields)