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.
a. Find the equilibrium solutions. 


y′(t) = 6 - 2y

Verified step by step guidance

1. Identify the given autonomous differential equation: \(y'(t) = 6 - 2y\).
2. Recall that equilibrium solutions occur where the derivative \(y'(t)\) is zero, meaning the function \(y(t)\) does not change over time.
3. Set the right-hand side of the differential equation equal to zero to find equilibrium points: \$6 - 2y = 0$.
4. Solve the algebraic equation for \(y\) to find the equilibrium value(s): \$2y = 6\( which implies \)y = 3$.
5. Conclude that the equilibrium solution is the constant function \(y(t) = 3\), which corresponds to a horizontal line in the direction field.

Key Concepts

Autonomous Differential Equations

An autonomous differential equation is one where the derivative y' depends only on the variable y, not explicitly on the independent variable t. This means the rate of change is determined solely by the current state y, making the direction field time-invariant and simplifying analysis of equilibrium points.

Equilibrium Solutions

Equilibrium solutions occur when the derivative y' equals zero, meaning the function y(t) remains constant over time. For autonomous equations y' = f(y), equilibrium solutions satisfy f(y0) = 0. These solutions correspond to horizontal lines in the direction field where the system is at rest.

Finding Equilibrium Points by Solving f(y) = 0

To find equilibrium solutions, set the right-hand side of the differential equation y' = f(y) equal to zero and solve for y. For example, in y' = 6 - 2y, setting 6 - 2y = 0 yields y = 3, which is the equilibrium solution where the system does not change.