Direction field analysis Consider the first-order initial value problem y'(t)=ay+b,y(0)=A for t≥0 where a, b, and A are real numbers.
a. Explain why y=−b/a is an equilibrium solution and corresponds to a horizontal line in the direction field.

Verified step by step guidance

1. Recall that an equilibrium solution to a differential equation is a constant solution where the derivative is zero for all values of \( t \). This means \( y'(t) = 0 \) along that solution.
2. Given the differential equation \( y'(t) = a y + b \), set \( y'(t) = 0 \) to find the equilibrium solution: \( 0 = a y + b \).
3. Solve for \( y \) to get \( y = -\frac{b}{a} \). This value of \( y \) makes the derivative zero, so the solution is constant and does not change with \( t \).
4. Since \( y = -\frac{b}{a} \) is constant, its graph is a horizontal line in the \( t y \)-plane, which corresponds to a horizontal line in the direction field.
5. In the direction field, the slope at every point on this line is zero, confirming that \( y = -\frac{b}{a} \) is an equilibrium solution.

Key Concepts

Equilibrium Solution

An equilibrium solution to a differential equation is a constant solution where the derivative is zero. For y'(t) = ay + b, setting y' = 0 gives y = -b/a, meaning the solution does not change over time and remains constant.

Direction Field

A direction field is a graphical representation showing the slope of solutions to a differential equation at various points. At an equilibrium solution, the slope is zero, so the direction field shows horizontal line segments, indicating no change in y.

Initial Value Problem (IVP)

An IVP specifies a differential equation along with an initial condition, such as y(0) = A. This condition determines a unique solution curve passing through the point (0, A) in the direction field.