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When working with autonomous differential equations like \( \frac{dx}{dt} = x - 4 \), finding the critical points is an essential first step. A critical point is a value of \( x \) where the rate of change \( \frac{dx}{dt} \) is zero. In simpler terms, it's where the function temporarily stops moving up or down.
To find these points, we set the equation \( f(x) = x - 4 = 0 \). Solving \( x - 4 = 0 \), we find that the critical point is \( x = 4 \). At this point, the system is balanced. Taking the time to identify these points helps us later when we carry out further analyses.

Once critical points are identified, we need to understand their stability: whether they will remain balanced upon small disturbances or not.

• If the system returns to the critical point after a slight change, it's stable.
• If it diverges further away, it's unstable.

In our equation \( f(x) = x - 4 \), analyze the intervals around \( x = 4 \).
For \( x < 4 \), \( f(x) = x - 4 < 0 \), indicating a decrease in \( x \). For \( x > 4 \), \( f(x) = x - 4 > 0 \), indicating an increase. Thus, the critical point at \( x = 4 \) is unstable, as solutions move away from it, not returning when nudged.

A phase diagram visually represents the behavior around critical points. It helps to summarize stability analysis results effectively.
In this diagram, mark the critical point \( x = 4 \).

• For \( x > 4 \), draw an arrow pointing right, showing that \( x \) is increasing.
• For \( x < 4 \), draw an arrow pointing left, showing that \( x \) is decreasing.

This creates a simple visual representation of instability, as the arrows diverge from \( x = 4 \). Phase diagrams are powerful tools for grasping the system's long-term behavior at a glance.

Solving the differential equation explicitly gives us a clear mathematical description of how \( x \) changes over time. For our equation, \( \frac{dx}{dt} = x - 4 \), we accomplish this by separating variables.
Rearrange to \( \frac{1}{x - 4} dx = dt \) and integrate both sides: \( \int \frac{1}{x - 4} \, dx = \int \, dt \).
This results in \( \ln|x - 4| = t + C \), where \( C \) is a constant.
Solve for \( x \): \( x = 4 + Ce^{t} \). The constant \( C \) is determined by initial conditions.
From this explicit solution, we see how \( x \) evolves. As \( t \to \infty \), solutions support the instability determined by earlier analysis; they indicate that if \( C > 0 \), \( x \to \infty \), and if \( C < 0 \), \( x \to 4 \), showcasing the divergence of solutions.